Question about Pentode reverb drive circuit

[nickb]

What is the DCR of the driver coil? This acts just like the 470R as an extra series resistor in this highly damped series resonant circuit. (The resonant frequency could be varied by using a different coupling cap.)

About 60 ohms, quite small.

[Helmholtz]

About 60 ohms, quite small.

O.K., thanks

[SoulFetish]

These are not relevant results as
1) You are showing the plate voltage not tank current
2) You are driving it into clipping

Back down the drive to keep in in the linear region, add a small resistor on the ground side of the tank and monitor the voltage across that.


I see the same if I crank it into clipping, but that is not how we are operating, or at least shouldn't be. it Even then there is no significant difference with and without the 470 ohm.



[ATTACH=CONFIG]50161[/ATTACH]


A Norton equivalent gives the same result. Mr Thevinen was great relieved to learn of that Since we are using idealized sources here the frequency response is different from when driving with the tube, but the result is the same.

[ATTACH=CONFIG]50160[/ATTACH]

I find this pretty compelling evidence that any difference between the two is negligible. For simplicity, I’ll probably omit the resistor.

[SoulFetish]

What’s the formula to calculate the resonant frequency in this case? I figure in order to bring the resonant requency even down to (about) 121Hz, the coupling cap would need to be 10uF. Firstly, Im not sure thats right. Secondly, bringing the resonant frequency down to twice the mains frequency is probably not a good ideal?

[nickb]

It's the usual f= 1/(2 x pi x sqrt( L x C)).
Roughly speaking, there's no point in being too precise as inductance is a rather variable quantity in practice and especially so with a tank or so I should think, with 1uf we're at 460Hz, so 10uF get us to the sqrt(1/10) of this i.e. 145Hz, so you were about right . I really don't see the merit in going bigger than 1uf as you don't change things much. OTOH you don't want to go any lower capacitance.

Then there's the 82K resistor. It's not really doing anything other then adding a path for cap leakage current to ground with no tank. If the leakage were high enough to be dangerous then the 82K isn't doing to help. So, you might consider a film cap rather than an electrolytic for safety reasons.

If you were to make it lower, even 10k won't change the frequency response much but will provide a bit more anti-zap protection. Lower it a bit more and you can get some useful HF attenuation e.g 2.7K gets you a corner frequency of about R/2/pi/L = 3.6Khz. Something else to play with, huh?

[Helmholtz]

What’s the formula to calculate the resonant frequency in this case? I figure in order to bring the resonant requency even down to (about) 121Hz, the coupling cap would need to be 10uF. Firstly, Im not sure thats right. Secondly, bringing the resonant frequency down to twice the mains frequency is probably not a good ideal?

As usual the formula for the resonant frequency is 1/fres= 2*pi*(L*C)^0.5. With L=120mH and C=10µ you get fres=459Hz. Because of the strong damping by the 15k series resistance you see no resonance peak but rather a bandpass response with a flat top. Increasing the coupling cap's value by a factor of 10 would lower the fres by around a factor of 3.
If you look at the current frequency response of Nick's simulation (blue curve), the actual lower corner frequency is already as low as ca. 100Hz. I don't think more bass drive is desirable.

Edit:
Didn't see Nick's post above. As usual, composing a reply in a foreign language takes me a little longer .

[SoulFetish]

It's the usual f= 1/(2 x pi x sqrt( L x C).
Roughly speaking, there's no point in being too precise as inductance is a rather variable quantity in practice and especially so with a tank or so I should think, with 1uf we're at 460Hz, so 10uF get us to the sqrt(1/10) of this i.e. 145Hz, so you were about right . I really don't see the merit in going bigger than 1uf as you don't change things much. OTOH you don't want to go any lower.


Then there's the 82K resistor. It's not really doing anything other then adding a path for cap leakage current to ground with no tank. If the leakage were high enough to be dangerous then the 82K isn't doing to help. So, you might consider a film cap rather than an electrolytic for safety reasons.

If you were to make it lower, even 10k won't change the frequency response much but will provide a bit more anti-zap protection. Lower it a bit more and you can get some useful HF attenuation e.g 2.7K gets you a corner frequency of about R/2/pi/L = 3.6Khz. Something else to play with, huh?

Yeah, not a bad idea!
But I don’t even use electrolytics is the power supply.

[nickb]

But I don’t even use electrolytics is the power supply.

I know, you crazy..........

But, I think maybe you are a perfectionist. That's a hard road to walk.

[Helmholtz]

I think it's time to give it a listening and evaluate different options.

[SoulFetish]

I think it's time to give it a listening and evaluate different options.

I’m scared.

I know, you crazy..........

You’re probably right, the evidence is overwhelming.

[Helmholtz]

Since we are using idealized sources here the frequency response is different from when driving with the tube, but the result is the same.

Why did you omit the resonant/coupling cap in the basic circuits L3/L4?

[nickb]

Since we are using idealized sources here the frequency response is different from when driving with the tube, but the result is the same.

Why did you omit the resonant/coupling cap in the basic circuits L3/L4?

I er....er... left them out deliberately to see if you were paying attention? Well done, you were.

Sorry about that. Here it is corrected.

[Helmholtz]

Thanks a lot. Obviously the resonant bandpass is far wider - reaches down lower - than I expected. And the bass drop in the complete circuit starts at a much higher frequency than the 3dB frequency of the cathode circuit (25Hz).

[66 Kicks]

These are not relevant results as
1) You are showing the plate voltage not tank current
2) You are driving it into clipping

Here is the same setup with the same control grid input at 1KHz. 1V/Div., 5.7Vpp, 2.85Vp, 2.85Vp/641r = 4.5mAp

[nickb]

Here is the same setup with the same control grid input at 1KHz. 1V/Div., 5.7Vpp, 2.85Vp, 2.85Vp/641r = 4.5mAp

You are looking the the plate voltage and this is what is misleading you.

The tube produces lots of 2nd harmonic and the tank load is reactive. So, when you drive around the resonant frequency you see little plate voltage swing at the fundamental due to the low impedance at resonant load. However the 2nd harmonic is now much bigger and so you see a very distorted waveform at the plate.

If you look at the current in the tank you will see it's just fine. Use as small a resistor as you can in series with the tank to ground so as not to disturb things too much.