Help me understand something on this gain stage circuit

[markmalin]

Hi, all!


A friend sent me this schematic for an overdrive pedal.  It's pretty straight forward, but can someone explain the 2 diodes and the 10uf/50v capacitor after the output of the second gain stage?  It's between the plate of the 2nd gain stage and the cathode of the first.  I'm not sure I understand what's going on.

Thanks!
Mark

[PRR]

The cathode cap on V1b makes the diode network somewhat pointless.

Breadboard it and try.

Also I bet 2 cents that if you image-search you can find the original page where that image is presented.

[dbishopbliss]

I was building a chip amp once and there was something similar. I was told it was for a voltage drop. Not sure if that was correct, but maybe.

[pdf64]

Yes, it’s a very convoluted and expensive mxr distortion, a waste of a valve and power supply really.
The connection of the diodes back to the 1st stage’s cathode is a red herring, a connection to 0V common would work pretty much the same.

[markmalin]

So having the diodes in there connected to the cathode of V1a is electrically the same as tying that 10uf/50v electrolytic to 0 volts? (i.e. ground?).  Would the voltage on the cathode of V1a fluctuate as the tube "self biases"?  Is this causing a pull (fluctuation) on the output signal??

[shooter]

Is there a psudo NFB from the circuit?

[PRR]

> So having the diodes in there connected to the cathode of V1a is electrically the same as tying that 10uf/50v electrolytic to 0 volts?
> Is there a psudo NFB from the circuit?

No and no.

[markmalin]

I was talking with a friend who's an EE.  Here's his take on this - I thought you guys would find this interesting:


"The inverted diodes form a classic symmetrical clipping circuit to create distortion, not overdrive. The 10uF / 50v cap serves as a coupling cap that passes the AC signal to the inverted diodes on its way back to the cathode of the previous tube. In other words, you have Negative Feedback which controls (limits) the overall gain.Although a clipped sample of the AC signal is sent to the previous stage, the 10uF / 50v cap must also block the DC that typically resides on that cathode. The 10uF cap is doing its job of passing an AC signal and blocking DC."
Mark

[shooter]

The inverted diodes form a classic symmetrical clipping circuit to create distortion, not overdrive.

that's how I "saw" them, There are examples in amps, I'm just not going to hunt the schematics down.  they're typically between gain stages and go to ground.

[sluckey]

The inverted diodes form a classic symmetrical clipping circuit to create distortion, not overdrive.

that's how I "saw" them, There are examples in amps, I'm just not going to hunt the schematics down.  they're typically between gain stages and go to ground.

Me too.

[markmalin]

The inverted diodes form a classic symmetrical clipping circuit to create distortion, not overdrive.

that's how I "saw" them, There are examples in amps, I'm just not going to hunt the schematics down.  they're typically between gain stages and go to ground.

I just had never seen that before and didn't understand what its purpose was.


Thanks for the input (everyone).  I understand it now :)

Mark 

[pdf64]

I was talking with a friend who's an EE.  Here's his take on this - I thought you guys would find this interesting:


"The inverted diodes form a classic symmetrical clipping circuit to create distortion, not overdrive. The 10uF / 50v cap serves as a coupling cap that passes the AC signal to the inverted diodes on its way back to the cathode of the previous tube. In other words, you have Negative Feedback which controls (limits) the overall gain.Although a clipped sample of the AC signal is sent to the previous stage, the 10uF / 50v cap must also block the DC that typically resides on that cathode. The 10uF cap is doing its job of passing an AC signal and blocking DC."
Mark

I don’t understand the distinction between ‘distortion’ and ‘overdrive’?
Surely the decoupling of the input stage’s cathode would act to break a negative feedback loop?

[dwinstonwood]

Scroll down to the "Clipping of Both Half Cycles" heading on this page:
https://www.electronics-tutorials.ws/diode/diode-clipping-circuits.html

Is that what's going on?  And, then the cap is there to only let the AC signal through the clipping diodes?

[Keppy]

Scroll down to the "Clipping of Both Half Cycles" heading on this page:
https://www.electronics-tutorials.ws/diode/diode-clipping-circuits.html

Is that what's going on?  And, then the cap is there to only let the AC signal through the clipping diodes?

Yes and yes, with one caveat.


If the diodes are run to signal instead of to ground (or another low impedance steady voltage), then instead of clipping at +/- .7v, they will clip at +/- .7v of voltage gain. In other words, instead of a hard clip at a predetermined voltage, they will clip when the output of the second stage gets more than .7v AC away from the stage 1 cathode. This preserves some of the original signal dynamics and adds some "softness" to the clipping.


Like PRR said though, with that big bypass cap on the cathode, I'm not sure if it actually makes a difference in this circuit. That cap makes the cathode look a whole lot like a low impedance voltage source. Someone must have thought it was worth the trouble, though, since they didn't just connect the diodes to ground.

[chocopower]

Maybe this help.
That pedal is a Guyatone TD1 (branded westbury)
I have the TO2 version. The same, but with bass and treble controls.


http://www.geofex.com/FX_images/westbury.gif


https://music-electronics-forum.com/forum/effectification/guitar-effects/38207-need-help-tuning-bass-response-in-guyatone-to-2-please-theory-inquiries-as-well

[sluckey]

There is no NFB or "gain" involved in the diode clipper. The voltage on V1b cathode is a pure dc voltage. No signal there due to the big 22µF cap that holds the cathode at AC ground. V1b cathode is just being used as a small DC reference voltage for the clipper diodes. This simply changes the point at which clipping occurs on the signal at the top of the clipper diodes.

[markmalin]

Maybe this help.
That pedal is a Guyatone TD1 (branded westbury)
I have the TO2 version. The same, but with bass and treble controls.

http://www.geofex.com/FX_images/westbury.gif

https://music-electronics-forum.com/forum/effectification/guitar-effects/38207-need-help-tuning-bass-response-in-guyatone-to-2-please-theory-inquiries-as-well

Thanks.  That's a good thread.  The guy sent me some youtube video links of the TD1 and I was surprised at how much "distortion" they were able to get out of these two gain stages.  I think the diode clipping explains that.  I was looking at the schematic wondering how close it was to a typical 2 triode circuit and the diodes were confusing.

[pdf64]

... V1b cathode is just being used as a small DC reference voltage for the clipper diodes. This simply changes the point at which clipping occurs on the signal at the top of the clipper diodes.

I don’t think it’s even that, as the cathode voltage is applied the same to both diodes. So the waveform will be 1.2Vp-p.   

[shooter]

I think the diode clipping explains that.

that's what happens when you take a pretty sine wave and force it to be square 

[pdf64]

I’d get rid of the clipping diodes and facilitate some proper distortion by boosting the input with a FET or regular transistor, if that’s feasible.
The 1M pot for the output vol is just yucky, it’ll cause treble boost at low settings and treble cut at medium to high settings.

[markmalin]

I’d get rid of the clipping diodes and facilitate some proper distortion by boosting the input with a FET or regular transistor, if that’s feasible.
The 1M pot for the output vol is just yucky, it’ll cause treble boost at low settings and treble cut at medium to high settings.

What value pot would you put in there for the master?

[pdf64]

I think about 100k (lin) should be the max, if it needs to be higher than that (eg to retain gain) then a buffer should perhaps be considered.

[shooter]

should the 300k be dropped to  ~~30k?

[pdf64]

I’d leave it alone to see how it works in practice, and then tweak as necessary.