P - Channel Switching - Page 1
Forum section: electronics / beginners

author: TheDood
author_rank: Regular Contributor
timestamp: November 22, 2019, 09:15:22 am
content: I'm trying to understand P - Channel switching.Is it possible to use a P - Channel enhancement FET in the middle of a cct? If I apply 5V to the Base of the NPN transistor, will the load before the FET (R1) act on how the FET is switched? How about after (R2)? With N - Channel FETs, if there's a load after the FET, the Gate V will have to be greater than the total cct VCC (VCC + Vgs). I'm not sure if I'm comprehending P - Channel, it seems operation is less dependent on where the FET is placed? As long as the P - Channel FET Gate is tied to GND, the P - Channel FET should be turned on?This is just a mock schematic to gain a principle understanding.

author: Dave
author_rank: Super Contributor
timestamp: November 22, 2019, 09:22:18 am
content: Quote from: TheDood on November 22, 2019, 09:15:22 amNo. It's the exact same story as with an N-channel transistor, except the voltages are reversed. You need to establish a high enough gate-source voltage to turn the transistor on (the voltage being negative with a P-channel transistor). If you're putting the load between the power rail and the transistor, you're effectively reducing the VGSwhen the transistor begins to conduct, partially closing it.Your schematic is drawn incorrectly, by the way. The arrow inside the transistor symbol should point towards the source terminal, not drain.

author: Peabody
author_rank: Super Contributor
timestamp: November 22, 2019, 05:00:10 pm
content: As Dave said, setting the gate to ground will turn the P-channel on ONLY if the voltage at the source pin is high enough to produce a Vgs high enough to turn on the mosfet. In your drawing, having the load at R1 would reduce the source voltage, which would tend to shut down the mosfet even if the gate is at ground.Typically, P-channels are used to switch the load's connection to the positive rail, and N-channels are used to switch the load's connection to ground. That way the P-channel's source will always be at the positive rail, and the N-channel's source will always be at ground.

author: TheDood
author_rank: Regular Contributor
timestamp: November 23, 2019, 03:36:25 am
content: Thanks guys, attached is the reference schematic I was modeling after (except I added a 2nd load), taken from here:https://youtu.be/vRiF_GghBY8Is there a way to switch FETs placed between loads without using charge pumps (are these what charge pumps are used for)?

author: TheDood
author_rank: Regular Contributor
timestamp: November 23, 2019, 04:05:50 am
content: Could you switch string B, utilizing a different component or IC, in place of FET M2, so that you didn't need to use a charge pump to offset the voltage (am I comprehending charge pump use correctly)? I appreciate the help guys, thanks.

author: pigrew
author_rank: Frequent Contributor
timestamp: November 23, 2019, 04:49:37 am
content: Quote from: TheDood on November 23, 2019, 03:36:25 amThe schematic looks almost right, except one must be very careful not to apply an overvoltage to the gate, and it won't work for your string C. Most MOSFET are limited to about 15 V from gate to source, and only fully turn "on" with Vgs <-6 V (depending on model...). The BJT current source needs to more accurately drive the current being placed across the resistor. One way to be to add an emitter resistor on the BJT.It will not be possible to turn off the C string since one would need -6 V to apply the appropriate bias to the gate of the PMOS and you don't have a negative supply. This is where charge pumps come in.... They can be used to create voltage above or below the power supply rails you have, so you could would have a negative supply.... Then again, it'd be better to just use nFETS (they generally have better characteristics than the pFETs do).If you try to turn off string C with your pFET by applying Vg=0V=GND, then the source will go to approx the threshold voltage of the transistor (~4V), and the transistor will heat up since it has non-zero voltage and current simultaneously. The LEDs will be greatly dimmed, and if your FET has a big enough heatsink, it'll should be OK but inefficient. Using PNP BJTs would be much better in this regard since the V_BE would only be ~0.9 V instead of the Vt of the MOSFET.