Push-Pull Primary Impedance
Plus a few transformer notes

What I intend to show is that the effective impedance (also called the composite impedance) seen by the pair of tubes in a push pull amplifier is 1/4 the end to end plate impedance of the transformer. From this we'll see that the load line seen by each tube in a pair of push-pull tubes depends on whether or not the tubes are biased class A or class B:

In class A, the load line for a single tube is 1/2 the end to end plate impedance.
In class B, the load line for a single tube is 1/4 the end to end plate impedance.
Other sites that talk about this:
Pearl's copy of the MIT's Theory on Push-Pull
Load Lines Made Simple AX85_M225  <--This example crosses the class AB transition. Because the example they use does not stay class A, I don't fully agree with this push pull example.
Steve's Bench on Composite Load Lines
Note: All voltages, currents and powers on this page are instantaneous unless I say they are rms.

The Basic Transformer Equations
Last Update 26-Sep-04


First we need to learn the basic ideal (loss less) transformer equations:

1.    The volts per turn on a leg of a transformer are constant. This rule is independent of the direction the current is flowing with respect to the dot.

1.1    V1 / N1 = V2 / N2

    or

1.2    V1 / V2 = N1 / N2

1.3   V2 = V1 * N2 / N1

2.    The ampere turns into the transformer must equal the ampere turns out of the transformer. This is simple energy conservation. (Winding 2 must have the same volts/ turn as winding 1, but winding 2 does not need to have the same amps as winding 1.)
For a two winding transformer this means
2.1   I1 * N1 = I2 * N2
Which becomes
2.2    I1 / I2 =  N2 / N1
            or
2.2    I2 / I1 = N1 / N2
            or
2.3   I1        = I2 * N2 / N1
If N1 has more turns than N2, we get less current on I1 (winding 1)
for a given current on I2 (winding 2).
For more than two windings we get this equation
2.4    I1 * N1 = I2 * N2  + I3 * N3 . . .
3.      The effective impedance of a transformer winding is V / I:
Z1 = V1 / I1
Z2 = V2 / I2
Now divide equation 1.2 divided by 2.2 to get the impedance ratio between two windings:
    ( V1 / I1 ) / (V2 / I2) = ( N1 / N2 ) / ( N2 / N1 )
                            = ( N1 * N1 ) / ( N2 * N2 )
                            = ( N1 / N2 )^2
3.1  Z1 / Z2                = ( N1 / N2 )^2
If N1 has more turns than N2, winding 1 has a higher impedance than winding 2.
4.     In a low loss signal transformer, we can usually ignore non-ideal behaviors. However, lets discuss a few non-ideal transformer behaviors. Winding 1 will be the drive winding for this discussion.

4.1    The volts / turn is set by the voltage on the drive winding after IR drop (Current Winding 1 * Winding 1's resistance) is subtracted from the drive winding's voltage.

1.1a    ( V1 - I(1) * R1 ) / N1 = V2 / N2
1.2a    ( V1 - I(1) * R1 ) / V2 = N1 / N2
1.3a     V2 =  ( V1 - I(1) * R1 ) * ( N2 / N1)
4.2    The actual voltage on the output winding is the corrected volts / turn from 4.1 less the IR drop on the output winding.
1.3b    V2 = [ ( V1 - I(1) * R1 ) * ( N2 / N1) ] - I(2) * R2
4.3    The amps into the drive winding is ideal reflected load current from 2.1 above + the magnetization current + the core loss current.
2.3a   I1 = I2 * N2 / N1  + V1 /{ j * 2 * pi * F * Lpri } + [ Core loss watts / V1 ]
The "j" is to remind us that this is a vector addition of currents. Remember, the core loss will be in phase with a reflected "resistive" load current. In this next equation, rms currents and voltages are used.
I1 = sqrt [ ( I2 * N2 / N1 + [Core loss watts / V1] )^2 + ( V1 /{ 2 * pi * F * Lpri} )^2  ]

The Mathematical Push-Pull Impedance Example
Last Update 30-Dec-04
 
If the secondary turns are N2 and the full primary turns are 2 * N1, the impedance ratio from the full primary to the secondary is (2 * N1 / N2 )^2 or 4 * (N1/N2)^2. The impedance ratio for one half the primary is just (N1/N2)^2.  The impedance of 1/2 the primary winding is 1/4 the impedance of the end to end primary.
If the output load is 8 ohms and N1/N2 is 25, then 8 ohms * (N1 / N2)^2 is 5000 ohms. The impedance across the entire primary will be four times the impedance across half the primary. Four times the half primary impedance is 20 kohm. This transformer would be rated at an impedance ratio of 20k:8.
If the tubes are biased into class A, they will share the load when driving the primary. When these two "parallel" tubes share a load line (a composite load line), the pair of tubes together see an impedance of (N1 / N2)^2. However, because there are two tubes sharing the load, the load line for each tube is twice the composite load line or 2 * (N1 / N2)^2.
With 400V across the entire primary, 20 mA will be flowing through the primary for an input power of 8 watts.
With 2 * N1 / N2 = 2 * 25, the ideal secondary voltage will be 400V / (2 * 25) = 8V
The ideal secondary current is 20 mA * (2 * 25) = 1 amp
The output power is 8 V * 1 A = 8W.
One tube will swing half the voltage (200V), but swings only 20 mA. 200V/ 20 mA is 10 kohm. 10 kohm is 1/2 the full primary winding impedance.
If tube 1 (P1) is removed, the impedance seen by tube 2 on primary 2 (P2) is (N1 / N2)^2. This is 1/4, not 1/2, the end to end primary impedance. This makes sense, the output power to the load hasn't changed, the P1 winding isn't contributing to supporting the output load and P2 must support the full output load.
If P1 is removed and P2 is to drive 200V across N1, P2 must now conduct 40 mA not 20 mA to deliver the same power to the load. This is to meet the amp-turns requirement of the output winding N2 reflected back to the primary with a 25:1 turns ratio.
I(P2) = 1A * 1 / 25 = 0.040A = 40 mA.

The impedance seen by the tube 200V/ 40 mA which is 5000 ohms.

The same is true for P1 when tube 2 is removed. P1 sees (N1 / N2)^2 or 1/4 the end to end primary impedance. This is the class B bias case. In class B only one tube is active at a time.
If this is confusing, keep reading. I've got pictures below to help out.



Moving Windings Around
Last Update 26-Sep-04


The purpose of this exercise is to get us use to how the current, voltage and impedance behaves as the windings are move around. In of these examples 1 through 4, there is 8 watts going into the transformer.

Starting with picture 1, we have a 400 V source consisting of two series 200V sources. This 400 V source is providing 20 mA into an effective load of 20 kohm. Notice that at the center tap (B C), there is zero total current flowing. The wire shown between the voltage source and the transformer does not have to be there.

In picture 2, we split the common center tap into two floating windings. The total impedance is still 20 kohm and each 200V source is driving 10 kohm ( 200V/ 20 mA)

In picture 3 the two sources are shown floating in parallel. Each source still sees a load of 10 kohm (200V / 20 mA), however, the composite load is 5 kohm (200V / 40 mA)

In picture 4, one source is removed and the composite load is kept the same. This one source sees a load of 5 kohm (200V / 40 mA). The single 200V source sees a lower impedance load because the other 200V source is not helping support the composite load of 5 kohm.

The composite load in picture 4 is 1/4 the load in picture 1. This is because the load in picture 4 has half the turns as the load in picture 1. This should be no surprise because we knowing the rule on reflected impedances:

Z1 / Z2 = (N1 / N2 )^2,
Z1 / Z2 = (1 / 2)^2 or 1/4.
Picture 4 is equivalent to a class B push pull amplifier where only one source (one tube) must support the entire load. In a class B amplifier, the load line on the active tube (the voltage) source is 1/4 the plate to plate impedance shown in picture 1.

Picture 3 is equivalent to a class A push pull amplifier. The composite load is the same as picture 4, but each tube only has to supply 1/2 the current for the same voltage swing. So each individual tube in a class A push-pull amplifier sees a load of 1/2 the plate to plate impedance (20 kohm) or twice the composite impedance (5 kohm). Remember, the load line on each individual tube is twice the impedance of composite load line because both tubes are supporting the load.
 


The Graphical Push-Pull Class A Impedance Example
Last Update 26-Sep-04


Lets look at the primary of an ideal transformer driven by an ideal tube at biased class A both at when the output is at idle and when swinging full output power:

Plate 1 (P1) sees a peak swing of  200V and 20 mA for an impedance of 200V / 20 mA = 10 kohm
Plate 2 (P2) sees a peak swing of  200V and 20 mA for an impedance of 200V / 20 mA = 10 kohm
From plate 1 to plate 2 we see a peak swing of 400V and 20 mA for an impedance of 400V / 20 mA = 20 kohm
On a Push-Pull transformer driven by tube biased to class A, each individual tube sees 1/2 the full primary impedance as its load line.

Lets look at this from a different slant,

P1's change in voltage is (450V - 50V) = 400V.
P1's change in current is (40 mA - 0 mA) = 40 mA.
So the load impedance P1 sees is 400V / 40 mA = 10 kohm.

Push-Pull to Single Ended Graphical Example.
Last Update 26-Sep-04


The load sharing may become more apparent if we convert the push-pull to parallel single ended (push-push).

The two tubes in parallel swing 200V peak and 40 mA peak total.

Z composite load    = 200 V / 40 mA = 5 kohm
Each individual tube is swinging 200V peak and 20 mA peak.
Z each tube            = 200 V /  20 mA = 10 kohm
Again if we pull tube 1, tube 2 must support the full (composite) load of 5 kohm.


Summary Class B Push-Pull Operation
Last Update 26-Sep-04


In class B operation, only one tube is supporting the load across N1. So the impedance ratio seen by the tube is ( N1 / N2 )^2. This is  1/4 the full primary impedance. Class B sees 1/4 the end to end transformer impedance because only half the turns on the transformer are driven. Class B has half the load impedance as class A because only 1 tube is active.



Current and Voltage Water Analogy
Last Update 2-Oct-04


I've noticed several posts on the internet where voltage and current are added incorrectly. I have a water analogy that may help people keep this straight.

Voltage is like water pressure or PSI
Current is like gallons per second in a pipe.
Resistance is the pressure drop in the pipe for a given gallons per second.
In series transformer windings the current is the same in all series windings. Like hoses hooked together, the gallons / second into the hose is the same as the gallons / second out of the hose. At the inlet to the hose, the pressure is higher than the outlet.

If two hoses are tied to the same inlet, the pressure at the inlet is the same for both hoses. If these two hoses are also tied at the outlet, the pressure is the same on the outlet side. If the hoses are the same diameter and length, you get twice the gallons / second through two hoses with the same pressure drop as one hose.

------------ Transformers ------------

The nearest example of a transformer in water terms is fluid coupling transmission. Ideally power into the transmission equals power out. So a high pressure inlet at low gallons/second can be used to develop a low pressure outlet at high gallons/ second.

------------ Capacitors ------------

The nearest example of a capacitor in water terms is water tank with a hose bib at the bottom of the tank. No matter how fast you start to draw the water out of the hose bib, the pressure stays the same at the hose bib when the valve is first opened. As the tank is drained (capacitor discharged), the pressure drops.
------------ Inductors ------------
The nearest example of an inductor in water terms is inertia. Once you start the water moving in a pipe, it takes pressure to make it stop. To make it stop instantaneously, you need infinite pressure.


Just for Kicks
What is the rms output power in the example used?
Last Update 2-Oct-04
Just for kicks:
The pair of tubes dissipate 2 * (250V * 20 mA) = 10 W at idle.
If each tube can swing to 50V across it, we can get 200V swing on each tube.
At the conditions shown, the peak output power is 400V * 20 mA = 8 W peak.
If the output is a sinewave, the RMS output power is 8 W / 2 = 4W rms.
The efficiency is 4W out / 20 W idle = 20%.
Math Reminder: With a sine wave, the peak voltage is sqrt(2) times the RMS voltage. This means the with a sinewave, the peak power is twice the rms power.


First edition 26-Sep-04, last update 30-Dec-04
I don't change the update date on individual sections for minor corrections. I only change the date  for content changes.