Thor mu amp stage/section

[WGTP]

Great pictures and helpful info.  On the other forum, there was a gentleman that mis-drew his SRPP as well, I thought.  It was kindly pointed out to me that you can use it that way as well, which produced a more symmetrical distortion with slightly less gain.  In some applications it produces more gain.  

X Bananov focuses some on the value of the resistor between the Jfets and the Source resistor being Jfet dependant, I think.  I don't understand all this stuff and it's worse when it is in Russian.

Anyway, I have a Mu, not SRPP, as the 3rd stage of a distortion on the breadboard.  I have found using a 470k resistor (might need to bypass with 470pf cap) before it, a 470k bias resistor and a 4.7k Source resistor to ground produces a very smooth compressor like low distortion sound.  Lowering the 4.7k resistor turns it back into a raging gain stage distortion monster.   I have also seen a diode used there in place of the resistor, but that seems to be a high gain deal.  

[sault]

What about the gateresistor of the first mu-amp stage? In different mu-amp designs you can find different values (the mini-booster uses 10M!). What happens when you change that value, the same thing as with the 2nd stage?

Okay, let's make sure that we're talking about the same thing - I didn't see a 10M resistor in the mini-booster schematics that I looked at. So, let's look at dthurstan's diagram. If you mean R3, the 1M resistor going to ground, that is what is known as a "pull-down" resistor. In practical terms, its purpose is to keep the "pop" noise down when you click a pedal on/off.

For a second stage, it functions as a load. Lower values load down your gain stage. I showed that a little with the first two graphs... with a 1Meg resistor going to ground (ie a very high load) the gain was 30-ish dB. That's 0.1 volt to over 3! Lowering that load sucked up quite a bit of gain... by the time it got down to 10k, it was at 22 dB... we just cut our gain by 2/3rds! So keep it reasonably high (a few hundred k) if you want to keep that amplification.

If you're talking about R21, that's what I addressed with my SPICE simulations below. Higher values (ie 1M and up) give high gain, but are very non-linear. Lower values (470k and below) lose quite a bit of gain but are much more linear.

First I thought just to use 1M for the gateresistor, but got confused by all the different values. What value should I choose to keep the mu-amp clean, at least for the first 3/4 of a turn on the gainpot?

I'm sorry, I get carried away sometimes.

Referencing the schematic below, keep R3 high (1Meg is just fine). The higher the value of R21 the higher the mu-amp's gain and distortion will be - below dthurstan chose 220k, which is probably the lowest you should go. It should be the "cleanest" you're going to get without throwing all of your gain away. If we look at my schematic on the last page, "R6" is the load resistor following the mu-amp. Higher values will preserve your gain (ie 1Meg), low values will cut it down significantly (below 100k especially).

As dthurstan noted, this configuration works well after a passive tone stack - enough gain to make up for the filter's insertion loss without getting too crazy. The push-pull of the mu-amp (okay, SRPP, whatever) gives a different harmonic response than using two jfets separately.... no even-order harmonics, and fewer higher-order until you start to push it harder, IIRC.

Does that help at all?

[Rutger]

Thanks alot Sault, it really helps! Your explanations are very clear

I was talking about the load- or pull-down resistor (R3). I keep mixing up those terms, sorry (It's also confusing that Americans use other (english) terms than Europeans). I saw a schematic where it was 10M (http://www.geofex.com/Article_Folders/modmuamp/modmuamp.htm), in the schematic of the Thor it is 220k.
So I should keep it like 1M to keep the gain up, thanks I think I'll stick with the 470k for R21 of the 2nd mu-amp stage.

I was planning on using a volumepot after the mu-amp, from what I understand now it needs to be something like a 500k or 1M pot?

I think I'll put a simple tonestack (Bridged T-filter) before the mu-amp, to simulate the tipical mid-dip of a fender tonestack, would be nice to experiment with.

[amptramp]

X Bananov focuses some on the value of the resistor between the Jfets and the Source resistor being Jfet dependant, I think.  I don't understand all this stuff and it's worse when it is in Russian.

Anyway, I have a Mu, not SRPP, as the 3rd stage of a distortion on the breadboard.

Most people use µ-amp amd Shunt Regulated Push-Pull (SRPP) as synonyms.  Consider a tube or depletion-mode JFET version of the SRPP.  You have an upper device, a lower device and a resistor from the drain/plate of the lower device to the source/cathode of the upper device.  If the gate/grid of the upper device is returned to its own source/cathode, it becomes a current source (high load resistance) in the JFET version or a reasonably low load resistor in the tube version.  This gives you an amplifier stage with high gain for a FET or moderate gain for a triode. This is not SRPP.

If the upper stage gate/grid is returned to the drain/anode of the lower stage on the bottom of the resistor, the behaviour of the circuit is entirely different.  The drain/plate current flows through the resistor and provides a negative bias with more current through the bottom device tending to turn off the top device.  Let's say you have JFET's or triodes with a 2000 µmho transconductance, that is, a change ove 1 volt at the input provides a change of 2 mA of current.  What size of resistor would cause the current in the upper device to change in the opposite direction by the same amount as the change in current in the lower device?  We already know that a 1 volt change in the input to the lower device causes a 2 mA change in current.  This 2 mA across the series resistor must cause the same amount of voltage change in the opposite direction in the upper stage. The 2 mA change must force the voltage to change by 1 volt across the resistor, so: 1V/2mA is 500 ohms.  This shows how the resistor must be the resiprocal of the upper device transconductance in value to get the same current change in the opposite direction for upper and lower devices - in other words, balanced push-pull operation.

But this is a totem-pole circuit - the current in the upper stage minus the current in the lower stage has to be delivered to the load or the stage will simply slam up against the upper or lower power rails.  Thus, you need a finite load (which may include a resistor for the upper gate or grid if it is AC-coupled to the lower drain/plate).  We have seen this already in the waveforms on the previous page and now you know why.