LFO + FET question (more out-of-phase stuff)

[midwayfair]

I'm trying to figure out if something would work.

I have this LFO (specifically A):


The output where the LEDs is connected settles at 4.5V. They blink out of phase simply by changing the voltage rail they're referenced to, with the 4.5V being either the "ground" or "supply rail" for each LED.

Let's say I wanted to use FETs like in the EA Tremolo instead of the LEDs.



When RG Keen did his Harmonic tremolo (see here: http://www.geofex.com/FX_images/FakeFenderVib3.pdf), it looks like he hooked up the source FETs to a 4.5V voltage reference and then fed them an out-of-phase signal.

The oscillation with the FET works by dropping the resistance when the voltage swings negative, and increasing its resistance when the voltage swings positive.

So here's the question(s):
If I hook up one FET's source to ground (as in the EA Tremolo) and the other FET's source to +9V instead of ground, and then feed them the output of the LFO that centers at 4.5V, will I get out-of-phase modulation of the two bands similar to what RG did in his harmonic tremolo? Or would I instead need to use P-channel devices? Or do I absolutely need to use out-of-phase LFO signals to get the devices to change their resistances counter to each other?

[GGBB]

If I hook up one FET's source to ground (as in the EA Tremolo) and the other FET's source to +9V instead of ground, and then feed them the output of the LFO that centers at 4.5V, will I get out-of-phase modulation of the two bands similar to what RG did in his harmonic tremolo?

That will only offset the gate-source value. It will not change the direction of the swing of the gate-source value, which is what matters. You need to have one copy of the LFO moving toward cutoff while the other copy is moving away, and at the same time the gate-source value for both needs to be above cutoff (NPN). The only way you can have one gate-source value approach cutoff while the other moves away through changing the source voltage is to move the source enough so that cutoff+source for the "out of phase" signal is above the gate voltage at all times. +9V may in fact do that (but probably not symmetrically), however in that case the fet will always be off if it is NPN because if Vgs-off + Vs > Vg then Vgs-off > Vgs.

So I think you need either a matched NPN and PNP pair with both sources at ground, or out of phase signals.

(edited for typos)

[midwayfair]

If I hook up one FET's source to ground (as in the EA Tremolo) and the other FET's source to +9V instead of ground, and then feed them the output of the LFO that centers at 4.5V, will I get out-of-phase modulation of the two bands similar to what RG did in his harmonic tremolo?

That will only offset the gate-source value. It will not change the direction of the swing of the gate-source value, which is what matters. You need to have one copy of the LFO moving toward cutoff while the other copy is moving away, and at the same time the gate-source value for both needs to be above cutoff (NPN). The only way you can have one gate-source value approach cutoff while the other moves away through changing the source voltage is to move the source enough so that cutoff+source for the "out of phase" signal is above the gate voltage at all times. +9V may in fact do that (but probably not symmetrically), however in that case the fet will always be off if it is NPN because if Vgs-off + Vs > Vg then Vgs-off > Vgs.

So I think you need either a matched NPN and PNP pair with both sources at ground, or out of phase signals.

(edited for typos)

Sigh, I was afraid of that. Thanks! Jubal81 made the cogent point elsewhere that if it COULD be done the way I did it, it would have been already.

[Transmogrifox]

Actually, you would be ok doing something equally as simple if you just rearrange some connections.

The trick is one FET source is referenced to the LFO while the gate is held fixed.  Then the gate of the other FET is referenced to the LFO while the source is held to a fixed reference.

As the LFO goes positive it turns off the FET with the fixed gate voltage while simultaneously turning on the gate of the other.

In the example of an EA tremolo, you simply abandon any reverence for the DC reference on the JFETs.  Depending on the FET gate turn on, you could tie the gate of FET 1 to ground through a 1 Meg resistor or something, then tie the drain to a largish capacitor (to pass audio band) in series with a large resistor.  The large resistor goes back to the LFO output which would range from 0V (pretty much on) up to 6V (pretty much off).

Then drain of FET 2 is referenced to 6V (typical Vbias divider with largish bypass cap) and the same 0V-6V LFO is connected to the gate of FET 2.  Then when LFO starts at 0V, this FET is mostly off while the increase toward 6V turns it on.  

In this configuration you get the same behavior.  Just think in terms of Vgs, not in terms of what V is actually moving.  Then the moving V connects to S on one and G on the other, and you still get the out of phase modulation from the same LFO.

//EDIT
Well, I just couldn't resist doing this in LTSpice.  Here is a proof of concept, simulated on wav file:
https://soundcloud.com/transmogrifox/stereo-trem-demo

From this:

[GGBB]

Very nice, but all this begs the question - wouldn't inverting the LFO signal be a lot easier?

[Transmogrifox]

Very nice, but all this begs the question - wouldn't inverting the LFO signal be a lot easier?

Perhaps the rest of the things related to the rEAgenerated tremolo make it look more complicated than it is.  

The whole concept I presented I think takes 2 extra resistors and 1 less capacitor to create the extra reference point.  It eliminates the AC coupling cap normally used in the EA trem replacing it with a DC bypass cap on the other side at a 2:1 improvement on parts count if you consider for the out-of-phase drive you would need 2x AC decouplers for the phase inverter approach.  In other words it replaces 2x capacitors with 1x and adds about 2 resistors.  No other semiconductor parts required.

Also from LFO schematic shown by the OP, this entirely eliminates the 2 buffer op amps used and tie the output from the depth pot goes to the control signals.

Inverted LFO takes a transistor (or op amp), probably 2 resistors, and probably 2 more capacitors to AC decouple unless you want to use the same number of components to make correct DC references.  So instead of eliminating a dual op amp package like what I presented would accomplish for the LFO suggested, the phase inversion would require keeping this op amp package and using it for a buffer + inverter (still not a bad solution).

What part of inverting the LFO is easier?

[midwayfair]

Transmogrifox, I can't thank you enough. I'll try it out as soon as I can actually clear the breadboard again (I'd have experimented myself but both of breadboards are occupied for a bit). The proof of concept sim is very helpful and I really appreciate that you kept the parts count down.

[Transmogrifox]

I'm glad you found it helpful. I'm interested to see what you brew up with this combination.

It might turn into something I can't help but build for myself

The proof of concept will certainly take some fine tuning to suit a specific application, but I get the impression you know what to do now that you have caught onto the idea.

Have fun.