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Bipolar "depth" control for cv
MUFF WIGGLER Forum Index -> Music Tech DIY  
Author Bipolar "depth" control for cv
stk
Bear with me, learning hmmm.....

I have a dc (or ac) source, and I want bipolar control over its amplitude.

Would I use a tl072 as a dual unity gain buffer (thus), one side in inverting mode, and then mix the output together with a simple pot (or another opamp)?

Thanks
Hi5


simplest design I have found yet. just make sure all the resistors are the same value and you will have a bi-polar attenuverter.
stk
Hi5 wrote:

simplest design I have found yet. just make sure all the resistors are the same value and you will have a bi-polar attenuverter.


Cheers for that.
I was more looking for _separate_ (ie just split, buffered & one of the two inverted - not premixed by the opamp as above) signals that I can feed in to my joystick. Although your schem will come in handy for something else.

(The joystick I'm putting together will be normalised to the power rails, but will accept an external input for each axis which will be attenuvertable).

Cos I'm at work and thus not able to try it out, would my original idea work? Thx.
Adam-V
What you need is a so-called attenuverter.

Have alook at the processor section at the bottom of this circuit: http://cgs.synth.net/modules/pic/schem_cgs75_vcs.gif

Cheers,
Adam-V
daverj
Yes, your original idea would work. A non-inverting amp from the jack to one side of the joystick pot and an inverting amp from the jack to the other side of the pot. You also need another non-inverting amp after the joystick pot.

Or you could use the circuit from Hi5 and use the joystick pot in place of the pot he shows.

If you are normalizing it to +5 volts, use a resistor divider on the switch pin of the jack. Don't tie it straight to power or you'll have a momentary short to power when you plug something into the jack.
stk
Cheers guys, got a model working (apologies to Hi5, that was actually almost exactly what I needed oops ).

One further question - is it considered "good" to buffer the inputs and outputs to make them play nice with other modules?

For eg in Hi5's example above, should one place a voltage follower on both in and out to isolate it from any other circuits?
stk
daverj wrote:
If you are normalizing it to +5 volts, use a resistor divider on the switch pin of the jack. Don't tie it straight to power or you'll have a momentary short to power when you plug something into the jack.


Ah I see, so much to learn smile
I'll be placing a trimpot as voltage divider between power and the normalised connection to fine tune the normalised voltage, so this would achieve the same thing, right?

Edit, thusly:

Edit 2, I think I'll go for a switch rather than normalising, seems more flexible that way.
daverj
Hi5's circuit already has a buffered output, so no need for an additional one there. With your original plan you would have needed a buffer on the output because the pot itself would be too high impedance as an output.

If the joystick pot is 100k then the circuit has 50k input impedance. That should be no problem, and doesn't require an input buffer. If the joystick pot was 10K then the input impedance would be about 9k and at that point an input buffer would probably be a good idea.

Yes, the trimpot should work fine as a voltage divider to supply your normalled voltage. That way there is always some resistance between the power rails and the tip of the plug you stick into the jack.

The way you drew it looks like separate jacks, but I am guessing that what you have in mind is to use a switched jack so that the power goes into the circuit when nothing is plugged in and it gets disconnected when you patch a module into that jack.

Is this in a +/- 5 volt system or a 0-10 volt system? The circuit you have will work fine with +/-5 signals. You'll have biasing problems if you are using it in a 0-10 volt system. I ask because of the +10 volt "input" voltage you are showing in your drawing.
stk
daverj wrote:
Hi5's circuit already has a buffered output, so no need for an additional one there.

I see. Reason I ask is, playing about here in 5Spice, I'm getting lower voltages when splitting the voltage source to both x and y axis. Putting a voltage follower after the voltage source fixes this (see attachment).
Edit- Although I did just work out I can use a single buffer before the switch (ie just after the trimmer) rather than two d'oh!

Quote:
The way you drew it looks like separate jacks, but I am guessing that what you have in mind is to use a switched jack so that the power goes into the circuit when nothing is plugged in and it gets disconnected when you patch a module into that jack.

Exactly (though I'm leaning more towards a manual switch now). I just drew it that way cos 5Spice complains when there's unconnected lines.

Quote:
Is this in a +/- 5 volt system or a 0-10 volt system? The circuit you have will work fine with +/-5 signals. You'll have biasing problems if you are using it in a 0-10 volt system. I ask because of the +10 volt "input" voltage you are showing in your drawing.

It's +/- 5 (Euro). The 10v input is just there cos I was playing about.

Thanks for all your help. I actually do have a copy of The Art of Linear Electronics here.. Packed up somewhere in 12 boxes of books sad banana
daverj
The voltage from the trimpot will change as you connect or disconnect it to the circuit. Rather than add a buffer amp on the input, I would simply use two trimpots. One for each channel. Adjust them so they supply the correct voltage when connected. It then doesn't matter that the voltage changes when disconnected.

Or simply select a pair of resistors to use instead of each trimpot. Once selected there shouldn't be any reason to adjust them again.
stk
daverj wrote:
Or simply select a pair of resistors to use instead of each trimpot. Once selected there shouldn't be any reason to adjust them again.


That makes sense to me and I've been sim'ing that in 5Spice, but I keep ending up with uneven/nonlinear ranges (-8.7 - +10, etc), whereas it behaves perfectly when buffered. Maybe it's the software seriously, i just don't get it

I've sketched it up as a perf layout with a 074, using one of the spare channels as the buffer.. It's a nice symmetrical layout hihi
daverj
The way I calculate it, without an input buffer you could put a 5.6K between the input and ground and a 7.05K between the input and +12v. These resistors wouldn't be directly on the input, but on the switched pin of the input jack. 7.05K isn't a standard value, so you would have to use two resistors in series or parallel to create that value.

BTW, with that 100K attenuator on the output for "Range X" you probably should add an output buffer. A 100K pot is not a good idea as a direct output. It'll work without the buffer but the level will change depending on what the input impedance of the next module is (and if you mult it to more than one module).
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