Need help spec'ing max AC signal for plate choke

[Deke609]

I want to try a nanocrsystalline ("Finemet") core plate choke in my Kaiju rebuild.  The winder needs a bunch of info, including max AC signal.  I originally asked for 600V out of laziness and desire to go overkill to be on the safe side.

The winder has written back that he can do 62H with a DCR of 300 (this will work in my rebuild) with an fsat of 11Hz @ 180V rms -- but opines that he thinks this is "overkill".

So I've taken a stab at calculating the max rms VAC signal that could appear at the plate of the 300B but I end up with a voltage higher than the voltage rating of the parafeed cap: about 685V rms. So I am clearly doing something wrong.  Please help! 

I've considered the entire chain: 2Vac output of DAC -> BeePre -> Kaiju.  Based on the BP's spec'ed gain of about 8.3 dB, I previously calculated that 2V input will give me a max output of 5.2V.  5.2V into the 5670 with a gain of about 34 gives me approx. 175V out of the driver, which when multiplied by the circa 3.9 gain of the 300B (with 330uF film bypass cap), gives me about 685V. All voltages are rms.

Note: I have a giant film cap across the 5670 bias regulator (I think 0.1F -- i.e, 100,000 uF) and a 330uF film cathode bypass cap, so I don;t think the gain changes much by frequency - but I obviously could be wrong about this.

What am I doing wrong?

MTIA, Derek

[Paul Birkeland]

Ohm's Law tells is that P=(V^2)/R

The amp is 8W, R of the opt. is 3000 ohms. When generating specs like this, always round up.

If you put 5V of signal into a Kaiju, it will be horribly hard clipping.  The 300B grid is biased at about 70V, you can't put more voltage in without a large circuit redesign.

[Deke609]

Thanks PB. Both points -- max power and max grid +ve swing -- make sense now that you point them out.

Using the power calculation, I get a max voltage of 154 using the stock 3K value. My OPTs are actually 3.3K, which only bumps that figure up to about 165V (assuming the same 8W power can be maintained). But I often listen without a parallel load resistor across the secondary -- i.e., with nominal 3.3K:64R loaded by 200 ohms phones that gives me approx. 10K primary impedance.  But I recall that the higher the plate load, the lower the power.

But I'm having trouble figuring out the power and voltage appearing at the plate with a 10K load b/c I don't know how the choke's ability to swing twice B+ figures into this, if at all. I thought I understood this stuff, but clearly I don't.  Help!

MTIA, Derek

[Paul Joppa]

The choke is designed to handle the DC current plus the AC voltage at the lowest frequency at rated power into rated load. Since the source impedance is a fraction of the load impedance, the maximum voltage is not very sensitive to higher impedance loads. It's good enough to assume the maximum voltage is a constant, within 20-30% or so - the effect of magnetic saturation is gradual enough that greater precision is pointless IMHO.

If you want to calculate it anyhow, the gain of a common-cathode triode voltage amplifier is widely documented - it's just mu * RL/(RL+rp)

For example, the PC-5 will handle 70mADC plus 154vRMS at 19 Hz (8 watts into 3K ohms). Gain is about 1.24 times greater (190vRMS) into an infinite load, which it will handle at 23Hz.

[Paul Birkeland]

But I often listen without a parallel load resistor across the secondary -- i.e., with nominal 3.3K:64R loaded by 200 ohms phones that gives me approx. 10K primary impedance. 

I have pointed out repeatedly both here and on another forum to you that this is not accurate.  I will again attach a measurement of a transformer that demonstrates this. 

[Deke609]

Many thanks PJ.

If you want to calculate it anyhow, the gain of a common-cathode triode voltage amplifier is widely documented - it's just mu * RL/(RL+rp)

Aha! I hadn't understood that about mu - uncritically, I just assumed it was the measure of voltage gain at the plate. But of course that's determined by the voltage divider formed by plate resistance in series with the plate load. That makes much more sense.

I have pointed out repeatedly both here and on another forum to you that this is not accurate.  I will again attach a measurement of a transformer that demonstrates this. 

Well, I figured that only applied to a truly unloaded secondary. But I am now reminded of a comment by PJ some months ago to the effect that with a normally loaded secondary you can usually ignore the primary inductance, but not so with a lightly loaded one. I think I've since *learned* why (well, I still need to figure out the theory/math behind it): with sufficient current through the secondary a tiny "shunt inductance" is reflected in parallel with the primary self inductance which renders the effective inductance of the primary negligible, leaving only the reflected load from the secondary -- i.e., the nominal/spec'ed primary impedance [edit: assuming the secondary has its spec'ed load]. What I don't know, but hope to figure out in the next few weeks, is the amount of current needed to achieve this. I would have thought that if the reflected "shunt inductance" is proportional to current through the secondary, then the smaller the current the better.  But PJ's previous comments suggest that this is not so. More to learn!

Edit - @PJ: Am I on the right track with this reflected "shunt inductance" stuff?

Many thanks both,  Derek