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PFET gates remain charged

tdsi

Sep 10, 2013
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I have a PFET circuit that is on when no voltage is applied and off when a PVI driving the gate is on. After the PVI turns off the PFET and the PVI then goes off, the PFET takes a long time to turn back on (if it does at all). There is a diode preventing feedback current going to the PVI. How can I discharge the gate voltage so that I have a fast turn-on/turn-off transition repetitively?

Thanks!
Jonathan
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Show us your circuit.

The general answer is that you need to provide both a charge and discharge path to the gate.

Unlike a bipolar transistor, just removing the voltage source from the gate of a mosfet will not turn it off, you need to actively (or passively) discharge it.

I am assuming that by "PFET" you mean P-Channel mosfet.
 

tdsi

Sep 10, 2013
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Thanks! I am referring to a P Channel mosfet. I was told that I can't put a pull down resistor (to the negative of the PVI) because it creates a voltage divider. I can't seem to figure out how to discharge it.
 

(*steve*)

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Please read the first line of my previous reply.
 

tdsi

Sep 10, 2013
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Sorry about that. Missed it. I have attached it as a pdf. Wasn't sure the best format to send it in.
 

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tdsi

Sep 10, 2013
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The circuit spans over two boards. everything from diode D15 to the right is on one board and everything else is on the other board. Space is very limited on the Mosfet board.
 

(*steve*)

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Your circuit seems to be very wrong. There seems to be no way to turn the mosfets on, and the diodes protecting the gates appear to short out your voltage source anyway.

What turns the mosfets on?

What are you trying to achieve?

A more typical circuit can be found in a datasheet for a similar device.

If this is not what you want to do, you will need to provide more explanation.
 

tdsi

Sep 10, 2013
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The circuit I am emulating is the normally closed contact of a form C relay. The normally open side is connected similarly to the one in the datasheet. The diodes work the opposite as for the N Channel FET. When the voltage is applied to the CM3 signal, as long as coil+ is off, the FET conducts because the diode reduces the voltage by 10Vdc. As soon as the coil+ goes high, the voltage increases turning off the FET. That is how the circuit works. The problem is once off, I can't get it to turn back on.
 

tdsi

Sep 10, 2013
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unless I momentarily touch my voltmeter probe to the gate or source of the FET. Then it turns back on right away.
 

(*steve*)

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You can't do "normally closed" with an SSR as power is required to ensure the mosfet remains in the ON state.

Your mosfets may be turning on due to stray voltages. You have not shown enough of your circuit for me to tell.

If you have external power you could make a logic level when asserted turn off the mosfets, but the total absence of power should leave them turned off.

(OK, if you had depletion mode mosfets it would work the other way, but you're not likely to find any in the voltage/current ratings you'd need)
 

tdsi

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This really is the whole circuit. 22-44 Vdc on the CM3 and a load on NC3 (or the other way around). The diodes between the gate and source are there for protection but also they ensure that the Vgs is around 10Vdc which should keep the mosfets on according to the datasheet. Unfortunately I don't have external power. That's why I am trying to make this work.
 

tdsi

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the coil- is connected to the source of the NO side of the circuit (similar to the datasheet). That side is working great. What I meant is that this is the whole circuit that is not working the way I intended.
 

tdsi

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Here is the complete circuit.
 

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(*steve*)

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Aha. Now we can see what's wrong.

You will note that the switching circuit on the left is similar to what I suggested (however it may not switch off very fast). The gates are biased against the the common source connections.

The circuit on the right has the gates biased against what?

A *major* problem is that you have unwittingly connected the NC and the NO connection together. It is only D15 that keeps them isolated. And D15 prevents you from doing anything effective in any case.

The first modification is to change the circuit so that one half of U1 is used for one channel, and the other half is used for the other channel.

This also means you can remove diodes, resistors and capacitors that you have presumably added to stop weird behaviour and shorts.

Once you do this, the left channel will continue to operate correctly, and you need to find a current source so you can power the NC mosfets when the input signal is not present.

If you know something about the load you're connecting this to then perhaps you can steal some power from that.
 

KrisBlueNZ

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Nov 28, 2011
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A few more comments in addition to what Steve has just explained.

The normally closed circuit is the problem. MOSFETs, whether P-channel or N-channel, require a gate-source voltage to be applied to them to make them conduct. The difference between them is just the polarity of that voltage; without applied voltage, neither type will conduct.

As Steve mentioned in an earlier post, depletion-mode MOSFETs do conduct by default, but they're not available with any significant power-handling capability.

So you cannot (at least, I don't know of any way to) make a drop-in replacement for an electromechanical relay that has two input connections and a normally closed output and nothing else. You need to get access to some kind of power source to provide bias to the MOSFETs in the normally closed path.

You can provide a power source at the input side, if input power is available, and transfer it to the MOSFETs for the normally closed path using the same method you are using for the normally open side. You would need a separate photovoltaic coupler, because (a) one needs to be energised when the other one isn't, and (b) the commoned sources of the MOSFETs in the NC and NO paths must be electrically separated. You would also need to arrange the input circuit so that the isolator for the NC circuit has a power source that's active when there is no voltage across the inputs that activate the isolator for the NO circuit.

In this case you could (and should) use N-channel MOSFETs for both paths.

Another alternative is to use voltage from the output side to activate the MOSFETs in the NC circuit. If the output is switching an AC circuit, it might work out better to use an SCR or triac. It will not conduct for the complete cycle, though. If you use an SCR or triac, you could control it with a simple optocoupler, which could be driven from the same inputs as the photovoltaic isolator; when the optocoupler is energised, it would short out the gate-cathode voltage for the SCR or triac and turn it OFF.

We could give you much more specific and helpful advice if you showed us how this relay replacement device will be used. In other words, a really COMPLETE circuit, including a description of the devices that need to connect on both sides of it.
 
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