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MPF102 as a switch with open and positive gate states?

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neazoi

neazoi

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Hi I want to use the mpf102 as a switch in a circuit and I would like to know,
if I leave the gate open will it be ON (closed switch)?
If I pot a positive voltage on the gate (5v or so) will it be switched OFF (open switch)?
 

No and no.
Look at its datasheet to see that if the source pin is 0V and the gate is -8V then it does not conduct and if the gate is the same voltage as the source pin then it conducts a little current.
 

Gate open unpredictable behavior as its a function of gate leakage, which
can enhance or turn off itself.

Fwd bias of gate will destroy this JFET. To turn it off large negative Vgs required.
Max of -8V or more negative needed.

1625445009042.png


Regards, Dana.
 

    neazoi

    neazoi

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I agree that the gate should never be disconnected.

The curves shown above with a current of 10mA when the Vgs is 0V is for a "typical' one. But some are only 2mA (minimum) and others are 20mA (maximum).
 

    neazoi

    neazoi

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Thanks a lot!
--- Updated ---

Is there any way I can do it with a single device?
I attach you the schematic.

I have designed this AF oscillator which varies it's frequency with a variable resistor.
I am building a small keyboard and it is shown in the schematic with keys and multiple variable resistors (one for each key).

Now I would like to do the next:
When no key is pressed, I want a specific tone to output from the oscillator.
But when any of the keys are pressed, this specific tone will be replaced by the tone of the key pressed.
When the key is depressed, the previous specific tone will be present again.

I thought it could be done with a single device but maybe a pair of 2n2222 (double inversion) could do it.

How can I do this function that I need?
 

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But when any of the keys are pressed, this specific tone will be replaced by the tone of the key pressed.
When the key is depressed, the previous specific tone will be present again.
Click to expand...
Are you asking for some kind of memory function? If the previously selected tone is required when the key is released you will need some method for it to remember the previous state.

Brian.
 

betwixt said:
Are you asking for some kind of memory function? If the previously selected tone is required when the key is released you will need some method for it to remember the previous state.

Brian.
Click to expand...
Not the previously SELECTED tone, it is a specific tone, so no memory required, just a way to switch between a key tone and a specific tone (eg 1500Hz).
Here is an idea that might work or might not.
An extra NPN is switched on when any of the keys are pressed. This causes the second NPN to switch off, i.e. disconnects the associated variable resistor from the oscillator. Notice the polarity of this second NPN.
So when a key is pressed the tone that this key represents is present.
When the key is depressed, the key tone disabled and the pilot tone (1500Hz) is present.

Would that work?
I just thought that it could possibly be done with a single device (PNP) or not?
 

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The problem with using a simple transistor is it would need to operate in saturation to be an effective and consistent switch. To do that you need to inject base current and doing so would raise the emitter voltage and stop the rest of the circuit working.

There is no simple solution but I think I would try replacing the 2.2K resistor with a pair of silicon diodes. The voltage across the 4.7uF capacitor is DC at approximately 1.1V. If you use two forward biased small signal diodes it will stabilize at around 1.2V but be able to sink more current through the switching transistor without it rising significantly. You might still have problems with creating the bias voltage for the switch transistors but you haven't given any indication of what drives them.

Brian.
 

    neazoi

    neazoi

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betwixt said:
The problem with using a simple transistor is it would need to operate in saturation to be an effective and consistent switch. To do that you need to inject base current and doing so would raise the emitter voltage and stop the rest of the circuit working.

There is no simple solution but I think I would try replacing the 2.2K resistor with a pair of silicon diodes. The voltage across the 4.7uF capacitor is DC at approximately 1.1V. If you use two forward biased small signal diodes it will stabilize at around 1.2V but be able to sink more current through the switching transistor without it rising significantly. You might still have problems with creating the bias voltage for the switch transistors but you haven't given any indication of what drives them.

Brian.
Click to expand...

If I find that the loading caused by the first switching transistor causes the oscillator to stop working or the frequency to be shifted in strange ways, I could then replace this transistor with a mosfet like the 2n7000. Wouldn't this work?
 

Difficult to be sure but you could try. However, the frequency is mostly decided by the 82nF capacitor and the resistance to ground formed by the potentiometers. Unfortunately, in that design they also change the DC levels and that's what stops it working.

You could try isolating the AC and DC paths by connecting the base of the first transistor through a resistor of about 5.6K to the top of the 4.7uF capacitor then adding a new capacitor of about 10uF to its base with the resistor network then going to ground on its far side. That keeps the DC bias happy and at the same time makes the switches go to ground where it's easier to control them.

Brian.
 

    neazoi

    neazoi

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