View Full Version : SWR attenuator schematic

N5FOG

12-31-2009, 12:55 AM

I'm trying to build a attenuator that will present a constant 50 ohm load to the radio and attenuate most of the power. But pass the rest of the power down the line to the SWR meter, tuner and antenna for tuning the tuner without the radio seeing any SWR.

I know the AH-4 uses a similar circuit so that the 706 transmits 10 watts to the tuner and sees a 1:1 SWR but the circuit passes a 1 or 2 watts to the antenna tuner part for SWR tuning purposes.

Anyone know if, a circuit like this has a name, and where I could find a schematic to build one or a web page with details on how to design one.

FOG

K9STH

12-31-2009, 02:19 AM

FOG:

It is called an attenuator! :rolleyes:

Glen, K9STH

10 db attenuator

http://www.nessengr.com/techdata/atten/attenschem.gif

N5FOG

12-31-2009, 02:49 AM

FOG:

It is called an attenuator! :rolleyes:

I understand attenuators and have built several for low drive amps so Glen I understand the concept of an attenuator and in my first line I said "I'm trying to build a attenuator". But in order to build a Pi or T network attenuator network you need to know the characteristic impedance which is where my question comes in.

If I build a standard 50 ohm Pi or T style attenuator is a load that isn't 50 ohms (which is what the antenna tuner when in the process of tuning will present) on one side of it, going to cause its impedance on the other side of the network that the radio is driving into to change and present a SWR ??

FOG

K9STH

12-31-2009, 02:58 AM

You asked if the circuit had a name! The name is attenuator! :rolleyes:

There are basically 2 types of attenuators. They are "T" and "Pi". 2U has given you a "T" type.

Here is a URL for designing either type for any impedance:

http://www.aubraux.com/design/rf-attenuator-design-tool.php

Glen, K9STH

AC0FP

12-31-2009, 03:20 AM

I understand attenuators and have built several for low drive amps so Glen I understand the concept of an attenuator and in my first line I said "I'm trying to build a attenuator". But in order to build a Pi or T network attenuator network you need to know the characteristic impedance which is where my question comes in.

If I build a standard 50 ohm Pi or T style attenuator is a load that isn't 50 ohms (which is what the antenna tuner when in the process of tuning will present) on one side of it, going to cause its impedance on the other side of the network that the radio is driving into to change and present a SWR ??

FOG

An attenuator by any other name is still an attenuator! Some of the Comet antennas are a good example.

I remember a VHF aircraft antenna in a blade style! It had a a 3:1 VSWR maximum over the aircraft AM band! What it was, a 3 dB power pad (pi) with about 10" of wire connected to the end that wasn't hooked to the transmitter! :p :o

If I remember my theory correctly, the isolation of the load to the source will be proportional to the amount of attenuation that is provided by the attenuator.

Pi network attenuators are often used in just such applications, to allow a constant impedance load to be applied to a signal generator, so the varying load of the circuit under test will not affect the output level of the signal generator. The signal generator will be seeing a relatively stable load of around 50 ohms, regardless of the impedance of the load. The tradeoff, of course, is the loss of signal level, in the case of a 10dB attenuator, you would have to boost the signal generator output by that amount, in order to get the same level of signal that would be seen across a load that matched the output impedance of the signal generator with no attenuator in the circuit.

Hope I made myself clear enough. 73, Jim

N5FOG

12-31-2009, 04:21 AM

You asked if the circuit had a name! The name is attenuator! :rolleyes:

Glen I think I made it clear in my first post my question was out of the scope of the normal use of an attenuator of just reducing power to a load of the same impedance as the source. Jim/AG3Y thank you, that's the exact bit of information I was looking for.

FOG

G3TXQ

12-31-2009, 09:01 AM

10 db attenuator

http://www.nessengr.com/techdata/atten/attenschem.gif

That's a 20dB attenuator!

Steve G3TXQ

G3TXQ

12-31-2009, 09:06 AM

Glen I think I made it clear in my first post my question was out of the scope of the normal use of an attenuator of just reducing power to a load of the same impedance as the source. Jim/AG3Y thank you, that's the exact bit of information I was looking for.

FOG

To add to what Jim said:

A 10dB attenuator will ensure that the VSWR seen by the radio is better than 1.22:1 for any load conditions, including an open circuit or a short circuit; a 6dB attenuator will ensure better than 1.7:1

Don't forget to do the calculations on which resistors in the attenuator will be dissipating most of the power; it will vary depending on the termination impedance. I would "play safe" and do the sums for a S/C and an O/C at the attenuator ouput.

73,

Steve G3TXQ

VK2TIL

12-31-2009, 09:57 AM

I think that the OP is asking about a "resistive SWR bridge".

These are especially popular in QRP circles but, with appropriate choice of resistor power ratings, could be used at almost any power level.

I Googled "resistive SWR bridge" and found a lot of hits; this is just one;

http://www.gqrp.com/resistive_swr_bridge.pdf

The essential principle is that, no matter what the mismatch, the transmitter will only see a maximum SWR that is manageable.

For example, even if the antenna is open or shorted, an SWR of perhaps 3:1 will be what the transmitter "sees".

Designs may be found by Googling or by consulting almost any QRP-oriented book; W1FB described bridges of this kind.

That's a 20dB attenuator!

Steve G3TXQ

Buy one at 10 db and get extra 10 db free :D

For extact 10 db, change the 40.9 ohm resistor to 25.974 ohm and the 10.1 one to 35.136 - those are extremely hard to figure out values for a home brew project.

Thanks, Steve, for the particulars. I've been away from this stuff for a while, but used to use such devices all the time at my last job. (Microwave transmitter prototype testing at Hughes )

AB9LZ

12-31-2009, 04:04 PM

I think that the OP is asking about a "resistive SWR bridge".

Ha! Yep, that's what he is looking for, I use one for a half wave end fed setup, it works great, just substitute your tuner of choice.

http://www.dxzone.com/cgi-bin/dir/jump2.cgi?ID=11595

73 m/4

KD0CAC

12-31-2009, 05:14 PM

New to all this , so not sure if what I'm thinking , will fit & help ?

I picked up a tool at our local Ham Radio Consignment shop .

A Palomar R-X noise bridge .

I am getting the impression that what your looking for would stay in line , where this unit is put in line just during the tuning process .

That's my post to keep an eye in the discussion .

K9STH

12-31-2009, 05:34 PM

FOG:

The "Pi" type of "matching pad" attenuator was used very extensively with signal generators using the "olde tyme" piston type internal attenuator. The impedance of the internal piston attenuator varies all over the place but it is generally "desired" to present a fixed impedance to the device under test.

The old Measurements Model 80 and Model 82, as well as the military TS-497 series, URM-25, and URM-26 series, are examples of generators that came "standard" with 6 dB "matching pads" (attenuators) for the purpose of "matching" the varying impedance of the generator to the fixed impedance of the unit being serviced.

Therefore, I guess that the "name" of the particular attenuator that you are looking for is "matching pad". However, it is nothing more than a "Pi" circuit attenuator.

Glen, K9STH

VK2TIL

12-31-2009, 08:56 PM

If I build a standard 50 ohm Pi or T style attenuator is a load that isn't 50 ohms (which is what the antenna tuner when in the process of tuning will present) on one side of it, going to cause its impedance on the other side of the network that the radio is driving into to change and present a SWR ?

Yes; the attenuator will only be "perfect" if there are 50-ohm loads on its input and its output ports.

Most attenuators we see are designed for the same Z (usually 50 ohms) on each port but there are asymmetrical attenuators; the "Min-Loss" pad used to "convert" 50 ohms to 75 ohms is a common example.

There are plenty of attenuator calculators on the 'net; most calculate symmetrical attenuators but this one handles asymmetrical ones;

http://chemandy.com/calculators/matching-pi-attenuator-calculator.htm

If you play with some numbers in this calculator you will find that the change in input impedance/SWR (as seen by a transmitter) is not great if there is a substantial mismatch at the output (ATU/antenna) end.

For example;

http://img689.imageshack.us/img689/5679/attk.jpg (http://img689.imageshack.us/i/attk.jpg/)

You will see that even putting a short on the output gives good VSWR on the input; that is what I think you are seeking.

This is a similar principle to the resistive SWR bridge that I referred-to earlier.

This is a good time to introduce the O/P to the term "return loss".

Return loss is just another way to say VSWR.

Only we give the ratio in Decibels instead of a decimal number.

A load that reflected 100% of the power back to the source (an open, or a short circuit) would have a VSWR of infinity, and a return loss of 0(zero)DB.

A load that absorbed 100% of the power would have a VSWR of 1:1, and a return loss of infinity.

Using a attenuator to protect a transmitter from high VSWR, or low return loss, the return loss will improve by 2 times the value (in db) of the attenuator.

A 10 db attenuator will have a 20 db return loss- the signal is attenuated 10db going from the tx to the load, and then the reflected wave, now -10DB or 10% of the transmitter power will be attenuated another 10DB going back thru the attenuator to the transmitter.

A 100 watt tx will have 10w at the load, and if the load reflects 100% of the 10watts (VSWR=infinity, return loss (of load) 0db)) than 1 watt will reach back to the transmitter.

If the VSWR or return loss of the LOAD improves from VSWR=infinity/zero DB r/l, then that improvement in DB will add to the 20 db r/l of the attenuator, and further improve the VSWR the transmitter "sees".

Rege

EDIT: A 10db attenuator will more than protect the finals on any transmitter I am aware of, if you look up the manufactures data sheet for the transistors in your transmitter, there will be a specification to the words of " the device is specified to produce rated power without damage a 1x.x VDC at YY:1 VSWR at any phase angle"

It will be straight forward to calculate the minimum value of attenuation needed to protect the finals.

Rege

New to all this , so not sure if what I'm thinking , will fit & help ?

I picked up a tool at our local Ham Radio Consignment shop .

A Palomar R-X noise bridge .

I am getting the impression that what your looking for would stay in line , where this unit is put in line just during the tuning process .

That's my post to keep an eye in the discussion .

You don't want to transmit into that Palomar, it's strictly a receive only unit.

The moise bridge is one step up from a VSWR meter in that it seperates the antenna/feedline system VSWR into a resistance and reactance readings.

It will help you pick the values of reactance to bring your antenna into resonance.

The O/P wants to use a attenuator to make sure that no matter how bad The antenna system VSWR is that the transmitter will not be damaged.

Rege

VK2TIL

01-01-2010, 04:13 AM

That calculator gives RL; as expected, it shows 40dB for my chosen-at-random 20dB attenuator.

As little as 3 or 5dB might be sufficient; I use those values for stabilising the impedance of things like mixer or amplifier inputs/outputs.

5dB = 10dB RL = c.2:1 VSWR. (Tables relating RL, VSWR & other measures are easily found with Google).

A 5dB attenuator seems about right for the OP's requirements; this would not be difficult to homebrew, at least for HF & low VHF.

The noise bridge is certainly a different animal, although one of my favourites for amateur work; I think that CAC understands that it's not an in-line device.

That calculator gives RL; as expected, it shows 40dB for my chosen-at-random 20dB attenuator.

40 dB may refer to voltage dB, which has a number that is twice power dB number.

Voltage dB = 20log(V1/V2)

Power dB = 10log(V1/V2)

Since P = V x V / R - 1/10 of voltage leads to 1/100 of power.

K9STH

01-01-2010, 06:02 AM

I don't know how they are calculating the dB, but a dB is a dB is a dB. It doesn't matter if you are talking about voltage or power. Now the ratio of power is different from the ratio of voltage. That is true. However, if you, for example, introduce a 20 dB attenuator into the circuit you will get 20 dB of power and 20 dB of voltage reduction. However, dB are logarithmic expression and power involves the squaring of the current and voltage does not involve the squaring of the current.

Go to

http://k9sth.com/uploads/A_dB_is_a_dB_is_a_dB.pdf

for the explanation which also includes examples and the mathematical calculations.

Glen, K9STH

I don't know how they are calculating the dB, but a dB is a dB is a dB. It doesn't matter if you are talking about voltage or power. Now the ratio of power is different from the ratio of voltage. That is true. However, if you, for example, introduce a 20 dB attenuator into the circuit you will get 20 dB of power and 20 dB of voltage reduction. However, dB are logarithmic expression and power involves the squaring of the current and voltage does not involve the squaring of the current.

Go to

http://k9sth.com/uploads/A_dB_is_a_dB_is_a_dB.pdf

for the explanation which also includes examples and the mathematical calculations.

Glen, K9STH

For a given ratio, power dB is not the same as voltage dB, as power ratio is calculated using the 10 log rule and voltage ratio the 20 log rule. So a ratio of 10 for power is 10 dB and for voltage 20 dB.

For a given dB, the ratio is not the same for power and voltage. For example 20 dB for power, the ratio is 100:1 and the same 20 dB for voltage is 10:1.

The difference is really about ratio of amplitude vs ratio of amplitude squre.

When you change voltage on a linear resistance, like a dummy load, the current will change as well. Otherwise, you can NOT say "a dB is a dB is a dB".

G3TXQ

01-01-2010, 09:11 AM

For a given ratio, power dB is not the same as voltage dB, as power ratio is calculated using the 10 log rule and voltage ratio the 20 log rule. So a ratio of 10 for power is 10 dB and for voltage 20 dB.

For a given dB, the ratio is not the same for power and voltage. For example 20 dB for power, the ratio is 100:1 and the same 20 dB for voltage is 10:1.

The difference is really about ratio of amplitude vs ratio of amplitude squre.

When you change voltage on a linear resistance, like a dummy load, the current will change as well. Otherwise, you can NOT say "a dB is a dB is a dB".

A dB is always a ratio of power, no matter how you calculate it or what label you give it! I think Glen was saying the same thing.

So, for example, 10dBuV is 10 times the power compared to 0dBuV, assuming a constant measurement impedance.

Steve G3TXQ

WB2UAQ

01-01-2010, 05:03 PM

Another way to get a controlled broadband way to knock down power for some applications is to use a coupler. The coupler that I have in mind uses a toroid that has a transmission line passing thru the center of the toroid (one turn primary) and 10 turns of magnet wire (secondary) applied around the toroid in the usual way.

A two toroid type can also be used (the type used for directional couplers in common directional power meters). 10 turns results in a 20 dB coupling factor.

The transmission line (s) that pass thru the toroid (s) are actually short pieces of coax cable where the shield is left open (normally on the output side in applications I work with). The shield prevents capactive coupling ensuring only magnetic coupling from the main RF path to the coupled output.

Think of these as a current transformer where the coupling factor is 20log (the # of magnet wire turns).

In the application you would terminate the main line into a dummy load and take the power off the "coupled" output. 100W would result in a 1W coupled output for testing or antenna tuning purposes.

Depending on what's in the junk box this might be easier to implement as a high power dummy load might already be on hand (such as an old Heathkit Cantenna). All that is needed are two common toroids (43 or 61), three connectors, some wire and a small box or even just a U shaped sheet metal channel. The toroids can be very small. A half inch diameter core can handle 100W to maybe 1 kW.

I built one of these just for experimental purposes to prove out the coupling factor concept. I am going to add one to the output of an experimental QRP transmatch based on KL7AJ's suggestions. 73 and Happy New Year to All.

A dB is always a ratio of power, no matter how you calculate it or what label you give it! I think Glen was saying the same thing.

So, for example, 10dBuV is 10 times the power compared to 0dBuV, assuming a constant measurement impedance.

Steve G3TXQ

From 1 uV to 10 uV, the voltage gain is 20 dB - 20 log rule

From 1 uW to 10 uW, the power gain is 10 dB - 10 log rule

G3TXQ

01-01-2010, 05:23 PM

From 1 uV to 10 uV, the voltage gain is 20 dB - 20 log rule

From 1 uW to 10 uW, the power gain is 10 dB - 10 log rule

Jim,

Look again at what I wrote:

10dBuV is 10 times more power than 0dBuV (into the same impedance)

73,

Steve G3TXQ

K9STH

01-01-2010, 06:13 PM

working with whole numbers:

1 watt to 10 watts = 10 dB

1 volt to 10 volts = 20 dB

Assuming impedance = 50 ohms

When P = 1 watt

P = (I^2)R

P/R = I^2

1/50 = I^2

.1414 = I

E = IR

E = (.1414)(50)

E = 7.07

When P = 10 watts

P = (I^2)(R)

P/R = I^2

10/50 = I^2

.2 = I^2

.4472 = I

E = IR

E = (.4472)(50)

E = 22.360

dB (for voltage) = 20 log10 V1/V2

db = (20)(log10 22.360/7.07

dB = (20)(log 10 3.162659)

dB = (20)(0.5000)

dB = 10

As can be seen, 10 dB is 10 dB is 10 dB!

Glen, K9STH

working with whole numbers:

1 watt to 10 watts = 10 dB

1 volt to 10 volts = 20 dB

So for a given ratio of 1 to 10, power db is 10 and voltage dB is 20.

As can be seen, 10 dB is 10 dB is 10 dB!

Glen, K9STH

For the same 10 dB, power gain is 10 and voltage gain is only 3.162659 or square root of 10.

So, given a 20 dB attenuator, the power drop is 100:1 and voltage drop is 10:1.

K8ERV

01-01-2010, 06:31 PM

Jim,

(into the same impedance)

73,

Steve G3TXQ

That's the key that many people miss. SAME.

TOM K8ERV Montrose Colo

Jim,

Look again at what I wrote:

10dBuV is 10 times more power than 0dBuV (into the same impedance)

73,

Steve G3TXQ

Why do you use V for power? V (Volt) is a measurement unit for voltage. W (Watt) is a measurement unit for power.

0dBuV is how many uV? 3.162659 uV or 1 uV? :D

K9STH

01-01-2010, 06:33 PM

That is correct. This difference in ratios is due to the "squaring" of the current between power and voltage.

P = (I^2)R

whereas

V = IR

It is this "squaring" of the current that confuses most people when comparing voltage to power when dB are expressed. That is why, several years ago, I wrote the short article showing the relationship of power to voltage by expressing them in actual calculations.

Clear as mud! Right! :rolleyes:

Glen, K9STH

That is correct. This difference in ratios is due to the "squaring" of the current between power and voltage.

P = (I^2)R

whereas

V = IR

Often, one started with a numerical ratio, like attenuating a 100 Watt signal to 1 Watt, then a 20 dB attenuator will do. If from 100 V to 1 V, one needs a 40 dB attenuator.

G3TXQ

01-01-2010, 06:52 PM

Why do you use V for power? V (Volt) is a measurement unit for voltage. W (Watt) is a measurement unit for power.

0dBuV is how many uV? 3.162659 uV or 1 uV? :D

Because dBuV is an engineering power reference unit - commonly used in EMC measurements and almost always within a 50 ohm measurement environment. It is equivalent to -107dBm, another absolute power reference unit. You will often find RF lab equipment calibrated in dBuV.

0dBuV is 1uV, and 10dBuV is 3.16uV

I try to avoid using the phrase "voltage decibels" - it's meaningless because a dB always refers to power. I find it leads to confusion, with some students trying to tell me that a passive 1:10 voltage transformer has a gain of 20dB!!

Steve G3TXQ

I try to avoid using the phrase "voltage decibels" - it's meaningless because a dB always refers to power. I find it leads to confusion, with some students trying to tell me that a passive 1:10 voltage transformer has a gain of 20dB!!

They are right, the voltage gain is 20 dB. :D:D:D

You are not a good teacher as you can't explain the confusion of 10 dB power gain (10:1) vs 10 dB voltage gain (3.16:1) and why there are 10 log rule and 20 log rule.

G3TXQ

01-01-2010, 08:02 PM

They are right, the voltage gain is 20 dB. :D:D:D

So what's the power gain, using your 10log rule?

Steve G3TXQ

VK2TIL

01-01-2010, 09:12 PM

I have some sympathy for AD2U's position; it might be said that he is correct in a mathematical sense.

If I have one chook and you have ten chooks, application of the decibel principle tells us that I have 0dBC and you have 10dBC; anyone with 100 chooks would have 20dBC.

Now, if our chooks turn into emus (another fine Aussie expression; http://dictionary.babylon.com/I_hope_your_chooks_turn_into_emus_and_kick_your_du nny_down ) I have 0dBE, you have 10dBE and that other bloke has 100dBE.

If our emus now turn into volts, we should have 0dBV and 10dBV; the third bloke has 20dBV. But that's not how it is in the world of decibels.

The square-law relationship between voltage and power was recognised in the early days of telecommunication and was accommodated by use of the "20" figure in decibel calculations involving voltage.

But it's really a "fiddle" I reckon; AD2U is "technically" correct.

G3TXQ

01-01-2010, 09:37 PM

I'm happy to stick with the ARRL's explanation. Here are some quotes from a QST article "Gaining on the Decibel":

dB = 20 log(Av)

Please notice that the foregoing only works when impedances are constant and is a derivation, not a definition, of dB.

The fallacy is that there is no such thing as "voltage dB" or "current dB". The decibel is a logarithmic expression of power ratio only.

I must emphasize this is the only meaningful definition of the Bel, and any attempt to apply the term to anything other than power ratios will get you into trouble

Full text - including an explanation of how the author lost a consultancy job by foolishly asking whether something was "voltage dB or power dB" - here:

http://www.setileague.org/articles/ham/dbpart1.pdf

For me, the transformer example illustrates how unhelpful it is to express voltage gain in decibel units. We all know that a passive transformer has 0dB gain. Any attempt to quote a gain other than zero, using dB as the units, is misleading.

Readers might care to look at the Wikipedia definition of the decibel, and in particular note this qualification when they derive the 20log rule: "when the impedance is held constant"

Steve G3TXQ

G3TXQ

01-01-2010, 09:48 PM

If I have one chook and you have ten chooks, application of the decibel principle tells us that I have 0dBC and you have 10dBC; anyone with 100 chooks would have 20dBC.

And that's how folk get into a mess - by using decibel units to express something other than a power ratio.

I live twice as far away from my local town as my cousin; does that mean I am 3dB further, or 6dB further than she is? What a nonsense ;)

Steve G3TXQ

VK2TIL

01-01-2010, 10:01 PM

Despite Tovey's famous admonition regarding "bleeding chunks", I quote from my earlier post;

The square-law relationship between voltage and power was recognised in the early days of telecommunication and was accommodated by use of the "20" figure in decibel calculations involving voltage.

So what's the power gain, using your 10log rule?

Steve G3TXQ

That depends on the signal source's output impedence and the load's impedence.

G3TXQ

01-01-2010, 10:17 PM

That depends on the signal source's output impedence and the load's impedence.

No - the gain of the transformer has nothing to do with the signal source's output impedance.

I'm surprised you can't see that a passive device must have a power gain of 0dB.

Steve G3TXQ

Readers might care to look at the Wikipedia definition of the decibel, and in particular note this qualification when they derive the 20log rule: "when the impedance is held constant"

Steve G3TXQ

From Wikipedia (http://en.wikipedia.org/wiki/Decibel):

The decibel (dB) is a logarithmic unit (http://en.wikipedia.org/wiki/Logarithmic_unit) of measurement that expresses the magnitude of a physical quantity (usually power (http://en.wikipedia.org/wiki/Power_(physics)) or intensity (http://en.wikipedia.org/wiki/Intensity_(physics))) relative to a specified or implied reference level. Since it expresses a ratio of two quantities with the same unit, it is a dimensionless unit (http://en.wikipedia.org/wiki/Dimensionless_unit). - quote ends.

It (dB) is about ratio, not power. When figuring out power ratio, 10 log rule is used and for voltage ratio, 20 log rule is used.

A passive transformer has 0 power gain, but it is often used to do proper impedence matching to facilitate maximum power transfer for best efficiency or best noise figure for a given system.

No - the gain of the transformer has nothing to do with the signal source's output impedance.

I'm surprised you can't see that a passive device must have a power gain of 0dB.

Steve G3TXQ

If the secondary winding is dead shorted, the power gain could be way less than 0 dB. :D:D:D

G3TXQ

01-01-2010, 10:35 PM

Despite Tovey's famous admonition regarding "bleeding chunks", I quote from my earlier post;

The square-law relationship between voltage and power was recognised in the early days of telecommunication and was accommodated by use of the "20" figure in decibel calculations involving voltage.

And I bet they also recognised that you only get the same relationship between voltage and power provided the impedances are the same.

Steve G3TXQ

G3TXQ

01-01-2010, 10:39 PM

If the secondary winding is dead shorted, the power gain could be way less than 0 dB. :D:D:D

Why? The power output will be zero and the power input will be zero - so power gain=0dB

I trust you're not trying to obfuscate the discussion by introducing a non-ideal transformer ;)

Steve G3TXQ

And that's how folk get into a mess - by using decibel units to express something other than a power ratio.

I live twice as far away from my local town as my cousin; does that mean I am 3dB further, or 6dB further than she is? What a nonsense ;)

Steve G3TXQ

When designing an IF stage, voltage gain is used as it is assumed that load impedence is significantly larger than signal source impendency.

dB is usually used in electrical engineering field to express some large numbers, for example a 1,000,000:1 voltage gain can be expressed as 120 dB.

Your example of distance "nonsense" is because you stuck at dBuV power referecne and forget that dB cab be used on something else, like voltage gain, frequency response boundary condition, etc.

Why? The power output will be zero and the power input will be zero - so power gain=0dB

In that case, the power gain is undefined since 0/0 is undefined :D:D:D

G3TXQ

01-01-2010, 10:54 PM

Getting late here, so I'll just leave you with this excellent paper on "Defining the decibel":

http://michael.e.gruchalla.org/WebpagePapers/Defining%20the%20Decibel.pdf

and point you to this quote from the paper:

The use of the decibel to express the voltage gain of an amplifier in a system of non-constant impedances is by far the most common misapplication of the decibel that you're likely to encounter

Now that Jim has invented a passive device that has 20dB gain, I need to see if I can't heat my house for free by daisy-chaining lots of these transformers together ;)

73,

Steve G3TXQ

G3TXQ

01-01-2010, 10:58 PM

In that case, the power gain is undefined since 0/0 is undefined :D:D:D

Ah - so it wont be "way less than 0dB" like you told us!

Steve G3TXQ

Getting late here, so I'll just leave you with this excellent paper on "Defining the decibel":

http://michael.e.gruchalla.org/WebpagePapers/Defining%20the%20Decibel.pdf

and point you to this quote from the paper:

Quote:

The use of the decibel to express the voltage gain of an amplifier in a system of non-constant impedances is by far the most common misapplication of the decibel that you're likely to encounter

That is the field of non-linear circuit analysis, which is the fundation of modern circuit analysis and computer based circuit simulation/design. A diode is a nonlinear device - when positively biased, it represents a small resistance and when biased below forward conducting voltage or reverse-baised, it is a large resistance, almost like a insulator. However, if reversely biased beyond its break down voltage, it will conducting again (zener diodes work in this region). When not properly protected, a diode biased beyond reverse break down voltage will be destroyed, becoming either a open circuit or a dead short piece of metal.

Now that Jim has invented a passive device that has 20dB gain, I need to see if I can't heat my house for free by daisy-chaining lots of these transformers together ;)

73,

Steve G3TXQ

It is there, just that you can't use it, since as soon as you use it, the impedence will not be the same.

Ah - so it wont be "way less than 0dB" like you told us!

Steve G3TXQ

It is undefined, so it could be either way less or way more - just you don't know.

If you have a 1000 Watt linear amplifier, you don't know what is its power gain until you apply proper input and termination. Without power, signal input, proper load, you won't know if it has a 10 dB gain (100W RF in/1000W RF out) or negative power gain (100 RF W in, nothing out).

G3TXQ

01-02-2010, 02:58 PM

That is the field of non-linear circuit analysis, which is the fundation of modern circuit analysis and computer based circuit simulation/design. A diode is a nonlinear device - when positively biased, it represents a small resistance and when biased below forward conducting voltage or reverse-baised, it is a large resistance, almost like a insulator. However, if reversely biased beyond its break down voltage, it will conducting again (zener diodes work in this region). When not properly protected, a diode biased beyond reverse break down voltage will be destroyed, becoming either a open circuit or a dead short piece of metal.

Jim, why are you introducing the topic of non-linearity - it has nothing to do with the discussion. Are you confusing non-linearity with the issue of non-constant-impedance between measuring points? They are quite separate issues, and non-linearity has nothing to do with the debate. An ideal 1:10 transformer is a linear device.

It is there, just that you can't use it, since as soon as you use it, the impedence will not be the same.

So let's place a 50 ohm load across the secondary of the ideal 1:10 transformer so that we dissipate power. The voltage across the secondary is still 10 times the voltage across the primary. Could you please tell me what the gain of the transformer is now? If you choose to express your answer in decibels, please remember that the definition of a decibel is that it is a logarithmic relationship between two powers.

Steve G3TXQ

Jim, why are you introducing the topic of non-linearity - it has nothing to do with the discussion. Are you confusing non-linearity with the issue of non-constant-impedance between measuring points? They are quite separate issues, and non-linearity has nothing to do with the debate. An ideal 1:10 transformer is a linear device.

You mentioned that pdf that deribes an amplifier, which is nonlinear by nature.

So let's place a 50 ohm load across the secondary of the ideal 1:10 transformer so that we dissipate power. The voltage across the secondary is still 10 times the voltage across the primary. Could you please tell me what the gain of the transformer is now?

voltage step-up ratio is 20 dB. :D

If you choose to express your answer in decibels, please remember that the definition of a decibel is that it is a logarithmic relationship between two powers.

Steve G3TXQ

That is your defininition. If it were two powers, why there are 10 log rule and 20 log rule?

I'm SOOOO CONFUSED ! ! ! ! :eek: :confused: :rolleyes: :p

I'm SOOOO CONFUSED ! ! ! ! :eek: :confused: :rolleyes: :p

haha g3txq threw in a 1:10 tasnformer - what this transformer did was nothing more than reflexing a 50 ohm load from the secondary side to a 0.5 ohm load on the primary side. So although the voltage was jacked up, the current was reduced propotionally, and the net result would be no power gain.

G3TXQ

01-02-2010, 04:52 PM

That is your defininition. If it were two powers, why there are 10 log rule and 20 log rule?

No, not my definition - every engineering textbook I have defines the decibel as:

decibel = 10 log (P1/P2)

In other words it is defined in terms of power. The text books typically then go on to a derivation as follows:

decibel = 10 log (P1/P2) = 10 log ((V1^2/R1)/(V2^2/R2))

They then introduce the crucial condition:

Iff R1=R2 then decibel = 10 log ((V1^2)/V2^2))

= 20 log (V1/V2)

Note that this expression is not the definition of the decibel; it is a derivation that is only valid if R1=R2. To apply it to a transformer where the voltages are measured at different impedance points, is to misapply it. This misapplication produces the nonsense that the transformer now has two different gains, measured in identical units.

These are not two expressions that are equally valid. The 10log(P1/P2) expression is universally applicable because it is the definition. The 20log(V1/V2) expression is a derivation which is only valid under certain conditions.

As the paper I referenced said: "The use of the decibel to express the voltage gain of an amplifier in a system of non-constant impedances is by far the most common misapplication of the decibel that you're likely to encounter"

You can't get away from the fact that when you use the decibel as a unit you are referring to a power ratio.

Steve G3TXQ

WB2UAQ

01-02-2010, 06:41 PM

Hi Steve,

I just want to re-arrange the log stuff a little bit. I have done this for explanation for those that lost sight of the fact that multiplication is the addition of logs.

(V1^2) / (V2^2) can be expressed as (V1/V2)^2 = (V1/V2) X (V1/V2). This allows the following:

dB = 10 log(V1/V2) + 10 log(V1/V2) = 20 log(V1/V2).

This helps explain where the factor "20" comes from if not covered earlier.

Also, please refer to an earlier comment of mine in this thread in regard to using a coupler to sample the power at a lower level. I used this method last night to measure the relative current at the output of my transmatch for an experiment that supports Eric's, KL7AJ, max current adjustment concept for selecting the best L and C's in a transmatch.

73, Pete

G3TXQ

01-02-2010, 06:53 PM

Pete,

Thanks - that's a useful intermediate step for anyone trying to follow the maths.

I, too, like the coupler approach and use it in a number of sensing applications around the shack. There's a couple of extra things to bear in mind when using it for this application:

* Just like the resistive attenuator, it doesn't fully isolate the TX from wide impedance swings at the input to the tuner because they will be reflected back into the current transformer primary.

* I would want to check that a core which handled the power OK in a current transformer which is terminated in 50 ohms, is also safe with a wide range of impedances across its secondary winding.

73,

Steve G3TXQ

No, not my definition - every engineering textbook I have defines the decibel as:

decibel = 10 log (P1/P2)

In other words it is defined in terms of power. The text books typically then go on to a derivation as follows:

decibel = 10 log (P1/P2) = 10 log ((V1^2/R1)/(V2^2/R2))

They then introduce the crucial condition:

Iff R1=R2 then decibel = 10 log ((V1^2)/V2^2))

= 20 log (V1/V2)

Steve G3TXQ

Typically doens't means always.

How do you describe an OpAmp's voltage gain of 100? The input impedence is usually 10k Ohm or much higher and the output impedence is less than 100 Ohm.

Usually, we will just say it has a voltage gain of 40 dB and be done with it. There are bunch of conditions implied here.

For new comers, it is confusing as they might not know all the conditions implied in using dB convention.

Hi Steve,

I just want to re-arrange the log stuff a little bit. I have done this for explanation for those that lost sight of the fact that multiplication is the addition of logs.

(V1^2) / (V2^2) can be expressed as (V1/V2)^2 = (V1/V2) X (V1/V2). This allows the following:

dB = 10 log(V1/V2) + 10 log(V1/V2) = 20 log(V1/V2).

This helps explain where the factor "20" comes from if not covered earlier.

That is a convenience with the assumption that impendence is constant. Once defined, we will use it on other places/conditions so that we don't have to note "Ah, by the way, this voltage gain is calculated by using 10 log rule not the 20 log rule".

The benefit of the 20 log rule is that, the voltage gain in dB will be the same as the power gain WHEN impedence is the same for V1 and V2. In many cases, the impedence may not be the same, then we still use voltage gain in dB calculated using the 20 log rule, just that the voltage gain won't necessarily be the same as the power gain.

G3TXQ

01-02-2010, 08:02 PM

How do you describe an OpAmp's voltage gain of 100? The input impedence is usually 10k Ohm or much higher and the output impedence is less than 100 Ohm.Usually, we will just say it has a voltage gain of 40 dB and be done with it.

You're right, folk have even misapplied the 20log expression to the voltage gain of an Op Amp. But you'll find that a manufacturer like National specs the voltage gain of its 741 Op Amp in sensible units of V/mV.

http://www.national.com/ds/LM/LM741.pdf

It's not rocket science; if you want to specify the ratio of two voltages where the impedances are not equal, use the appropriate units - for example V/mV - don't misapply and massage a unit which is a power ratio!

There are bunch of conditions implied here.

Actually, just one condition - you have to be measuring the voltages across the same impedance.

You've obviously disregarded my earlier references. Let me try just one more:

ARRL Antenna Book, 21st (latest) edition, Chapter 2 Antenna Fundamentals, Page 2-9, sidebar "Introduction to the decibel":

If the voltage ratio is given, the number of decibels is equal to 20 times the common logarithm of the ratio. That is:

dB=20log(V1/V2)

When a voltage ratio is used, both voltages must be measured across the same value of impedance. Unless this is done the decibel figure is meaningless, because it is fundamentally a measure of a power ratio

That's what you get when you apply the 20log expression to transformer primary and secondary voltages - meaningless decibel figures!

Steve G3TXQ

That's what you get when you apply the 20log expression to transformer primary and secondary voltages - meaningless decibel figures!

Steve G3TXQ

No, in that case, voltage gain is 20 dB and current gain is -20 dB and power gain will be 20 -20 = 0 dB.

G3TXQ

01-02-2010, 08:28 PM

No, in that case, voltage gain is 20 dB and current gain is -20 dB and power gain will be 20 -20 = 0 dB.

Jim, you have it your way - I'll stick with the ARRL view and my engineering text books.

Steve G3TXQ

Jim, you have it your way - I'll stick with the ARRL view and my engineering text books.

Steve G3TXQ

They made the 20 log rule so that under certain conditions, voltage gain expressed in dB is the same as power gain expressed in dB. In other conditions, the two numbers may not be the same.

If you stick with your narrowed view, you can't even use 40 dB over S9, as that is a voltage ratio.

G3TXQ

01-02-2010, 09:14 PM

If you stick with your narrowed view, you can't even use 40 dB over S9, as that is a voltage ratio.

S9 is a power level, not a voltage level; here's the IARU recommendation:

IARU Region 1 Technical Recommendation R.1 defines S9 for the HF bands to be a receiver input power of -73 dBm. This is a level of 50 microvolts at the receiver's antenna input assuming the input impedance of the receiver is 50 ohms.

40dB above S9 means that the input signal power is -33dBm

Steve G3TXQ

S9 is a power level, not a voltage level; here's the IARU recommendation:

40dB above S9 means that the input signal power is -33dBm

Steve G3TXQ

S9 is referenced at dBm level -73 at HF, -93 at VHF but one unit is at 6dB voltage difference (twice).

The signal is usually derived from AGC voltage.

G3TXQ

01-02-2010, 10:20 PM

S9 is referenced at dBm level -73 at HF, -93 at VHF but one unit is at 6dB voltage difference (twice).

Jim, you're missing the point!

I'm very happy to measure the voltage at some point in a receiver, and apply the 20Log rule to convert the voltage change to dB units .....

....... because the reference impedance does not change.

I don't know how many more ways to express it, or how many more references to give you: what you can't do is apply the 20Log(V1/V2) expression when the voltages are being measured across different impedances

Steve G3TXQ

I don't know how many more ways to express it, or how many more references to give you: what you can't do is apply the 20Log(V1/V2) expression when the voltages are being measured across different impedances

Steve G3TXQ

There is no problem when one only comparing voltages. When impedances are different, voltage ratio in dB will not be the same as power ratio in dB.

In mamy cases, people are only interested in voltage ratio, not power ratio.

The dB is just a convenient way to express large numbers and substitute multification with addtion when dealing with cascading circuits. It is just one more way to express a number.

G3TXQ

01-02-2010, 11:32 PM

In mamy cases, people are only interested in voltage ratio, not power ratio.

Then they should use the appropriate units, not units defined as a power ratio.

The dB is just a convenient way to express large numbers and substitute multification with addtion when dealing with cascading circuits.

Being convenient doesn't make something correct!

If you haven't already done so, do read the Ham radio article on the decibel - it gives an interesting historical perspective on how the 20Log(V1/V2) expression came to be so widely misunderstood and mis-applied:

http://michael.e.gruchalla.org/WebpagePapers/Defining%20the%20Decibel.pdf

Steve G3TXQ

Then they should use the appropriate units, not units defined as a power ratio.

Being convenient doesn't make something correct!

Steve G3TXQ

The decibel (dB) is a logarithmic unit (http://en.wikipedia.org/wiki/Logarithmic_unit) of measurement that expresses the magnitude of a physical quantity (usually power (http://en.wikipedia.org/wiki/Power_(physics)) or intensity (http://en.wikipedia.org/wiki/Intensity_(physics))) relative to a specified or implied reference level.

The following paragraph describes you and any one arguing that dB is only a power ratio:

"

An example illustrating the subtleties of the 20-log vs. 10-log rules is given by the so-called polarization ellipticity (http://en.wikipedia.org/wiki/Polarization_ellipticity), the minor-to-major-axis ratio of the polarization ellipse. It is an amplitude ratio, thus when reported in decibels, it follows the 20-log rule, αdB = 20log10α = 10log10α2. A reader unfamiliar with the applicability of each 20- and 10-log rules might find αdB inconsistent, arguing that it redefines the original α as a power ratio rather than an amplitude ratio. The crux is that the subscript dB might imply either of two different mathematical operations, thus it can only be interpreted unambiguously given additional information about the nature of the quantity being reported (power ratio or amplitude ratio). "

G3TXQ

01-03-2010, 10:02 AM

So, readers, here's a cautionary tale:

John decides to invent a unit for expressing the road speed of his vehicles. He decides on a unit he calls "Feet per Minute" and he defines it like this:

FPM = (Distance travelled in feet)/(Time in minutes)

Along comes Jack who spots that his vehicle's wheel circumference is exactly 6ft; so he is able to derive another expression which relates FPM to the rotational velocity of his road wheels, which he finds easier to measure:

FPM = 6 x (Number of revolutions)/(Time in minutes)

Jack is at pains to point out that this derived expression only works if the wheel has a circumference of 6ft.

Next Jim appears on the scene - he thinks that these new FPM units are real neat and "convenient". He looks at Jack's expression, ignores the caveat, and mistakenly believes that it is a definition of FPM. He uses FPM as a way of describing the rotational velocity of his wheels, rather than the road speed; and he applies Jack's expression no matter what the wheel circumference. Why he didn't simply use the appropriate RPM units no-one knows! Pretty soon all Jim's pals are doing the same thing.

Unfortunately no-one "lived happily ever after" in this sad story; because from then on, whenever anyone used FPM units they would get asked: "Do you mean the road speed? Or do you mean the rotational velocity of a wheel which - if it were 6ft in circumference, which it isn't - would result in the vehicle travelling at that speed, which it isn't?"

No wonder the ARRL warns:

When a voltage ratio is used, both voltages must be measured across the same value of impedance. Unless this is done the decibel figure is meaningless, because it is fundamentally a measure of a power ratio

Steve G3TXQ

SM0AOM

01-03-2010, 11:08 AM

It is a sign of "lack of engineering thinking" to argue about "voltage decibels", something I now and then had to point out to my students when teaching Electrical Measurements and Microwave Technology.

There should be no discussion about that the decibel is defined as a power ratio. The 10*log(P2/P1) formula is then always valid.

You can then express power as P = I^2*R or U^2/R, which leads to the much-misused formulas 20*log(U2/U1) or 20*log(I2/I1), which are only valid for the same termination impedance at both amplitude levels.

This is clearly spelled out in the decibel appendix of Terman's "Radio Engineering" whose first edition was written when the decibel was a fairly new concept.

"When the two powers P1 and P2 being compared are dissipated in equal resistances the power ratio is proportional to the square of the voltage ratio or square of the current ratio. Under these conditions it is possible to express a voltage or current ratio in decibels by using the relation db = 20 log(E2/E1) = 20 log(I2/I1). This relation must be used with caution, however, since it holds only when the resistance associated with E1 (and I1) is the same as the resistance associated with E2 (or I2)."

Later, an "extended decibel definition" came into use in especially audio work; F. Langford-Smith can be quoted from Radio Designer's Handbook:

"Much confusion has been caused by the incorrect or careless use of decibels to indicate the gain of an amplifier..."

He then goes on to state extended relations where the effects of unequal termination resistances are included to avoid errors:

Level difference in db = 20 log(E2/E1) + 10 log (R2/R1)

or

Level difference in db = 20 log(I2/I1) + 10 log (R2/R1)

These relations can be further extended to handle reactive terminations.

The "dBuV" has been discussed earlier. It is a perfectly legal way of obtaining the same calculation convenience for amplitude units such as field strengths [dBuV/m], because they are tied to power level references by implicit use of "characteristic impedances" who can be either cable, structural or (far-)field impedances. The "dBuA" unit, familiar to those who deal with conducted EMI matters, falls into the same category.

If anyone desperately needs a logarithmic ratio based on amplitudes,

I would recommend the Neper. It is defined as the natural logarithm

(ln or loge) of the ratio of the two amplitudes, N = ln(U2/U1).

This definition is also valid only for equal impedances.

The Neper comes out as a direct consequence of wave-mechanics descriptions of electrical networks or transmission lines.

73/

Karl-Arne

SM0AOM

It is a sign of "lack of engineering thinking" to argue about "voltage decibels", something I now and then had to point out to my students when teaching Electrical Measurements and Microwave Technology.

There should be no discussion about that the decibel is defined as a power ratio. The 10*log(P2/P1) formula is then always valid.

If all one concerned was the voltage gain, why does he have to drag in the power ratio?

If V2/V1 = 10, then (V2/v1) square is 100 - this is a valid dimensionless number.

So, log 100 is also a valid number, which is 2.

Then, 20 x log 100 = 40 is also a valid number.

For convenience, one can say the number is 40 dB, an amplitude ratio calculated using 20 log rule.

Only when one also considers the power ratio, the impedance comes into play.

Unfortunately no-one "lived happily ever after" in this sad story; because from then on, whenever anyone used FPM units they would get asked: "Do you mean the road speed? Or do you mean the rotational velocity of a wheel which - if it were 6ft in circumference, which it isn't - would result in the vehicle travelling at that speed, which it isn't?"

Steve G3TXQ

That depends on the reference, if you use the Sun as the reference, you need to take into consideration of the Earth's speed.

BTW, dB is a dimensionless unit, a pure number - speed is not, you need to specify FPM, or 5 km/s

No wonder the ARRL warns:

"When a voltage ratio is used, both voltages must be measured across the same value of impedance. Unless this is done the decibel figure is meaningless, because it is fundamentally a measure of a power ratio "

Steve G3TXQ

A ratio is dimensionless, so let's see this

V1 = 10V over 50 ohm, V2 = 100V over 50ohm

V2/V1 = 10, expressed in dB it will be 20 dB, right? Same impedance here.

Now let's say V3 = 10V over 10 kohm, V4 = 100V over 8 ohm.

So V4/V3 = V2/V1 = 10.

If V2/V1 = 20 dB, then V4/V3 = 20 dB as a voltage ratio.

If you insist V4/V3's voltage ratio is not valid in dB term, one can just say V4/V3's voltage ratio is the same as V2/V1's voltage ratio, which is 20 dB.

G3TXQ

01-04-2010, 09:29 AM

It's a mistake to think that, because a unit is dimensionless, it can be applied to any situation where it might be convenient.

For example, when dealing with camera lenses we define the "f-number" to be the ratio of the focal length to the diameter; it's a dimensionless unit of inches/inches. On the other hand, when dealing with the geometry of my TV picture I talk about its "Aspect Ratio", defined as the ratio of its width to its height; that also has dimensionless units of inches/inches.

But it would be a total misapplication of the terms, and confusing, to talk about the Aspect Ratio of a lens or the f-number of the TV picture.

Using dB to express voltage gain is a bigger mistake: not only is it a mis-application of the term, it is also has the wrong originating dimensions. The dB is defined as a power ratio, and therefore has units of Watts/Watts.

When you tell me a 1:10 transformer has a voltage gain of 20dB, you're effectively telling me that "the ratio of Vout to Vin is 100 Watts per Watt".

If Vin and Vout happen to be measured across equal impedances, the phrase just about has some credibility. If the impedances are not equal, I have to interpret it as:

"the ratio of Vout to Vin is such that it would result in 100 times more power if the voltages had been measured across equal impedances, which, by the way, they haven't"

I can't imagine a more convoluted way of telling me that the voltage gain is 10!

As the ARRL warns:

When a voltage ratio is used, both voltages must be measured across the same value of impedance. Unless this is done the decibel figure is meaningless, because it is fundamentally a measure of a power ratio

and Terman said:

When the two powers P1 and P2 being compared are dissipated in equal resistances the power ratio is proportional to the square of the voltage ratio or square of the current ratio. Under these conditions it is possible to express a voltage or current ratio in decibels by using the relation db = 20 log(E2/E1) = 20 log(I2/I1). This relation must be used with caution, however, since it holds only when the resistance associated with E1 (and I1) is the same as the resistance associated with E2 (or I2).

Steve G3TXQ

It's a mistake to think that, because a unit is dimensionless, it can be applied to any situation where it might be convenient.

A rato is a pure number.

For example, when dealing with camera lenses we define the "f-number" to be the ratio of the focal length to the diameter; it's a dimensionless unit of inches/inches. On the other hand, when dealing with the geometry of my TV picture I talk about its "Aspect Ratio", defined as the ratio of its width to its height; that also has dimensionless units of inches/inches.

But it would be a total misapplication of the terms, and confusing, to talk about the Aspect Ratio of a lens or the f-number of the TV picture.

The F number implies diameter / focal length of a lens. It is similar to dBm, where the m has a specual meaning for the reference.

Using dB to express voltage gain is a bigger mistake: not only is it a mis-application of the term, it is also has the wrong originating dimensions. The dB is defined as a power ratio, and therefore has units of Watts/Watts.

Text books start with power ratio, so that they can extend to voltage ratio and 20 log rule. Watts/Watts is dimensionless.

When you tell me a 1:10 transformer has a voltage gain of 20dB, you're effectively telling me that "the ratio of Vout to Vin is 100 Watts per Watt".

It is 10, not 100.

Only when impedances are the same you can say the power ratio is 100. If the two impedance differ from each other, the power ratio won't be 100, BUT the voltage ratio is still 10.

If Vin and Vout happen to be measured across equal impedances, the phrase just about has some credibility. If the impedances are not equal, I have to interpret it as:

"the ratio of Vout to Vin is such that it would result in 100 times more power if the voltages had been measured across equal impedances, which, by the way, they haven't"

[/QUOTE]

But the ratio of 10 still hold, just that when the impedances are not equal, the power ratio is not 100.

I can't imagine a more convoluted way of telling me that the voltage gain is 10!

Because you can't image situations where voltage ratio isn't square root of power ratio.

As the ARRL warns: ...............

Steve G3TXQ

That is only for guys who don't understand dimensionless units and could not figure out 10 is 10 in a dimensionless world.

G3TXQ

01-04-2010, 04:42 PM

The F number implies diameter / focal length of a lens. It is similar to dBm, where the m has a specual meaning for the reference.

Incorrect - it's a common mistake which beginners make to think that "dBm" is a dimensionless ratio like the f-number of a lens; in fact it has the dimension of mWatts. For example 20dBm is an absolute power measurement of 100mWatts.

Text books start with power ratio, so that they can extend to voltage ratio and 20 log rule.

No - they start with power ratio because that is the definition of the decibel. Anything else is a derivation. Can you show me any engineering textbook which starts with a 20Log(V1/V2) definition of a decibel?

Watts/Watts is dimensionless.

Correct! But as I explained, just because a unit is dimensionless doesn't mean it should be applied outside of its specific definition. The f-number of a lens is a dimensionless quantity, but we wouldn't specify it in dB; or perhaps you would - now would that be the 10Log(focal length/diameter) or 20Log(focal length/diameter) ....... ?

It is 10, not 100.

Nope - it's 100Watts/Watt, because that's what 20dB means.

That is only for guys who don't understand dimensionless units and could not figure out 10 is 10 in a dimensionless world.

Actually it's for guys that like to see engineering units applied properly.

As the ARRL warns:

When a voltage ratio is used, both voltages must be measured across the same value of impedance. Unless this is done the decibel figure is meaningless, because it is fundamentally a measure of a power ratio

and Terman said:

When the two powers P1 and P2 being compared are dissipated in equal resistances the power ratio is proportional to the square of the voltage ratio or square of the current ratio. Under these conditions it is possible to express a voltage or current ratio in decibels by using the relation db = 20 log(E2/E1) = 20 log(I2/I1). This relation must be used with caution, however, since it holds only when the resistance associated with E1 (and I1) is the same as the resistance associated with E2 (or I2).

Steve G3TXQ

Incorrect - it's a common mistake which beginners make to think that "dBm" is a dimensionless ratio like the f-number of a lens; in fact it has the dimension of mWatts. For example 20dBm is an absolute power measurement of 100mWatts.

20 dBm is 20 dB over mW, 20 dB is dimensionless, all the meaning is in the m. This is just like in 100 Watts, the number 100 is just a pure number.

20 dB means a particular type of mathmatical manipulation.

No - they start with power ratio because that is the definition of the decibel. Anything else is a derivation. Can you show me any engineering textbook which starts with a 20Log(V1/V2) definition of a decibel?

dB is a ratio, it could be power ratio, or it could be amplitude ratio.

The f-number of a lens is a dimensionless quantity, but we wouldn't specify it in dB .......

Because that is not conventional method to communicate.

Using dB to express voltage gain brings all the benefits of logarithmic manipulation.

G3TXQ

01-04-2010, 06:38 PM

20 dBm is 20 dB over mW, 20 dB is dimensionless, all the meaning is in the m.

Exactly - so you were mistaken when you said that the F-number - which is dimensionless - is similar to the dBm, which has dimensions of power

dB is a ratio, it could be power ratio, or it could be amplitude ratio.

Why stop there? Why not use it for expressing the ratio of two bandwidths?

No - it's defined as a power ratio. Then, I guess you couldn't find any engineering text book which started with a 20Log(V1/V2) definition ?

Using dB to express voltage gain brings all the benefits of logarithmic manipulation.

So does using units like Nepers which were designed for what you are trying to do - defined as a logarithimic ratio of two Voltages.

As the ARRL warns:

When a voltage ratio is used, both voltages must be measured across the same value of impedance. Unless this is done the decibel figure is meaningless, because it is fundamentally a measure of a power ratio

and Terman said:

When the two powers P1 and P2 being compared are dissipated in equal resistances the power ratio is proportional to the square of the voltage ratio or square of the current ratio. Under these conditions it is possible to express a voltage or current ratio in decibels by using the relation db = 20 log(E2/E1) = 20 log(I2/I1). This relation must be used with caution, however, since it holds only when the resistance associated with E1 (and I1) is the same as the resistance associated with E2 (or I2).

Steve G3TXQ

Exactly - so you were mistaken when you said that the F-number - which is dimensionless - is similar to the dBm, which has dimensions of power

F stands for focal length.

dB (not dBm) has no dimension of power.

No - it's defined as a power ratio. Then, I guess you couldn't find any engineering text book which started with a 20Log(V1/V2) definition ?

It (dB) is not a power ratio - this is what wikipedia has for dB

The decibel (dB) is a logarithmic unit (http://en.wikipedia.org/wiki/Logarithmic_unit) of measurement that expresses the magnitude of a physical quantity (usually power (http://en.wikipedia.org/wiki/Power_(physics)) or intensity (http://en.wikipedia.org/wiki/Intensity_(physics))) relative to a specified or implied reference level. Since it expresses a ratio of two quantities with the same unit, it is a dimensionless unit (http://en.wikipedia.org/wiki/Dimensionless_unit).

Can you read ? (usually power (http://en.wikipedia.org/wiki/Power_(physics)) or intensity (http://en.wikipedia.org/wiki/Intensity_(physics))) <== but may be something else.

So does using units like Nepers which were designed for what you are trying to do - defined as a logarithimic ratio of two Voltages.

Why bother? Just because you can't understand (V2/V1) square is as valid as (V2/V1)?

The ARRL warnings are for those who can't understand 10 = 10, :D:D:D

So does using units like Nepers which were designed for what you are trying to do - defined as a logarithimic ratio of two Voltages.

I don't use Nepers, if you insist dB is power ratio, you use Nepers :D:D:D

G3TXQ

01-05-2010, 08:57 AM

F stands for focal length.

Nope! "F-number" is a ratio of focal length to lens diameter

It (dB) is not a power ratio - this is what wikipedia has for dB .....

Interesting that you would prefer Wikipedia as an authoritative source rather than ARRL or engineering textbooks! But if you look further down the same Wikipedia page you'll see that it clearly states:

Since the decibel is defined with respect to power, not amplitude .....

Steve G3TXQ

G3TXQ

01-05-2010, 09:14 AM

This will be my final posting on this topic!

ARRL handbook:

The number of decibels corresponding to a given power ratio is given by:

dB = 10 log(P2/P1)

Note that the decibel is based on power ratios. Voltage or current ratios can be used, but only when the impedance is the same for both values of voltage or current. The gain of an amplifier cannot be expressed correctly in decibels if it is based on the ratio of the output voltage to the input voltage unless both voltages are measured across the same value of impedance.

ARRL Antenna Book, 21st (latest) edition, Chapter 2 Antenna Fundamentals, Page 2-9, sidebar "Introduction to the decibel":

If the voltage ratio is given, the number of decibels is equal to 20 times the common logarithm of the ratio. That is:

dB=20log(V1/V2)

When a voltage ratio is used, both voltages must be measured across the same value of impedance. Unless this is done the decibel figure is meaningless, because it is fundamentally a measure of a power ratio.

Radio handbook, William Orr, W6SAI:

The decibel is defined

Ndb = 10 log(Po/Pi)

..... then later:

NdB = 20 log (Eo/Ei) or 20 log (Io/Ii)

where the subscript o denotes the output voltage or current and i the input voltage or current. Remember, this equation is true only if the voltage or current gain in question represents a power gain which is the square of it and not if the power gain which results from this is some other quantity due to impedance changes.

An introduction to Radio Frequency Design, Wes hayward, W7ZOI:

Gain is usually presented in dB. Formally, G=10log(Pl/Pin). If the source and load terminations are equal, G= 20 log(Vl/Vin). Logarithmic units should be used only with care when applied to voltage gain.

The Radio Communication Handbook, RSGB:

The number of decibels N representing the ratio of two power levels P1 and P2 is 10 times the common logarithm of the power ratio thus:

The ratio N = 10 log (P2/P1) decibels.

If it is required to express voltage (or current) ratios in this way, they must relate to identical impedance values.

Radio Engineering,Terman:

When the two powers P1 and P2 being compared are dissipated in equal resistances the power ratio is proportional to the square of the voltage ratio or square of the current ratio. Under these conditions it is possible to express a voltage or current ratio in decibels by using the relation db = 20 log(E2/E1) = 20 log(I2/I1). This relation must be used with caution, however, since it holds only when the resistance associated with E1 (and I1) is the same as the resistance associated with E2 (or I2).

Ham Radio Magazine:

The use of the decibel to express the voltage gain of an amplifier in a system of non-constant impedances is by far the most common misapplication of the decibel that you're likely to encounter

QST Magazine:

The fallacy in this is that there is no such thing as voltage dB or current dB. The decibel is a logarithmic expression of power ratio only

Each of those references define the decibel as a power ratio and caution against using it for voltage ratios in circumstances where they are being measured across different impedances. You obviously do not accept Terman, ARRL, RSGB, William Orr, Wes Hayward, QST or Ham Radio as authoritative, so you're unlikley to be persuaded by me!

The simple fact is that a dB is a power ratio. If you choose to specify a voltage gain as 20dB you're saying that the voltage gain is equivalent to a power gain of 100. That is never true unless voltages are measured across equal impedances.

73 and a Happy 2010,

Steve G3TXQ

No, not my definition - every engineering textbook I have defines the decibel as:

decibel = 10 log (P1/P2)

In other words it is defined in terms of power. The text books typically then go on to a derivation as follows:

decibel = 10 log (P1/P2) = 10 log ((V1^2/R1)/(V2^2/R2))

They then introduce the crucial condition:

Iff R1=R2 then decibel = 10 log ((V1^2)/V2^2))

= 20 log (V1/V2)

Note that this expression is not the definition of the decibel; it is a derivation that is only valid if R1=R2. To apply it to a transformer where the voltages are measured at different impedance points, is to misapply it. This misapplication produces the nonsense that the transformer now has two different gains, measured in identical units.

These are not two expressions that are equally valid. The 10log(P1/P2) expression is universally applicable because it is the definition. The 20log(V1/V2) expression is a derivation which is only valid under certain conditions.

As the paper I referenced said: "The use of the decibel to express the voltage gain of an amplifier in a system of non-constant impedances is by far the most common misapplication of the decibel that you're likely to encounter"

You can't get away from the fact that when you use the decibel as a unit you are referring to a power ratio.

Steve G3TXQ

I've seen good techs think if they put four identical 25 watt attenuators in series they can dissipate 100 watts. This only costs an attenuator or two. And I've got a 54 Ohm dummy load that USED to be a 50 Ohm, unidirectional 20 dB attenuator, where the user transmitted into the wrong (low power) end. (I checked the attenuators in our EMC lab to be sure they were symmetrical.)

Cortland

KA5S

The simple fact is that a dB is a power ratio. If you choose to specify a voltage gain as 20dB you're saying that the voltage gain is equivalent to a power gain of 100. That is never true unless voltages are measured across equal impedances.

73 and a Happy 2010,

Steve G3TXQ

The dB is from deci Bel

Bel means log (A/B), a mathmatical manipulation.

Deci means 10, as 1 meter = 10 deci meter.

So dB means 10 log (A/B)

Saying dB must be power ratio is just like saying mathematics is all about 1+1=2, nothing else.

As for your quote from references - Do you have a functioning brain?