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## bias meter shunt?

Hello all - time for another very simple question here. I "think" I understand, but wish to confirm with the group to be SURE I am doing this whole thing in a SAFE manner - that will work.

From what I understand reading/studying W8JI's http://www.w8ji.com/metering_amplifier.htm webpage and all the comments and reading I've done here the ONLY connection between the B- (taken at neg rail of capacitor bank) and chassis ground is the diode, low voltage capacitor, AND the bias meter current shunt? This shunt will normally be in the tenth's of an ohm - but have seen resistor ranges from 1 ohm to 200 ohms between B- and chassis ground.

In low voltage current meters I tend to use custom wound shunts in the hundredths of an ohm to reduce voltage drop and have small resistance 1mA meters (maybe only 50 to 100 ohm meters).

Is there a major safety (at this voltage ALL safety is "major") - issues with using a 50 ohm 1mA meter with a 60 milliohm shunt vs adding a 400 ohm resistor in series with meter to make a 450 ohm meter to use a .6 ohm resistor for shunt?

Thank you to all for comments and sharing knowledge.

73 de Ken H>

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On my new amplifier I plan on using a 1 ohm resistor between the B- and chassis ground for the grid current. That way you can use a 1 mA meter with a series resistor to read grid current. Using ohm's law 1 volt drop across the meter represents 1000 mA of current. For a maximum meter reading for a grid current of 500 mA you would use a 500 ohm series resistor. If you need a higher or lower reading then adjust the series resistor accordingly. Basically, using a 1 mA meter the voltage reading is "1000 ohms per volt".

For W8JI:

To read grid current I definitely agree that the B- has to be isolated from chassis ground and that another meter shunt has to go between the low side of the voltage and capacitor bank. In the case of my particular high voltage transformer I will be using a full wave and not a full wave bridge and that means the center tap of the transformer. If I were to use a full wave bridge I would get more than 6000 volts!

Glen, K9STH

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Originally Posted by K9FV
Hello all - time for another very simple question here. I "think" I understand, but wish to confirm with the group to be SURE I am doing this whole thing in a SAFE manner - that will work.

From what I understand reading/studying W8JI's http://www.w8ji.com/metering_amplifier.htm webpage and all the comments and reading I've done here the ONLY connection between the B- (taken at neg rail of capacitor bank) and chassis ground is the diode, low voltage capacitor, AND the bias meter current shunt? This shunt will normally be in the tenth's of an ohm - but have seen resistor ranges from 1 ohm to 200 ohms between B- and chassis ground.

In low voltage current meters I tend to use custom wound shunts in the hundredths of an ohm to reduce voltage drop and have small resistance 1mA meters (maybe only 50 to 100 ohm meters).

Is there a major safety (at this voltage ALL safety is "major") - issues with using a 50 ohm 1mA meter with a 60 milliohm shunt vs adding a 400 ohm resistor in series with meter to make a 450 ohm meter to use a .6 ohm resistor for shunt?

Thank you to all for comments and sharing knowledge.

73 de Ken H>
Ken,

There are infinite combinations of shunts and multipliers.

The reason I put the voltage at the meter shunt up around 1/2 volt, is this allows a single diode to be used as a clamp to protect the meter. A standard rectifier can allow as much as .8 volts or more under a high current fault condition.

If you use a 50 mV meter the maximum potential for the meter in a fault would be more than .8 / .05 = 16 times full scale!!! Ouch.

The circuit also becomes more sensitive to resistance errors in connections and to ground loop errors.

The resistor is also non-standard.

I think my metering page explains why, for high current applications, we want a 50 mV or similar shunt and why for low current applications we generally want a higher basic shunt voltage. You have 3000 volts to deal with. Why would you care if you lost .3 volts or so from a shunt drop when normal sag is 300 volts under load? :-)

If you want to protect the meter from pinning from overload, how can you do that with a 50 mV shunt? If you want to add a reliable grid trip circuit, how can you do that with a 50 mV FS shunt?

73 Tom

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Tom, Glen - thanks for the input. I had read over the stuff several times but the reasoning behind just didn't click with me until now. NOW (finally!!) I see why the low value shunt is good for low voltage, but for high voltage it's MUCH better to use a higher value shunt.

That is the approach I'll use. Thanks to all again.

73 de Ken H>

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Originally Posted by HFRF
In 40 years I have NEVER had a shunt in an amplifier open. I have had a box of 1 ohm 100 watt resistors for shunt use. Never seen one open yet. By using the "volt meter method" which uses a 1 ohm resistor shunt with a 1000 ohm series resistor going to a 1 ma meter calibrated to read 1 amp (or other similar ranges), the likelihood that the meter will fry is about zero. I have seen many diodes used as clamps in high voltage circuits literally disintegrate during failures. Forget the diodes, use big resistors.

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Originally Posted by HFRF

Apparently, one has to spell every detail out or those that need to will jump on it for criticism value.
Using a 50-100 watt resistor may stop the shunt resistor from opening, but it is a little silly to use a 50 watt resistor for a resistor that dissipates only 0.01 watts in normal operation. (Based on a 1 ohm resistor with 100 mA grid current.)

Let's look at the problems objectively instead of emotionally. Let's assume two ways....one where the builder is smart enough to limit surge with a proper effective resistance total in the capacitor path of 15-20 ohms and the other where the builder is a fool and uses oil filled caps or exceptionally low ESR electrolytics with no fault limit resistance:

1.) If the HV faults to ground, voltage across the shunt will be full supply voltage distributed by ratio of shunt to fault path resistance. With a 1 ohm shunt and a 15 ohm fault path resistance, we would have 1/16th of the supply voltage across the shunt during a fault. That would be 3000/16 = 187.5 volts across the meter, any bias components, and anything else in parallel with that fault path from the capacitor negative lead to the chassis.

The peak dissipation in the shunt resistor system would be 35,156 watts.

Does that sound good to you for a worse case design?

2.) If the builder does not include additional or planned ESR in the fault path, and the builder uses oil filled or low ESR electrolytics, the fault path resistance might be just a few ohms. Let's assume 3 ohms.

Now fault current is 3000/4 = 750 amps. Shunt voltage is 750 volts across the meter. Shunt peak momentary dissipation is 750*750 = 562 kW

I certainly don't object to someone using a 50-100 watt resistor in a system where normal dissipation is around .01 watts if that is what they want to do, but pretending that large resistor clamps the negative rail and protects the meter or biasing or other components that might be in the fault path very clearly is wrong.

A normal small diode will handle a few hundred amps surge without opening. It will clamp someplace less than 2 volts or so, and although the junction might go shorted the meter will be safe.

Second, a heavy but low voltage (let's say 50 volt) capacitor shunting the diode will arc and short and provide additional clamping at 100 volts or so. This is still less voltage than the unprotected 50 watt 1 ohm in worse case conditions.

I certainly don't mind alternatives or a little back-and-forth debate, but dismissing the use of a clamp that starts to hold at .6 to .7 volts and offering a system that will allow peak voltages of 150-1000 volts or more across the metering path (and possibly bias path) is not the best solution, and certainly not an improvement.

Every Ameritron amp since around 1985 has had diode negative rail clamps, and has been ESR limited for HV in some form (sometimes though selected component ESR values). There hasn't been a single recorded HV fault opening of a diode (I checked just now) in the service history of several thousand amplifiers, except a few cooked open from tube filament to grid shorts. There haven't been a single grid shunt failure or meter opening from a HV fault.

Not only are they used there, I use that method in some pretty large commercial amps and never have had a problem.

I don't think the idea a 100 watt resistor is a viable better solution has any merit at all. The 100 watt resistor, other than wasting space, isn't the issue. The issue is the voltage division between capacitor path ESR and shunt resistance. We really always should have SOMETHING to clamp that path. 100-1000 volts across a 1/2 volt rated meter is never good.

73 Tom

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Originally Posted by HFRF
Your usual snotty, arrogant, dismissive, and condescending attitude toward anybody who dares to disagree with your opinions, understanding of the facts, beliefs, or etc. is childish. You need to grow up and stop telling everybody in this world they are wrong except for you of coarse. Your need to dig in and fight to the death mentality is just pathetic.
I guess that means you have no valid technical justification for letting the meter see several hundred volts in a fault, so the only thing left is name calling. :-)

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Originally Posted by HFRF
Who cares about a cheapo \$30 meter?
I don't understand why you insist on leaving the diode out. Most people probably won't want a meter to open if they can prevent it with a thirty cent diode.

I'll ask the question again in a different way. What will happen when the rfc in the grid path opens in the grid circuit (Drake / Heathkit), or a small resistor in the plate or grid circuit (Drake) opens or a small meter shunt opens which is in many amps produced over the years. These parts don't open when the amp is in standby.
Oh yes they certainly do! If a tube has gas, which is all too common these days with Chinese tubes, the grid chokes, mica caps, and meter shunts will often explode.

They open when the amp is keyed and a 100 watts is being dumped into the cathode.
They can and do open any time, regardless of drive or not. When the arc happens depends entirely on the condition of the tube.

Now I'll ask again, what do you think will happen when the grid or plate current path opens while the cathode is being driven?
Drive has nothing to do with it. The most common fault occurs when the tube ionizes inside. That connects the anode to the grid primarily, but can also connect anode to cathode if the fault is severe enough or if the grid circuit blows open.

Do you think some stupid semiconductor is going to prevent all the failures of parts in the grid / plate circuit path?
I don't know how smart the semiconductor is, but it will not prevent the arc. Nothing can prevent the arc.

The job of the semiconductor is to clamp the voltage from the B- rail to the chassis. This prevents the shunts from seeing high voltages and being damaged, and prevents the meters from being damaged. It also greatly reduces negative voltage that appears on the cathode of the tube during the fault as the positive rail is pulled to chassis through the anode-grid path. This reduces chances of bias circuit and exciter damage.

Why all the great desire to remove protection? It makes no sense because nothing gets better by leaving the \$0.30 diode out and adding a \$5 50 watt resistor.

Whats going to happen when the cathode rises to plate voltage potential when the cathode circuit opens?
That has nothing to do with the diode being there or not, but if the grid it tied to ground solidly the grid acts like a shield inside the tube. HV from the fault reaching the cathode is greatly reduced by the shielding of the grid.

If the grid is floated on chokes or capacitors the arc to the grid simply pulls the grid above chassis, and the grid to cathode path ionizes and arcs. This dumps full supply voltage on the cathode which then can dump supply voltage into the bias system and also back through the input circuit to the exciter.

That's a problem with having the grids float, and it absolutely happens. This is why the grid resistors and caps were removed from the AL572 and 811H amps, and the grids are now directly grounded with no resistors.

Do you want nothing more than a \$.50 capacitor between the 3000 volts at the cathode and the RF in coax connector? Do you want 3000 volts in your Icom 7800? Whats going to happen to the tube with 100 watts at the cathode with no grid or cathode continuity?
I explained all that. Leaving to diode out makes that problem worse, not better. Floating the grid above ground on capacitors makes the problem worse, not better.

The grid grounding is the line of defense from anode to cathode directly.

The diode is the line of defense from the negative rail going high in a fault, and dumping negative back through the bias to the cathode.

For some reason I can't understand you want to remove a layer of protection for meters, shunts, bias, and the exciter just to save \$0.30. To me, that makes no sense.

Whats going to happen if R1 to R4 on the Ameritron ALXX metering board open?
The virtually never open with the diode in place.

Whats going to happen if rfc1 opens on the capacitor filter board opens?
Absolutely nothing. First it rarely if ever opens, and second the diode any bypass cap take over.

The design of the Ameritron amplifiers are a tube disaster waiting to happen if the wrong \$.50 part should open.
That's just absolutely nonsense! You are sitting there telling me all the components that PROTECT the circuit from negative rail rising are dangerous! This tells me you either do not understand the circuitry, or you have never even thought through it.

This is a pretty simple DC path we are talking about. Take some time to look at it and I think you will see the errors in your statements.

73 Tom

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Tom, I'm looking at the triode board schematic to understand these components and their functions better. I'm thinking of an 8877-type amp here, with an indirectly heated cathode. He has the big diode from cathode to ground that clamps the voltage across the grid shunt resistor to about 0.8 volts. But he also has a pair of little 1N4000-series diodes in series across the big diode (opposite polarity). I'm trying to understand what happens In the case of an amp where the floating grid choke blew open and the grid-cathode space was ionized. The ~187 amps of fault current is going to flow from the cathode towards the B- through the plate shunt/meter but it is protected same as if there was a plate-grid short. The little diodes keep that cathode from going above 1.6 volts, right? But don't we have an RF choke from the cathode to the metering circuitry (required, so rf drive can only go to the cathode and not into the dc circuits) that is susceptible to being blown open? Maybe it's a moot point, since the 8877 grid is FIRMLY grounded right to the chassis, no grid chokes.
And on another issue, you are right, an amp sitting there in standby can certainly have an arc. It has happened to my Drake L4B. It takes out the little series resistor in the B+ lead. I have now taken those finals and done a gettering job on them (300 ma, 1000 volts with +24 on the grid for 4 hours) and installed a 20 ohm power R in the B+ lead. So far so good. But of 8 suspect 3-500s that someone loaned me to see if I could recover them from being gassy only one came back. The rest still had creeping plate current if I took them over the above levels, even after 3 hours. I also had a flash when I fired up my 8877 for the first time with a "supposedly" good 8877. Now it has a 3CPX1500 with all the protection you have been talking about here.

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Originally Posted by K2XT
Tom, I'm looking at the triode board schematic to understand these components and their functions better. I'm thinking of an 8877-type amp here, with an indirectly heated cathode. He has the big diode from cathode to ground that clamps the voltage across the grid shunt resistor to about 0.8 volts. But he also has a pair of little 1N4000-series diodes in series across the big diode (opposite polarity).
The supply is DC, so normally one diode in the correct direction is all that is needed. The other diodes won't hurt or help.

I'm trying to understand what happens In the case of an amp where the floating grid choke blew open and the grid-cathode space was ionized. The ~187 amps of fault current is going to flow from the cathode towards the B- through the plate shunt/meter but it is protected same as if there was a plate-grid short.
If the grids are not grounded, it sets up all sorts of issues. First, the cathode has to float to some higher voltage to bias the tube off. This limits any clamping we can do there.

In the case of an ungrounded grid where significant current diverts to the plate current shunt, and additional diode across the plate current shunt would be advisable. In the Ameritron amps I added MOV's to amps with resistors in the grid path, but recently I eliminated all grid resistors.

The little diodes keep that cathode from going above 1.6 volts, right?
I'd have to see the schematic. It is impossible to clamp the cathode very low in a GG amp because of RF voltage and bias (including cut off bias). It is possible to protect the meter with a diode across the plate current shunt.

But don't we have an RF choke from the cathode to the metering circuitry (required, so rf drive can only go to the cathode and not into the dc circuits) that is susceptible to being blown open? Maybe it's a moot point, since the 8877 grid is FIRMLY grounded right to the chassis, no grid chokes.
You won't have much of anything making it to the cathode then.

And on another issue, you are right, an amp sitting there in standby can certainly have an arc.
Happens every day. :-) Virtually every new 3CX3000F7 I used to buy would pop on initial application of HV, then it would be fine.

73 Tom

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