OK. I am having a brain fart here. In a class AB amp with 2 tubes in push pull, let's say the idle current for each tube is 50mA, and the max current is 150mA. Current flows in each tube for a little more than half the cycle, alternating back and forth between the tubes.
That being the case, if the B+ supply for the plates is after the choke, then that choke and power transformer should be minimumly rated for the current of ONE of the tubes at max signal draw, plus screen draw and preamp section draw... right? OOOOOOOR, do I need to take into consideration additional current from 2 tubes at the crossover point? Both tbes will never be simultaneously drawing max current ... right?
Thanx for your help in my brain farting state.
That being the case, if the B+ supply for the plates is after the choke, then that choke and power transformer should be minimumly rated for the current of ONE of the tubes at max signal draw, plus screen draw and preamp section draw... right? OOOOOOOR, do I need to take into consideration additional current from 2 tubes at the crossover point? Both tbes will never be simultaneously drawing max current ... right?
Thanx for your help in my brain farting state.
The HT capacitor at the output transformer will average the currents in the output stage. If you really wanted to be precise, you could invoke a little calculus and work out the area under the graph to give you an average current draw. But as loudspeakers are nothing like resistive, your calculation will be wildly wrong, so you needn't bother. The peak current of 150mA in one valve plus screen current and driver current will probably be a little high, but will allow a margin for error.
I would calculate for 150mA + screen current... + the other stages.
If you are using CLC or CRC smoothing you'll have to multipy the figure by about 1.5 to get the PT current.
If using LC smoothing, no fudge factor is required.
If you are using CLC or CRC smoothing you'll have to multipy the figure by about 1.5 to get the PT current.
If using LC smoothing, no fudge factor is required.
THANX
Awesome, that's what I thought. Thanx a lot for the help gentlemen.
I had originally mistakenly spec'd a pwr transformer for dual KT88's running in UL class AB at 560v, and mistakenly figured 2x max current + preamp/buffer current and came up with a HUGE current requirement... around 400mA, so that is why i asked for the check, HOPING that I was wrong and not having to run a 15lb transformer ;-)
Awesome, that's what I thought. Thanx a lot for the help gentlemen.
I had originally mistakenly spec'd a pwr transformer for dual KT88's running in UL class AB at 560v, and mistakenly figured 2x max current + preamp/buffer current and came up with a HUGE current requirement... around 400mA, so that is why i asked for the check, HOPING that I was wrong and not having to run a 15lb transformer ;-)
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