P-FET Gate protection diode. - Page 1
Forum section: electronics / beginners

author: Samariton
author_rank: Newbie
timestamp: June 14, 2024, 12:33:46 pm
content: I feel a little stupid for asking this but analogue electronics is one of my weaknesses and I am attempting to rectify that. I have uses a P-FET for reverse polarity protection and I have included a gate protection diode but something has been bugging me. I know that the zener will clamp the Vgs to it's voltage but why doesn't it clamp the source voltage to its zener voltage?I have attached an image of a circuit not dissimilar to my own (hopefully it has attached) and my understanding is that when a zener is reverse biased with the appropriate voltage (10V and the example) it will start conducting. So why is Vload 30V and not 10V?Edit: I was ignoring the resistor. I knew it would.be something obvious I was overlooking. Thanks to those that answered.

author: chilternview
author_rank: Frequent Contributor
timestamp: June 14, 2024, 12:43:34 pm
content: Because the internal diode conducts.You've wired it the wrong way round, the source of the PMOS FET should be connected to the most positive voltage (your input) and the drain to the more negative (your load). Then the internal diode is reverse biassed.

author: ArdWar
author_rank: Frequent Contributor
timestamp: June 14, 2024, 12:47:19 pm
content: Are you confused with forward operation or reverse one?The trick here is to pick zener with voltages much higher than PMOS VGS(TH)so the PMOS will be fully on before any clamping action comes into play. But it's practically moot anyway because during forward operation the source will follow input voltage with minimal drop due to PMOS body diode, even without gate drive. Zener is there just to prevent gate from getting "pulled down too far".

author: Samariton
author_rank: Newbie
timestamp: June 14, 2024, 12:49:31 pm
content: It's not the wrong way around. The FET is protecting against reverse polarity. It conducts through the body diode which raises the source voltage to 30V and, because the gate is 0V, the fet turns on. In reverse polarity, the FET is off. My question is to do with why the voltage across the load is 30V and not 10V as my understanding is that the Diode would be conducting at voltages above 10V. Just to clarify, the voltage is supposed to be 30V, I just don't understand why it is 30V and not being clamped by the zener diode.

author: Samariton
author_rank: Newbie
timestamp: June 14, 2024, 12:51:58 pm
content: Quote from: ArdWar on June 14, 2024, 12:47:19 pmUnder normal operation, IE correct polarity, the source voltage is 30V and the gate is 0V but I'm not understanding why the zener isn't clamping the voltage at 10V. In this condition it shouldn't but I don't follow why that is the case.

author: ArdWar
author_rank: Frequent Contributor
timestamp: June 14, 2024, 01:01:30 pm
content: Quote from: Samariton on June 14, 2024, 12:49:31 pmThat's because zener is connected in series with a high resistance. So instead of clamping output voltage with respect to ground, it clamp its anode with respect to output voltage.Note that zener simply "try" to clamp its cathode to be VZvolt above its anode. It does that by conducting more current when its terminals get more than the breakdown voltage. In practice that means either cathode getting "dragged down" or anode getting "dragged up, whichever is easier (or blows up when neither are easy).Here in your circuit, the current loop for zener is basically Vin -> PFET -> Zener -> resistor -> GND. When Vin (and Vout) gets above zener voltage, zener will conducts more current. That current causes voltage drop to develop in the resistor that lowers the voltage drop across zener until it balances out.Try drawing zener with its idealized equivalent circuit, it should be obvious.

author: voltsandjolts
author_rank: Unknown
timestamp: June 14, 2024, 01:04:51 pm
content: The Zener has two pins. It will do it's best to clamp the voltage across it's two pins at 10V. While clamping, current flows through it, which then must flow through the series resistor, creating a voltage across that resistor.

author: Samariton
author_rank: Newbie
timestamp: June 14, 2024, 01:08:23 pm
content: Quote from: ArdWar on June 14, 2024, 01:01:30 pmThanks, I will do that!

author: Doctorandus_P
author_rank: Super Contributor
timestamp: June 15, 2024, 03:20:44 am
content: 1. You have a power supply of 30V.2. The body diode of the FET conducts, so you will have 29.4V on the source.3. The zener conducts, so you have 19.4 volt on the gate.4. Because of the gate voltge the FET opens fully.5. So now there is 10V on the source, 20V on the gate and 20V/100k = 200uA through the resistor.