Mu Amp - Diode vs. Source Resistor

[WGTP]

I have noticed on some of the Russian designs, that the source resistor is replaced with a diode.  What's up with that? 

[Bill Mountain]

I have noticed on some of the Russian designs, that the source resistor is replaced with a diode.  What's up with that?  

The diode can set bias just like a resistor.  I've seen it in a few tube designs also.  I love playing around with emitter diodes in transistor boosts as well.

I don't know the science but since diodes conduct at certain voltages it acts like a bypassed source/cathode/emitter resistor which biases the tube/jfet/transistor.

MerlinB talks about it on page 34 here:

http://valvewizard1.webs.com/Common_Gain_Stage.pdf

[diydave]

With a diode at the source, you get a Vgate-source, V-cathode or V-emiter of 0.6 volts, without de 'hastle' of calculating a certain biasresistor with a certain current that's running through your fet, tube or transistor.
Can be interesting, but I do believe there is a certain drawback.

[R.G.]

Can be interesting, but I do believe there is a certain drawback.

At least one of the drawbacks is that with a fixed voltage, there is no DC feedback with current changes to stabilize the current through the thing.

[WGTP]

Thanks for the responses.  Since I can't read Russian, and struggle with English, I assumed the point of using the diodes was to somehow embelish the "tubishness" of the tone.  Any ideas about that?  I have tried an LED, but they have a funny release at the end of the notes.  A single 1N4818 SI is cool and a 1N34A GE seems to have the highest gain.  Does the "knee" of the diode effect things any?  I have also used them in parallel with resistors from 2.2k to 4.7k in value.  With vintage ears, it's hard to tell much difference, other than different levels of distortion due to gain changes. 

[R.G.]

It's important to isolate changes, and to think about what each change does to both the AC and DC conditions.

Resistors, LEDs, silicon diodes and germanium diodes all affect both AC and DC differently. Resistors have the same impedance to both AC and DC. In setting DC conditions, they create a DC bias voltage equal to the DC current through them. This bias voltage sets up the AC gain and operating point of the FET. The resistor also affect the AC gain of the FET by adding some AC feedback by the way it responds to AC signal currents flowing through it. This is why bypassing a source (or emitter, for BJTs) resistor with a capacitor increases gain - it removes the AC feedback coming from the resistor without affecting the DC conditions.

Imagine you had a magic battery that could be set to any voltage, but otherwise was just a battery. If you used this instead of a source resistor, you could dial in any source voltage you wanted, thereby setting the DC operating conditions for the FET to get the right gain, operating point, etc. The FET would operate at the maximum gain for that operating point and drain load, because there would be no feedback from the source to lower gain. The FET's DC operation sets the current through the battery based on that source voltage, which is fixed.(As a side note to more advanced circuiteers, as the FET drifts, the current drifts because the battery doesn't help compensate things.)

The sound from this would depend entirely on what the fixed source voltage did to the operating point of the FET, and the rest of the FET circuit.

Inserting a diode of any kind is similar to presetting the battery. Each diode conceptually acts like a fixed voltage drop at any current, so long as it's operating at a current well above the conduction knee.  It has a very small, but measurable AC resistance. If either the peak AC signal current or the FET characteristics make the current through t he diode dip to the knee region, things change radically. The incremental AC resistance rises dramatically and the voltage across the diode changes quickly from nearly fixed to plunging toward zero. The FET, which was headed for cutoff to get to the diode knee, cuts off even faster.

The only real differences in diodes in this application are the big differences in forward voltages and the much smaller differences in AC resistance in the conduction region. The forward voltage differences have a very big effect on the FET's operating point and AC gain, as well as the signal voltage where it drops out. This will have audible differences, of course.

Paralleling with a resistor means that you're getting a bias voltage on the source that changes. From zero current to the diode knee voltage across the resistor, the diode doesn't conduct at all, and the resistor acts like the diode isn't there. When the diode starts conducting, it limits the voltage across the resistor to a (mostly) fixed value, so the resistor conducts a fixed current, and doesn't have any further effect on the FET operation. The diode then controls what happens above that current. This combination has the effect of moving the diode cutoff to a higher current, and making the operation below diode cutoff voltage much less abrupt to signals.

What this all sounds like depends heavily on how big the signal is. Small signal - under maybe 50mV on the source terminal - will have much less abrupt changes in gain because they're not transiting through the diode cutoff region much, and the change will be more a gain change at that point. Big signals will sound like they're being clipped at the diode conduction knee.

[WGTP]

Thanks for the responses.  So it appear the trick is to use find the correct diode value working in conjunction with the overdriven Jfet to produce the desired clipping affect?  

[tca]

The idea is probably this.

[DDD]

AFAIK the diode (instead of resistor in the source circuit) is intended just to bias FET, and nothing more, no "tube sound", etc.

[WGTP]

Some of the diodes placed throughout the Russian Schematics I've seen also appear to limit the signal getting to the second or third stages to keep them from over-driving too hard.  Also some single diodes are placed to ground in the signal path to create asymmetrical distortion...

Thanks for your input. 

[Davelectro]

Seems like we're talking about two different concepts here...

The source diode is for biasing purposes. The one between gate and source emulates grid current (in conjunction with two carefully selected resistors).

[WGTP]

Sorry, the original part got explained, so I sort of went off topic.