PP vs two tubes parallel (SE) - Why more watts?
Moderators: pompeiisneaks, Colossal
PP vs two tubes parallel (SE) - Why more watts?
Could I get some input on this explanation -
As explained by PDF64
If operated in class A at the same conditions, 2 in parallel should put out the same power as 2 in push-pull.
Given the same power supplies etc, the push-pull version would have lower hum from the B+ (as B+ ripple, being common mode, would be largely cancelled out in the balanced output stage) and for the same frequency response could have a smaller cheaper lighter output transformer, as the static mag fields from either side of the primary cancel out.
Class AB allows the power tubes to be run hot for part of the cycle because they are also in cut off (ie cooling down) for part of the cycle.
So the time averaged plate dissipation can be within the limiting value.
Given the same power supplies etc, the push-pull version would have lower hum from the B+ (as B+ ripple, being common mode, would be largely cancelled out in the balanced output stage) and for the same frequency response could have a smaller cheaper lighter output transformer, as the static mag fields from either side of the primary cancel out.
Class AB allows the power tubes to be run hot for part of the cycle because they are also in cut off (ie cooling down) for part of the cycle.
So the time averaged plate dissipation can be within the limiting value.
Re: PP vs two tubes parallel (SE) - Why more watts?
What that means is that in a push-pull output stage, the tubes are biased cold enough so that when a big-enough signal is hitting the grids (i.e. from the phase inverter), then during the part of the cycle where the output tube grid signal swings really negative, this puts that output tube into 'cutoff' mode. At this point there is no current going through the side of the OT primary which that output tube is driving.
Therefore the reflected load of the OT is halved (because only the other half of the OT primary is driving the secondary - and that 1/2 has only 1/2 the amount of primary turns that the whole primary has), and at this point the load line therefore steepens (i.e. becomes a 'B' load line).
During this part of the signal cycle, a whole lot more plate current goes through that side of the OT primary and delivering more current through the OT to the speaker, for this part of the cycle. (When the phase reverses, the other side of the OT is doing this trick and vice versa).
So we have the output stage operating in the 'B' load line mode, increasing the peak current (from what is would be with the 'A' load line). Therefore the amp puts out more power when running in this mode. Now during the part of the cycle where each tube is operating on the 'B' load line, the tube is operating over the peak-rated plate dissipation. But you can do this without harming that tube, because for the opposite part of the cycle that same tube is in cutoff and there is no plate current, therefore the plate dissipation averages out over time. This 'Class AB' situation only happens when you have a sufficiently big enough driving signal at the output tube grid in order to drive each respective output tube into cutoff (in opposite phases) within part of the signal cycle.
If the input signal driving the signal grids isn't big enough, then both output tubes stay operating within the Class 'A' load line.
Therefore the reflected load of the OT is halved (because only the other half of the OT primary is driving the secondary - and that 1/2 has only 1/2 the amount of primary turns that the whole primary has), and at this point the load line therefore steepens (i.e. becomes a 'B' load line).
During this part of the signal cycle, a whole lot more plate current goes through that side of the OT primary and delivering more current through the OT to the speaker, for this part of the cycle. (When the phase reverses, the other side of the OT is doing this trick and vice versa).
So we have the output stage operating in the 'B' load line mode, increasing the peak current (from what is would be with the 'A' load line). Therefore the amp puts out more power when running in this mode. Now during the part of the cycle where each tube is operating on the 'B' load line, the tube is operating over the peak-rated plate dissipation. But you can do this without harming that tube, because for the opposite part of the cycle that same tube is in cutoff and there is no plate current, therefore the plate dissipation averages out over time. This 'Class AB' situation only happens when you have a sufficiently big enough driving signal at the output tube grid in order to drive each respective output tube into cutoff (in opposite phases) within part of the signal cycle.
If the input signal driving the signal grids isn't big enough, then both output tubes stay operating within the Class 'A' load line.
He who dies with the most tubes... wins
Re: PP vs two tubes parallel (SE) - Why more watts?
As mentioned in the other thread, the difference is in the voltage and current swing. In class A PP, you purposely limit the voltage swing, so its output equals to that of the parallel SE.
However, in class AB PP, a few things happen - 1) the voltage swing limit is lifted, 2) the load line changes from class A to class AB (or nearly class B as a quick approximation). The two conditions then enable class AB PP to deliver a higher output power than the parallel SE.
However, in class AB PP, a few things happen - 1) the voltage swing limit is lifted, 2) the load line changes from class A to class AB (or nearly class B as a quick approximation). The two conditions then enable class AB PP to deliver a higher output power than the parallel SE.