Push-Pull Output Transformers - Part III, The Final Countdown:

Discussion in 'Amateur Radio Amplifiers' started by KD2NCU, Sep 28, 2017.

1. K7JEMHam MemberQRZ Page

If we assume that Is is right at the choke feed to the amplifier, and that the 28v source is a true power supply, with zero ohms of AC reactance, then the current flowing at that point will be a steady state direct current, equal to whatever average current is flowing through the transistors. The choke is an inductor, it has reactance. This allows a circuit to have DC on one side, and AC + DC on the other. In a DC circuit, the instantaneous current flow is equal to the average current flow over many cycles, assuming the transmitter is operating at the same power output.

The inductor (choke) has some inductive reactance at the frequency of operation. We don't know the value of inductor, and we don't know the frequency. But it really doesn't matter if we assume that it was properly designed for optimum results.

From a DC perspective, the amplifier will look like a resistive load, not like a pulsating current load. A scope placed across the DC feed will show this to be true.

2. KD2NCUHam MemberQRZ Page

Wait, what happened? I thought you wanted to use circuit analysis. ??? Don’t like the answer it’s giving you? What, specifically, do you disagree with regarding the circuit analysis I provided other than the answer? You don’t believe Kirchhoff’s current law? “Gasp” May he smite you with a thousand nodes! Do you find a flaw in my reasoning and analysis other than you just don't like the answer? Can you explain how the two currents can be unequal? (Using circuit analysis, of course) If they’re unequal, where is the imbalance in current going or coming from? You have this electronic “stuff” inside the dashed box in the diagram I showed. You have one wire going in (from the DC supply), and one wire coming out (going to the drain). Just stay with circuit analysis for a second and tell me how the two instantaneous currents can ever be different.
For those two currents to be anything other than exactly equal in magnitude and shape at all times the number of electrons going in would not be equal to the number of electrons going out. For example, at the start of Q1’s cycle, its current is almost zero and Q2 is off. If the current from the DC supply is equal to the average of the collector current then there are a hell of a lot more electrons going into that red dashed box than there are going out. Where are all those extra electrons going? Mid cycle, the collector current is greater than the average. Where are all those extra electrons coming from? Are the wires swelling up larger than normal when the collector current is lower than the average and then shrinking smaller than normal when the collector current is smaller than the average?

C'mon now, you challenged my circuit analysis man-hood, I analyzed this simple 2 node circuit using Kirchhoff's current analysis and you're just brushing it aside. Now it's your turn to put your circuit analysis man-hood on the table and show how (using circuit analysis) those two currents can ever be different. Whatever explanation you come up with MUST obey Kirchhoff's current and voltage laws or the gods of Kirchhoff shall smite you down with a horrible vengeance.
BTW, since you asked, check my Bio yet? Nice picture of Zeus, don't you think?

Last edited: Oct 5, 2017
3. K7JEMHam MemberQRZ Page

You're wrong on so many levels here.

But a question I have for you, that has never been answered is how can the current be pulsing at a DC point, when that point is an AC short? The amplifier is fed with pure DC to the input of the choke. The current at that point is pure DC, it is not pulsing at the rate of the RF. Once you can acknowledge this, then maybe we can go onto some other points.

So, do you really think that the DC supply is providing current that pulsating? Or is it providing a steady source of DC?

4. WA1GFZHam MemberQRZ Page

KD2NCU,
Your analysis (T2) I1=I2 only in a perfect world with 100% coupling between phases of T2. Take a real amplifier and add a third winding and attach it to a scope. You should not see anything on that winding. I tried it and found the coupling efficiency is lower than 100% in every configuration I tried. This is why some amplifiers derive a feedback signal off t2. Also the FETs have current overlaps if operating in class AB. This could also effect balance. The FET that is turned off T2 winding is sourcing the load while the FET that is drawing current is sinking off the load and t2. This could be a source of imbalance. There is a load imbalance between phases of T2 if you think about it. I found T2 needed to be large to not have core offset issues. At least as big as the output transformer. Yet when I combined functions I could get good results on a common core. I have an old Harris amp board that used 8 MRF150s. T2 and T3 shared the same core. The MRF amplifier I modified did it with one core and fed DC through the coax center conductor. My final design I fed the DC through the shields. My second harmonics dropped to around -35dBc on all bands. The system starts to fall apart when the coupling efficiency drops below about 96%. gfz

5. WW1WWHam MemberQRZ Page

You forgot V=L dI/dT. The collector voltage is a half sine but the supply voltage is pure DC. The energy is stored in the magnetic field of the inductor. Also in a real amplifier Q2 is not out of the picture. In the off state it presents between 200 and 5000pf of capacitance depending on the device.

6. WA1GFZHam MemberQRZ Page

The only energy stored in the inductor is the offset between phases because in a perfect world the fields would cancel. This is the reason the T2 in an EB104 can almost melt lead. Then consider dT in an amplifier that covers 160 through 6 meters. Then look at how L changes over frequency when wound on say type 43 material. Then look at the losses in the core. All this stuff skews the simulation so the phases don't match. Yes the 200pf of output C drags on the circuit. gfz

7. KD2NCUHam MemberQRZ Page

K7JEM, why would you think a DC supply cannot provide pulses of current?
Figure 1: Let’s open and close the switch of figure 1 rapidly at a constant rate and let’s leave it closed for 1 msec and open for 3 msec.
When the switch is closed the current through the resistor will be 1 amp.
Given the duty cycle, the average current in the resistor will be 0.25 amps.
What does the current flowing from the DC source look like?
•1 amp pulses matching the resistor current exactly?
•A constant flat DC current of 0.25 amps?

Figure 2: Let’s look at a class B emitter follower amplifier right on the edge of conduction at the emitter base junction, base bias resistors not shown.
Let’s give it just the right input signal so that the base current is a full wave rectified sinewave of current with peak value of 1 mA and a minimum value of 0 mA. ie; a series of full wave rectified sine pulses with peak value of 1 mA.
Let’s say the transistor has a Beta of 50.
So the collector current will also be full wave rectified sinewave of current with a peak value of 50 mA, an exact replica of the base current with 50 times the amplitude.
What will the supply current be?
•A full wave rectified sinewave of current with peak value of 50 mA matching the collector current exactly?
•A flat DC current equal to the average value of the collector current or 0.637 x 50 mA = 31.85 mA?

In both cases, the DC supply current is an exact replica of the resistor or collector current. Otherwise, what changed the pulses into flat DC current? Do you disagree with this?
A DC supply, as you correctly stated acts like a short circuit to AC. So AC current flows very freely through or from it.
You are confused about how the DC supply acts.
It will work very hard to keep the DC voltage constant.
However, it does not care one bit what the current is and does absolutely nothing to try to smooth or filter an AC current or somehow turn it into a flat DC current.
The DC supply will provide whatever current, AC, DC, pulses, sinewaves, square waves, etc. that the circuit connected to it asks for and it will not care a bit. Do you (and maybe WW1WW) think differently?

To answer your question, the DC source in the example we’ve been discussing, just like all other DC sources, will work to keep the DC voltage constant, and will not give a @#\$% what the current is and will provide whatever current the connected circuit asks for. You and WW1WW keep talking about “Pure DC” as though it has some mystical magical properties. I think that you think a DC source provides a fixed DC voltage and that it is only capable of providing a fixed constant current or something like that or that it somehow magically turns a pulsating current into a smooth DC current? Why don’t you and WW1WW go ahead and explain what you mean by “Pure DC” and what properties are conferred on something when you claim it is Pure DC, and how you think a DC supply works.

Ok, let’s now talk about the fact that Q2 is not out of the picture, stored energy, capacitance etc. looking at some actual numbers.

Let’s say during an on cycle, each transistor current is a half sine pulse of, say 16 amps. What you said up above is that the current from the DC supply into the center tap of the coil is constant flat DC current equal to the average of a full wave rectified sinewave peaking at 16 amps or 0.637 x 16 amps = 10.192 amps of dead flat DC current.

So just at the instant that Q1 starts to conduct, let’s say it’s only risen to say 1 mA, the current into the center tap is 10.192 amps. So we have an extra 10.191 amps going …. uh, where? You really think that Q2 capacitance, stored energy and all that other gobbledygook accounts for the excess 10.191 amps? Really? And it’s not just for an instant. The collector current is smaller than the average for quite a while. Dudes, that’s one huge boatload of electrons that you can’t account for. Stored energy doesn’t account for these electrons. WW1WW are you actually saying that electrons can be converted to magnetic flux or stored energy or whatever and later be converted back to electrons? So we store the extra boatloads of electrons in the magnetic field until the collector current rises above the average and then we start converting the magnetic field back to electrons to make up the shortfall when the collector current is larger than the average. And then we convert the trons back to magnetic energy again during the last part of the cycle when the collector current is lower than the average. And that accounts for up to 10 amps of mismatch between the collector current and the current from the DC supply. Really guys? This actually sounds reasonable to you guys?

WW1WW: "You forgot V=L dI/dT. The collector voltage is a half sine but the supply voltage is pure DC." You will have to put your V=L di/dt into an equation or some kind of context. Throwing out "You forgot V=L dI/dT" doesn't mean @#\$%&. Anyone can throw out drivel like that. What point are you making? And tell me exactly what you mean by Pure DC. Exactly what properties does this "Pure DC" have? As though throwing out "but the supply voltage is pure DC" proves something. Elaborate a bit. What are you saying that proves? You really think that capacitance of the off transistor accounts for a mismatch of 10 amps?

I agree that this is all based on the perfect world assumption and that there actually will be some imbalance and so on. But it does sound like WA1GFZ agrees that if there were perfect balance the flux would cancel and there would be zero inductance presented to the current coming from the DC source? (no stored energy, no flux = no inductance) And you should have mentioned leakage flux/inductance as well which will show up as some small inductance in series with the current entering the center tap. So to the extent someone screws up the design or construction of the coil, spaces the wires unevenly or incorrectly, uses the wrong size of wire, and screws up the bias of the transistors so there is more overlap than is needed or unbalanced overlap, and leakage flux, and mismatched transistors, the less perfect everything else is. But even in the imperfect world, there is no way you can account for the mismatch of 10.191 amps and you cannot convert electrons and store them in a magnetic field so the current going into the center tap from the DC bus still has to be overwhelmingly a replica of the collector/drain current of the conducting transistor half sine pulses, not a smooth constant DC as has been claimed. All the other stuff is negligible in comparison to 16 amp half sine current pulses and the up to 10 amp mismatch between the collector current at the start of its conduction cycle and the average value of the DC supply current.

Last edited: Oct 6, 2017
8. KD2NCUHam MemberQRZ Page

A couple of points I don't agree completely with:
1. Being able to attach a third winding and being able to use that as a feedback loop doesn't necessarily prove that there is a lack of balance or lack of perfect coupling. The bifilar winding is a parallel transmission line wound around the core. There is no practical way you can put a third winding on the core to pick up a feedback signal without CAUSING an unbalance. If you were to put the third winding exactly parallel and equidistant from the two existing windings that were perfectly balanced, this would not CAUSE an imbalance but you wouldn't pick up anything either because you would be coupling equally to two wires with equal and opposite currents whose effects on the third wire are cancelled by one another. If you put the third wire anywhere other than exactly parallel to and equidistant from the two existing wires, you are coupling more to one wire than the other so even if they were perfectly balanced, you are closer to one wire than the other and their effects on the third wire are no longer equal and opposite. Additionally, since you stuck this third wire closer to one than the other and coupled to it, you have changed what's going on in the closer wire. ie; your third wire is causing an imbalance. ie; you cannot measure something without changing it. Somewhat like the fact you can't stick a probe or wire close to one leg of a dipole to measure something without upsetting the balance of the dipole to some extent.
2. I don't think overlaps impact the validity of I1=I2 much. The transmission line wound on the core is a common mode choke and will try to suppress any current components that are not equal and opposite in the two wires thereby forcing them in the direction of I1=I2. Overlaps and differences between conduction of the transistors don't change the fundamental action of the common mode choke at the center of the bifilar coil. It sill tries to make I1=I2 but they just end up at slightly different values during Q1 as compared to during Q2. I think the validity of I1=I2 depends far more on the quality of the design and construction of the coil and the core material than anything else.
3. I'm not sure I agree with "The FET that is turned off T2 winding is sourcing the load while the FET that is drawing current is sinking off the load and t2. This could be a source of imbalance. There is a load imbalance between phases of T2 if you think about it." then again, I'm not exactly sure what you are saying. I'm going to take the off transistor completely out of the picture because anything it's doing is negligible compared to all the other large currents. So I think you might be saying that in the diagram below, the lower coil is more loaded than the upper coil or something like that?

When Q1 cycles, it begins to put a voltage across the upper coil with the polarity shown. As long as the core is not saturated, this induces an equal voltage across the lower coil with the polarity shown. In addition, the coil attempts to make the current in the lower coil equal and opposite to the current in the upper coil. So the device is acting exactly like an autotransformer.
In the perfect world, T2 acts like a perfect transformer and none of what you mentioned matters. To the extent that T2 is designed with the right number of turns, the right wire size and spacing, and built correctly with the correct core and so on, the closer it's actual performance will come to the ideal transformer. And generally, transformers can made that very closely approximate ideal transformers, say within a few percent. Off the top of my head I would say that the worse T2 is designed and fabricated, the more likely it would be that the stuff you mentioned would matter, so now I guess I'm saying I think I sort of agree with you after all a little bit maybe some at least partly with part of it but it's late as hell and it hurts my head to think about it right now so I will leave it at that for now and think about it more later. Good discussion. Good night.

Last edited: Oct 7, 2017
9. K7JEMHam MemberQRZ Page

Without getting into a convoluted discussion, it is sometimes easier to look at an actual working circuit and talk about what is going on. There are lots of circuits being used by lots of different manufacturers, but I found a simple circuit designed by a ham to work at 144 MHz. It is one transistor, so we don't have to worry or wonder what is going on in the other side of the circuit. Here is a picture of the schematic and board layout:

As can be seen, there is a single transistor running at 13v that is shunt fed through an inductor (L2). The DC current to run this amp should be somewhere around 6 amps at 13v to make an output of 50 watts. This should all be clear, and without debate at this point.

L2 is around 45 ohms of inductive reactance, plus or minus some amount, but close to that at 144MHz. If the DC feed were pulsating at 144 MHz, it would see 45 ohms to the collector of the transistor, due to reactance. This reactance is there all the time at that frequency, it does not vary with amplitude of signal, or current through the coil. But 45 ohms of inductive reactance will limit the current that can be delivered, if it is indeed pulsing at 144 MHz. The most current we would expect to see would be 13.8/45 = .3 amps. That would be the peak when the transistor was full on. When it was off, we wouldn't see any current flow.

The capacitors C9 and C10 have a combined capacitive reactance of .1 ohm at 144 MHz, which further smooths the "pulsating DC". Any current that makes it though the inductor is pretty much shunted to ground by those caps.

So, if we assume that the amp is running at 50 watts out, and has a near sine wave output, the voltage at the collector of Q1 will be 26 volts peak to peak, or less. That is about 9.2v RMS of 144MHz AC at the collector. The RMS AC current flowing back towards the power supply could be calculated as 9.2/45= .204 amps. Which is a long way from 6 amps. But there are also the capacitors to consider, and they represent a .1 ohm short circuit to ground at 144MHz. This will shunt the vast majority of current away from the PS. The actual current going through this capacitor would be the calculated current at that point minus a very small amount of leak through. The amount getting through to the PS would be .1/45 * .204A or .45 mA. The next two stages of RFC would drop that down even further.

So, with 6 amps of current flowing from the power supply as DC, there is a maximum of .00045 amps of AC, or less. Pretty much "pure DC".

WW1WW likes this.
10. WW1WWHam MemberQRZ Page

It would be interesting to hear how KD2NCU explains how this Real circuit functions. How can a transistor that is only capable of sinking current to ground produce a perfect sine wave?