Hey-Hey!!!,
I asked this question elsewhere, and got no satisfactory answer. I have a PP driver stage, triodes, common cathode, with both grids driven with signal form the phase splitter upstream. The stage is loaded with a center tapped choke. What sort of load does this pair see? I got answers from a fourth of the choke's inductance( half the turns ) to the whole thig plus things like the other troide's plate resistance.
Can a few of you help me sort this out?
regards,
Douglas
I asked this question elsewhere, and got no satisfactory answer. I have a PP driver stage, triodes, common cathode, with both grids driven with signal form the phase splitter upstream. The stage is loaded with a center tapped choke. What sort of load does this pair see? I got answers from a fourth of the choke's inductance( half the turns ) to the whole thig plus things like the other troide's plate resistance.
Can a few of you help me sort this out?
regards,
Douglas
One tube will see 1/4 the end-to-end inductance, because L is proportional to the square of the turns; but since there is another tube effectively in parallel, helping it out, each tube sees only 1/2 the inductance. Thus for a 5k reactance at 20Hz, it must see about 40H, so the total choke must be 80H.
Tim
Tim
well,
it seems as if you didn't look into the responses you received on the asylum, so why not build the circuit and tell us what happens... all you need is: the load, the Rp, and the -3db point and you will know the answer.
i'll say it again... the load seen by each common cored inductively loaded class A pp triode is 1/2 the total load seen by the pair. this assumes linear devices with no losses, it is the ideal, and in audio its what we strive for.
and before this gets out of hand... inductance is but part of the load in the class A pp pair.
dave
it seems as if you didn't look into the responses you received on the asylum, so why not build the circuit and tell us what happens... all you need is: the load, the Rp, and the -3db point and you will know the answer.
i'll say it again... the load seen by each common cored inductively loaded class A pp triode is 1/2 the total load seen by the pair. this assumes linear devices with no losses, it is the ideal, and in audio its what we strive for.
and before this gets out of hand... inductance is but part of the load in the class A pp pair.
dave
Hey-Hey!!!,
I read every single one, and the refrenced links. There were as many differnet answers as answers posted( almost), not too mention the rest of the responses.
So I asked again. I can see what is going on with the circuit, that's why I'm building it( and having it built). I wanted a closed form way of looking at what is going on.
All in all, I am glad to see there are folks who are willing to help me see what is going on.
regards,
Douglas
I read every single one, and the refrenced links. There were as many differnet answers as answers posted( almost), not too mention the rest of the responses.
So I asked again. I can see what is going on with the circuit, that's why I'm building it( and having it built). I wanted a closed form way of looking at what is going on.
All in all, I am glad to see there are folks who are willing to help me see what is going on.
regards,
Douglas
Hey Joel,
It's not a grid choke( said with Austrian accent ). The inductor is for an anode load on a PP driver stage. Either 2x 12B4A, 6CK4 or single 6BX7, 5687, 5998A...balance additionally assisted by CCS in common cathode load.
Here is part of that which is making question marks in the air over my head when I think about this.
If I hook up the grid circuit( of the power stage) to each end of the choke and leave off one of the triodes, I will still get some signal to/out of the hanging end of the choke, right? So, put in the triode, feed it a signal as perfectly out of phase with the one going to the other end of the coil and what sort of load do they( the triodes loaded by the choke) see?
While I am fairly good at actually making the PP amps sound good
knowing exactly how they work is another thing entirely...
regards,
Douglas
It's not a grid choke( said with Austrian accent ). The inductor is for an anode load on a PP driver stage. Either 2x 12B4A, 6CK4 or single 6BX7, 5687, 5998A...balance additionally assisted by CCS in common cathode load.
Here is part of that which is making question marks in the air over my head when I think about this.
If I hook up the grid circuit( of the power stage) to each end of the choke and leave off one of the triodes, I will still get some signal to/out of the hanging end of the choke, right? So, put in the triode, feed it a signal as perfectly out of phase with the one going to the other end of the coil and what sort of load do they( the triodes loaded by the choke) see?While I am fairly good at actually making the PP amps sound good
knowing exactly how they work is another thing entirely...regards,
Douglas
!!!!
A combination of your two - theories - is reality.
First of all, in theory, the free end of the choke will follow the other end, of course with an opposite polarity; however, in practice, loading, leakage inductance, DC resistance and parasitic capacitances will plauge the end result. Also it depends on the current capacity of the choke; if it can handle DC, then it will still function well enough, but if it can't and needs a balanced current instead, then all you'll get is distortion. LL and DCR vs. load contribute to reduce the free end's voltage output, while C will resonate with the LL, causing a resonant peak at some frequency, and a steady roll-off beyond that.
Tim
is reality.First of all, in theory, the free end of the choke will follow the other end, of course with an opposite polarity; however, in practice, loading, leakage inductance, DC resistance and parasitic capacitances will plauge the end result. Also it depends on the current capacity of the choke; if it can handle DC, then it will still function well enough, but if it can't and needs a balanced current instead, then all you'll get is distortion. LL and DCR vs. load contribute to reduce the free end's voltage output, while C will resonate with the LL, causing a resonant peak at some frequency, and a steady roll-off beyond that.
Tim
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