Check my FET switch design

[saturn]

Hi all,

I'm doing a design to check the capacitance of a capacitor... And here is a working design that I used some time ago...

However now my new project need to check 2 capacitors. I think I can change to a PIC that have 2 comparator. But I want to leave some flexibility as a provision for futher as I might need to check 3 or 4 capacitors later. So I have came out with the following idea:

Using a FET as a switch to connect one capacitor to the checking port at a time... Please refer to second schematic. Now my question, will this work?

Thanks and best regards.

 

[VVV]

Re: FET Design question

It can work if the C+ connection is connected to a positive voltage (+5V, for example) high enough such that the FETs can be turned off. Can the measurement system tolerate that?

If not, can you use actual relays? Those would allow you to use any measurement system.

 

[saturn]

FET Design question

Hello,

The C+ pin will be driven to +5V by GP5 (charging process), and it will be 0V when GP5 driven to low (discharge process). AN1 will always in input mode...

Well I see this is what you have told me... so I will guess I need a relay anyway.

And I see I'm not quite comfortable with FET... any tutorial on how I can use FET as switch, amplifier etc... My Google search yields some tutorials with in interior diagram/structure of a FET etc, but what I need is some design examples...

Thanks.

 

[VVV]

Re: FET Design question

Here is a typical FET switch. When the bipolar is turned on, +5V is being applied to the cathode of the diode and so the diode is off. R1 shorts the G and S of the FET, so the FET will be on and the signal can pass through it. With the bipolar off, the gate of the FET is pulled to -12V and so the FET is forced off, the signal not being able to pass through it.

To ensure the switch operates properly, there are restrictions on the input voltage range: the FET must stay on if the bipolar is on. So the maximum input voltage is about 5.6V. Anything higher and the gate appears negative to the source and begins to turn off the FET. Also, the FET must stay off when the bipolar is off. That means that the input voltage cannot go so negative that the FET can turn on. The 4393 has a turnoff voltage of about -3V, but other FETs can have voltages of -10V. So, to guarantee the fet stays off, the source cannot go to less than -12+3V=-9V, for the 2N4393, or lower than -6V for a 2N4392. So the input voltage can vary between about +/-5V without any problem.