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Single transistor VCA circuit
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richardc64



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PostPosted: Fri Feb 11, 2011 9:11 am    Post subject: Reply with quote  Mark this post and the followings unread

I second Dan's comments and those of MickeyDelp. In the Roland percussion units this VCA is used for snare and cymbal, and are followed by an hpf or bpf which diminish the thump. Perhaps if the input were a sine or triangle, to simulate a Tom or Bass, a bit of thump might be desirable.

JovianPyx wrote:
Ah, I see - R1 supplies a bias which would prevent the rectification, but not the thump because the assymetry with respect to zero volts is still there.


Or is the cause that pesky CV feed-through, since the envelope is applied to the collector, where the output is taken from?

btw, in the DR110 information I have, the noise/metal input levels for cymbals are shown as being 0.5vpp. In the sample I posted the level was approx 1vpp. The envelopes are around 5v.

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JovianPyx



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PostPosted: Fri Feb 11, 2011 9:29 am    Post subject: Reply with quote  Mark this post and the followings unread

Yes, when doing a synth board with a specific purpose like percussion one can be clever about the design and I would agree that the thump would be beneficial for a bass drum and pop/click might work for snare or other higher pitched percussion sounds. In that case, the super simple VCA performs two functions, to provide an envelope and to provide a transient into the audio path.

Quote:
Or is the cause that pesky CV feed-through, since the envelope is applied to the collector, where the output is taken from?


The thump comes from the assymetry about ground which originates as a DC component added to the input signal (which also occurs in the OTA type when not offset trimmed), but physically in this design it comes, as you say because the control signal is fed into the collector of the transistor (thus using it as a crude multiplier). If you look at the circuit, you can imagine that the control input is really the transistor amplifier's power supply. And I bet we all know what happens in an audio circuit when a power supply has a signal on it other than DC - that signal is copied onto the amplifier's output.

Heh, this reminds me of something I did when I was a teenager - I wanted to make a circuit that could double the frequency of a signal. I was playing with a sinewave (which is one of the waveforms for which this worked) and I had an JFET. I fed (through resistors) the input sine wave both into the FET gate and into the drain through a drain resistor (very similar to the BJT VCA circuit). Out came a signal that was no longer a sinewave, but it was double the frequency. Triangle input caused the output to be close to a triangle and double the frequency.

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Dan Lavin



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PostPosted: Fri Feb 11, 2011 1:52 pm    Post subject: Reply with quote  Mark this post and the followings unread

To somewhat second Scott,
I learned long ago, when it comes to synth circuits, if it seems too easy, it probably won't work...at least in my particular application!

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metal_head_82



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PostPosted: Wed Apr 06, 2011 12:28 pm    Post subject: Reply with quote  Mark this post and the followings unread

Just in order to revive this old thread:

When using DC-coupled signals - even a single transistor with a modulation signal applied to base can be used as VCA Wink

I did this recently in one of my projects. I used a 555 OSC and took a BC547 to gate the signal. The only problem is that gating is inverted...
Meaning gate on = signal off and vice versa...

But this is a damn simple "VCA". As soon as I finish this little bugger I'll post a link here.

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g.gabba



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PostPosted: Sun Mar 17, 2013 3:35 am    Post subject: Reply with quote  Mark this post and the followings unread

Sorry for bringing up the old threat, but I think a simple VCA is always interresting.

I found in the wonderfull YUSYNTH Archive the chronosynth.
The chronosynth has a very simple VCA.
I didnt had my Stuff here , so I cant breadboard it.
So I simulate this, and it seem to work.
I omit one Resistor and one cap. Please notice the input Levels.

GL
gabba


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inlifeindeath



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PostPosted: Tue Mar 19, 2013 11:49 am    Post subject: Reply with quote  Mark this post and the followings unread

what is the benefit in using a FET vs a small signal transistor?
I've seen this come up when using them as faux variable resistors as well.

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JovianPyx



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PostPosted: Tue Mar 19, 2013 12:23 pm    Post subject: Reply with quote  Mark this post and the followings unread

A FET gate-source junction is reverse biased, so its input impedance is very high, that is, it draws very little current.

A bipolar transistor uses a forward biased base-emitter junction so it's impedance is low, that is, it will draw substantially more current than the gate-source junction of a FET.

Any circuit that monitors or uses a control voltage would be better if it is high impedance so that it affects the control voltage a little as possible.

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elmegil



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PostPosted: Tue Mar 19, 2013 1:19 pm    Post subject: Reply with quote  Mark this post and the followings unread

What a beautifully clear explanation, thank you!

One thing I'd like to clarify further (for myself at least).

You are comparing gate->source to gate->emitter, when I'd have thought the equivalent, functionally would be gate->collector.

Is there gate->drain current flow? Why would that not be more comparable to gate->emitter?

Thanks

Pete
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JovianPyx



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PostPosted: Tue Mar 19, 2013 2:03 pm    Post subject: Reply with quote  Mark this post and the followings unread

elmegil wrote:
You are comparing gate->source to gate->emitter, when I'd have thought the equivalent, functionally would be gate->collector.

Is there gate->drain current flow? Why would that not be more comparable to gate->emitter?


Actually, there is no gate-collector (that's a bit of a morph between bipolar transistor and FET) Smile

The input signal on an FET affects only the gate-source diode junction.

The input signal on a bipolar transistor affects the base-emitter diode junction.

The actual input current drawn by the transistor will be determined only by those diodes. The base-collector diode has little if anything to do with input impedance - same goes with the gate-drain.

HOWEVER - it should be noted that many FETs can actually be operated "upside down", that is, reversing the drain for the source. This isn't as odd as it seems as many FETs are symmetrically constructed such that the gate placement is the same distance from the source contact as it is from the drain contact. Note also that a FET transistor is made of only two pieces of silicon whereas the bipolar transistor is made of three (hence NPN or PNP).

There should not be gate-drain current. This is because the diode junction is reverse biased for a FET and the source and drain contacts are at opposite ends of the same chunk of silicon, so it wouldn't matter which you chose, the junction is still reverse biased. Note that yes, technically, even though the junction is reverse biased there is a "leakage current", but it is still far far smaller than the forward bias current that flows through a base-emitter junction of a bipolar transistor.

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elmegil



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PostPosted: Tue Mar 19, 2013 2:35 pm    Post subject: Reply with quote  Mark this post and the followings unread

Embarassed

I meant base->collector and base->emitter, of course :/
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JovianPyx



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PostPosted: Tue Mar 19, 2013 4:24 pm    Post subject: Reply with quote  Mark this post and the followings unread

ah, there is no substantial base-collector current either. My assumption is that In that case, looking at the base-collector diode, it would be reverse biased if Vbc were the only influence. (that is probably a naive look at it) When considering input impedance of a transistor buffer, we care only about the impedance with respect to ground, so we're looking at just the one junction.
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akrearke



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PostPosted: Fri Dec 06, 2013 10:32 am    Post subject: Reply with quote  Mark this post and the followings unread

Raising this thread from the dead, I have some questions for the more experienced and better informed.
Let me describe what I'm intending. I want to control a bipolar signal like a vco output with a simple cv controllable vca that output the signal about ground like the original signal then invert that to the original phase.
So looking at these simple vca circuits I see that the action of the transistor shift the signal to positive side only. Okay that is fine now how would one shift the signal again to be about ground?
These are the solutions I am looking at: the inverting amplifier stage completes the inversion, but does it shift the signal (if powered by a dual supply)?
Or, perhaps using some such circuit as the mfos level shifters found on the soundlab modifications page?
I would prefer to have something as simple as the ms-20 vca, as my circuit needs 10 independent vcas.
Any thougts?
Thanks guys.
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elmegil



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PostPosted: Fri Dec 06, 2013 11:37 am    Post subject: Reply with quote  Mark this post and the followings unread

First thing that occurs naively (I don't know that I qualify as "more experienced" by any significant amount on this topic Wink ), would be tie the + input of the op amp to -V rather than ground? Or is that breaking something because it then looks like a 30V single-supply configuration?
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JovianPyx



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PostPosted: Fri Dec 06, 2013 3:36 pm    Post subject: Reply with quote  Mark this post and the followings unread

I don't see how putting the + input to -V helps.

In an inverter, the + input is usually grounded as a reference (this is what usually makes the summing node a sort of virtual ground). The opamp subtracts the - and + inputs to derive it's output which in an inverter always tries to cause the + and - inputs to be equal. If + input is at -V, I believe (in an inverter circuit) the opamp will try to drive the output hard to the negative rail.

I think what you may be looking for is a circuit called a level shifter.

Ultimately, however, I think that a more advanced VCA circuit may serve you better. Look at the LM13700 datasheet, there are circuit examples in most of them and they should include a VCA.

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elmegil



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PostPosted: Fri Dec 06, 2013 3:51 pm    Post subject: Reply with quote  Mark this post and the followings unread

Not what I meant:

-Vcc perhaps more clear? set the "virtual ground" at -15V instead of actual ground.
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JovianPyx



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PostPosted: Fri Dec 06, 2013 4:06 pm    Post subject: Reply with quote  Mark this post and the followings unread

elmegil wrote:
Not what I meant:

-Vcc perhaps more clear? set the "virtual ground" at -15V instead of actual ground.


Heh, well I still don't see it. Not sure what you mean by -Vcc. And which virtual ground ... the summing node or another virtual ground made of 2 resistors?

If the summing node is at -15v, (which means the + input is at -15v) you still have no negative headroom and with almost any input, the output will drive hard negative.

Maybe if you drew a schemo I'd see what you mean?

This is different than summing a constant negative DC voltage with the input signal to cause the level to shift.

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elmegil



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PostPosted: Fri Dec 06, 2013 6:56 pm    Post subject: Reply with quote  Mark this post and the followings unread

I'm messing up all my terminology. In this case by virtual ground I mean where, in the original circuit, ground is going to the positive input of the op amp, and by -Vcc I mean the negative rail.

I've drawn what I was talking about (which I figured was probably wrong, but not sure why.... more in a moment), and what I think you're talking about...but not quite.

if you want it to go to near the middle this way, you'd adjust the offset with 1/2 of the negative rail into the summing node, wouldn't you? Assuming balanced rails of course.


Edit: And then I go and forget "more in a moment"... should just go get some sleep I think. I'm not sure I see why what I was suggesting would cause the output to just slam to the negative rail.


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JovianPyx



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PostPosted: Sat Dec 07, 2013 4:11 am    Post subject: Reply with quote  Mark this post and the followings unread

The upper schematic is the one I believe will drive negative with most inputs.

The bottom one, however, looks like it could work. I'd probably take it a step further and take the resistor to the wiper of a pot with the other ends at ground and -V. This would make it adjustable. Or the resistor to -V could be a "rheotstat" wire pot.

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