will this work?

Hey all;

I'm trying to design a circuit that will ground a signal when the control voltage is applied, like this:

signal

o | | | o | ||-+ ||

Reply to
tempus fugit
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Yes.

A jfet is the correct one to use. And yes a J111 will need negative volts on the gate to turn off. I suggest that if you only have postive volts available then you use a p-channel jfet like one from the J175 family. Choose according to your available cut-off voltage.

Graham

Reply to
Eeyore

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Thanks again for the reply Graham (one of many of my posts of late!). Why would a JFET be better to use? I have successfully done this with a 2n2222, but it doesn't quite kill the whole signal (you can still hear it a bit). I was looking at a MOSFET like BS270 because it has a lower Ron than a BJT, and of course for the low turn on current.

Reply to
tempus fugit

The mosfet will work, 0 or +5 on the gate for off/on, but be aware that in the fet OFF state, the D-S substrate diode will conduct if the drain tries to swing any more than a few tenths of a volt negative. Jfets don't have this problem.

You could use a CMOS analog switch, 74HC4066 or something. Make a tee switch for really good on/off ratios.

Opto solid-state relays can be interesting, too.

John

Reply to
John Larkin

You *can* do it with a bipolar device but as soon as the signal amplitude exceeds ~ 400mV the transistor will conduct in the reverse direction. Also, as you've noticed, it doesn't clamp the signal totally. Neither will a jfet totally since it'll simply become a small resistance. The very best technique is to have a series and shunt fet. I'd choose a series fet alone for best results but it requires being a bit more 'clever' with driving the gate since there's signal on it.

I've never used a mosfet for this application and since no pro-audio company I know does, I suspect there may be a good reason for that but I've never considered the reason why. Asymmetric characteristics strikes me as possible.

Graham

Reply to
Eeyore

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Hmmm.....

Your point is well taken Graham. Maybe I'll have to try one and see if there are any problems. The signal is not actually going to pass through the MOSFET. I plan to connect both output (from guitar) and input (to amp) to the drain. SO when the MOSFET is off, there will be a direct connection from in to out, but when it is on, both will be grounded.

Thanks again

Reply to
tempus fugit

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Just had a thought....

What type of material is used in an analog switch (like

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I have been messing around with one of these, and it works nicely. I had originally planned to use this device, but the way I am switching has changed, so I can go to a simpler (and smaller) device.

I'm just wondering if it uses MOSFETs for switching.

Reply to
tempus fugit

negative

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Thanks for the reply John. I wonder what the output voltage of a guitar pickup is..... So a MOSFET should be fine for switching DC if the voltage is positive then?

THanks

Reply to
tempus fugit

You might also add a second N-channel MOSFET in series with the first, but with its source and drain reversed. You connect the gates together. This puts a second body diode in series with the first that prevents either from conducting when the gate is low. When the gate is high, both channels have low resistance. The limitation on negative off state signal voltage is then that which begins to turn the reversed device on with zero gate voltage and negative source voltage. But even with a logic level MOSFET the turn on threshold is normally above a volt. A big improvement on the .3 or .4 volts it takes to turn the body diode on. And that turn on threshold is added to the body diode turn on voltage before significant current passes.

Reply to
John Popelish

Typically around 1V peaks as an absolute maximum for passive guitars, larger for actives. Average voltage is much lower, maybe 0.1V - hard to say because it depends on a lot of factors.

nifty

Reply to
niftydog

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