Can anyone point me to a link to calculation of the impedance seen by the plate of the power tube in push pull stage of class AB amp. I read RDH4, when in class B region, you see 1/4 or total anode to anode impedance or R=1/4 RL(N1/N2)^2
But I cannot find an explanation given why R=1/2 RL(N1/N2)^2 when the stage is still in the class A region where both tubes are on.
Building a better world (one tube amp at a time)
"I have never had to invoke a formula to fight oscillation in a guitar amp."- Enzo
Above explanation is correct, this one reaches the same end arriving from a different angle:
Class AB is a confusing label, so letīs split it into a "pure Class B" section and a "pure Class A" one.
1) pure Class B: itīs the main component and for maximum power you should order your transformer based on this value.
In any case, apply the made up example to any datasheet you get:
suppose you have a 6L6, fed 430V, idle current 30mA ; when fully driven can pass 170mA, plate voltage will drop to almost 50V , just follow the red curve.
Simple Ohmīs Law, thereīs no mystery to it:
* Voltage swing: 430V-50V=380V peak
* current swing: 170mA-30mA=140mA peak
so optimum plate load impedance:
* Zp: 380V/0.14A=2700 ohms ... thatīs for a single plate. Za
Since itīs a push pull, plate to plate impedance Zaa is 4X that (because plate to plate transformer is 2X voltage=4X impedance) so Zaa=10800 ohms. Yup, almost 11k, not a typo.
Bear with me and youīll understand the full picture.
2) pure Class A zone: up to 30mA (or whatever idle current you choose): not sure how to draw a graph for this, but the concept is easy to grasp:
* at idle, both tubes are pulling 30 mA each from the power supply, obviously each at its own end.
* when one gets a more positive signal from the PI, it will start pulling more current ... which will eventually be fed to the speaker thanks to the transformer.
So far so good, whatīs expected and intuitive.
Suppose signal is enough to go from 30mA to 60mA , then positive current variation is 30mA .
Now to the counter intuitive part, which sounds crazy until it clicks:
* the other tube will get a more negative signal from the PI, and will start pulling less current from power supply.
Signal from PI will be same as the other one, just opposite phase.
Current will *lower* from 30mA to 0 mA, so negative current variation is 30mA.
Now the "click" part:
This negative current variation is also passing through the OT, it will be reflected onto the load, but since transformer winding phase is inverted , such current into the speaker will be positive and will add up with current supplied by the opposite tube.
It is the same as if the other tube had supplied 60mA instead of 30 mA
* you can say that in the Class A section both push pull tubes act as if they were in parallel and on the same side.
Again using Ohmīs Law, we know that for 2X current and same voltage swing we have half impedance, so for the Class A section only optimum Zaa (plate to plate impedance) will be 5500 ohms instead of 11000 .
All this applies to the above graph, with screen voltage 250V.
Leo (and his imitators) rised screen voltage to chilling 400V and above and in his amps 6L6 can supply WAY more peak current , so optimum plate impedance is way lower, but the "Class A load being half the Class B one" still remains.
* for finer tuning, you can now assume that "effective" current swing is now 170 mA : 140mA "real" and 30mA "reflected from the other tube" so redo Math using this new value.
Juan Manuel Fahey
One tube (of two) handles 1/2 of the whole signal, ie: 1/2 of 1/2 is 1/4.
...and, I assume that is actually TRANSFORMER and not TRANSISTOR.
Last edited by Old Tele man; 10-24-2017 at 09:59 PM.
...and the Devil said: "...yes, but it's a DRY heat!"
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