I am trying to figure out how to switch a signal off of one path and onto another by inserting a plug into a jack. I have been trying to make it work with a switching jack, but I don’t think they work the right way. The idea is to divert a control signal, off of a circuit that goes through a pot and onto a circuit that goes through a vactrol, and I want to to do it by plugging in a jack carrying a CV signal. The problem with switching jacks is that they break circuits rather than switching them, at least in the way I need. Should I be using a transistor that switches based on the presence of a CV input? I don’t really want to lose to any of the CV from a voltage drop.
Here’s a sketch of what I’m trying to do. Basically, I want to be able to use the signal level pot as a CV level pot when CV is connected. I figure increasing the resistance on the LED half of the vactrol will reduce the CV level just fine. But, I n order to be able to use the pot for both levels I need to divert the signal at SW1 to go through the vactrol, and I’d like to do that without adding too many more components, since I’m trying to use one pot for two things with the intent of reducing components in the first place.
Typically the way that’s done is to use a switched jack for the CV, with a DC +5 V or +10 V or whatever connected (through a resistor) to the tip switch. Then it’s always a CV attenuator, but with nothing plugged in the “CV” is a DC voltage, so it behaves like a signal attenuator.
It would be a bit more clear what you want with this circuit if you would draw in an output, it contains only input signals at the moment. Or am I to understand that the point called ‘signal in’ is also the output of this ‘double attenuator circuit’? If so, then why use a switch at all? If there is no CV RV1 will control the impedance, if there is a CV and RV1 is at max then the vactrol will control the impedance (and there are a lot of possibilities in between if you combine them). If you leave out the switch In stead of remembering to set the switch in the right position you only need to remember what the pot is for and to set that properly to switch from pot-to-vactrol control and vv.
Sorry about the confusion. This is just the variable resistor part of an attenuverter circuit, which sits between the circuit, which connects at “signal in”, and ground.
I definitely could have been clearer about the switch–I’d intended it to represent the effect I wanted to achieve, rather than and actual component. I meant to mention that in my post, but it was late and I was clearly not at the top of my game.
What I’d hoped to achieve is to switch the signal path as illustrated without having to add an actual switch, either through the use of a switched jack or some similar electronic wizardry. As @analogoutput pointed out, though, connecting a CV signal directly to a vactrol is poor practice, so I will probably scrap this ill-conceived notion and implement CV in the circuit in another way.
Well, a lot of the circuitry presented on these pages can be improved upon so I wouldn’t be bothered about that too much. Anyway I hope you will keep trying the ideas you have because they are likely to lead you to a better understanding of practical electronics.
@analogoutput I read your blog post and I have to wonder, in the filter on the op-amp, since the resistor and not the capacitor is connected to ground, why would it be a low-pass and not a high-pass filter?
The capacitor isn’t connected to ground. It is parallel to the LED+resistor, providing a low impedance path for high frequency signals. Low frequencies see a high impedance on that path. So the frequency response at the LED is suppressed at high frequencies.
Something along these lines might be what you’re looking for:
Yeah, that’s the diagram I was confused by. I’m used to using the capacitor location in the circuit to determine the type of filter (ie: cap to ground=LPF). Since the cap in the RC filter part of the diagram isn’t connected to ground and the resistor is (through another resistor, but they’re logically one resistor) I interpreted it as a HPF. That’s about all I know about filters though, and it looks like this is kind of a special case.
My initial instinct on reading the diagram, and seeing that it was labeled as a low-pass filter was to stand up and shout “INCORRECT!” and, on thinking that over, I decided to ask why I was confused, rather than continue to be confused or to look like an ass on the forum for calling you out.
ETA: Would you still want a resistor in series with the pot to provide a minimum input impedance and protect the circuit?
An RC filter looks kind of like a voltage divider with the cap in place of one of the resistors: cap and resistor in series, input at one end, ground at the other, output in the middle. As a ludicrous but useful oversimplification you can think of the cap as a frequency dependent resistor: High resistance at low frequency, low resistance at high frequency. Then when the cap is on top of the “voltage divider” and the resistor is on the bottom, at high frequency you get no attenuation and at low frequency you get big attenuation — high pass filter. Vice versa when the cap is on the bottom.
Here I think it’s correct to say you do have signal → cap → resistor → ground, but what would normally be the RC filter output is connected to the op amp - input and the vactrol connects to the top. If you compare the voltages at the - input and the top, the former is high pass filtered relative to the latter, but the former is also forced by the op amp to be the same as the + input, and that means the op amp output is low pass filtered relative to the input.
With a 100k pot the input impedance is 100k regardless of where the wiper is. (Not so with an inverting op amp configuration but it is here.) On the other hand it is true that if the wiper is at the CW end the op amp input is directly exposed to the CV input, and if somehow someone managed to connect maybe 24 V there it could fry the op amp. Unlikely, but adding a 1k resistor either before the pot or between the pot and op amp would limit current in that case and might make the difference between frying and not. I would probably not bother, since I don’t know how I’d put 24 V on a patch cable even if I wanted to, but it couldn’t hurt really. And if you put a diode to ground on that op amp input to protect the vactrol LED, you’d definitely need that resistor (before the diode). Putting the resistor before the pot will impose a minimum attenuation on the CV, but if it’s 1k and a 100k pot it’ll only attenuate by 1%. Putting it after the pot won’t attenuate it at all.
