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 aOther sites that talk about this:single tubeis 1/2 the end to end plate impedance.

In class B, the load line for asingle tubeis 1/4 the end to end plate impedance.

Pearl's copy of the MIT's Theory on Push-PullNote: All voltages, currents and powers on this page are instantaneous unless I say they are rms.

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

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 / N2or

1.2 V1 / V2 = N1 / N21.3 V2 = V1 * N2 / N1

For a two winding transformer this means2.1 I1 * N1 = I2 * N2Which becomes2.2 I1 / I2 = N2 / N1For more than two windings we get this equation

or

2.2 I2 / I1 = N1 / N2

or

2.3 I1 = I2 * N2 / N1If N1 has more turns than N2, we get less current on I1 (winding 1)

for a given current on I2 (winding 2).2.4 I1 * N1 = I2 * N2 + I3 * N3 . . .

Z1 = V1 / I1

Z2 = V2 / I2

( 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.

4.1 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)

1.3b V2 = [ ( V1 - I(1) * R1 ) * ( N2 / N1) ] - I(2) * R2

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 ]

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

With 400V across theIf 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.entireprimary, 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 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.If this is confusing, keep reading. I've got pictures below to help out.I(P2) = 1A * 1 / 25 = 0.040A = 40 mA.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.The impedance seen by the tube 200V/ 40 mA which is 5000 ohms.

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,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.

Z1 / Z2 = (1 / 2)^2 or 1/4.

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.

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.

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

Z each tube = 200 V / 20 mA = 10 kohm

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.

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 PSIIn 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.

Current is like gallons per second in a pipe.

Resistance is the pressure drop in the pipe for a given gallons per second.

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 ------------

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

Last Update 2-Oct-04

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.