# Main power input section P-CH MOSFET connection confirmation

In our Design we are using simple P-Ch MOSFET connection to protect reverse current flow direction.

Gate connected to ground so Source to drain current flow will happen through Protection diode as well. If its reverse current flow Protection diode will be reverse bias.

Q1. Is it in reverse current flow Source to drain reverse current will allow because Gate always connected to ground?

Q2. Is this MOSFET connection will blow the device once reached High spike voltage and current?

Regards, RK

In your configuration, when the power is applied, MOSFET cannot be turned on immediately, because there's no voltage at the Source pin. So, current will flow first through the body diode until the voltage at the Source pin reaches the voltage level that turns the MOSFET on (You can see the forward-biased diode in your schematic). Once the MOSFET has turned on, it will short that diode and the current will flow through $R_{DS-on}$ in normal operation.

1) Of course. But if you don't connect the MOSFET that way, load current will always flow through body diode and the reverse polarity protection cannot be achieved.

2) Of course. But besides that, $V_{GS}$ is a deadly limitation for MOSFET. If supply voltage is 24V then the MOSFET will fail, because $V_{GS-max}$ is 12V for your MOSFET. A zener protection is required here.

Another thing which must be considered is power dissipation: $P_M = I_D^2 \cdot R_{DS-on}$. So the load current is important here. Your MOSFET has an on-resistance of $R_{DS-on} = 15m\Omega$ and it's in a SOT1220 package. And if the load current is 5A then $P_M = 5^2 \cdot 0.015 = 0.3W$. Thermal resistance, $R_{th-ja}$, of that package with proper copper-pad for drain is 67°C/W, so temperature rise will be around 20°C which causes no problems. But if the load current is 7A then temperature rise will be around 50°C -causing your MOSFET to get as hot as 75°C under room temperature!

Of course reverse polarity protection can be achieved with P-Ch. MOSFETs, but I personally recommend N-Ch MOSFETs because they've higher performance and lower price compared to P-Ch ones. Here's a configuration I'm using:

simulate this circuit – Schematic created using CircuitLab

1. The current indeed flows in reverse through the MOSFET when the circuit is in normal operation. It has to be connected the way you show, otherwise the body diode will conduct when you want it to be blocking.

2. If you exceed the rated blocking voltage (12V) or forward current the MOSFET can fail. Or if you allow it to get too hot. You can allow higher negative voltages or higher forward currents by using a MOSFET rated for higher voltage or current (though it will likely cost more or have higher losses).

There is another failure mode. The gate-to-source voltage is rated at only +/-12V so if the input goes below ground by more than 12V you will be exceeding the absolute maximum Vgs and the gate oxide can fail. A zener and resistor can help prevent this.

Breakdown from Drain to source will not necessarily be destructive if the energy and current are limited, but gate-source breakdown will ruin the MOSFET unless it has internal protection zeners.

simulate this circuit – Schematic created using CircuitLab

If you have no fear a high voltage will be connected to the input, you may still wish to plop a cap on the input such as 1uF/10V to help protect against ESD.

• My input voltage is 5V/4Amps. Could you please provide the calculation to find Zener and Resistor values for my design? Commented Feb 3, 2017 at 11:45
• If the voltage never exceeds +/-Vgs(max) you don't need the Zener or the resistor. Otherwise it depends on the MOSFET rating and the maximum expected input voltage. In the case of your schematic above the MOSFET is rated at 12V max, so you can use a 6.2V zener and 1K to be safe in most situations. Commented Feb 3, 2017 at 12:29

The mosfet rev pol protect is fundamentally valid .I have always used a gate zener and resister .Spehro states this .I have used N chan like Rohat said .N chan all things being equal will have less RDS on .There is still one way for the FET to blow up :Imagine that there is large Bus capacitance C2 storing lots of energy and having low ESR and short heavy cable going to say a large lead acid battery .If there was a very sudden volt drop due to a large load being applied elsewhere like say a large invertor.Your proposed circuit will allow current to backfeed through the conducting FET.This current could blow the FET .If you monitor DS volts by whatever means you can shut the gate down if current tries to flow backwards .I have done this with a transistor and a diode .I have seen others use a comparetor .The potential to blow the fet is probably worse with the modern SMD packages .In the old days we were using TO220 packages and sometimes to247 to get the RDs on low enough to resoundingly beat a low volt shottkey like an old phillips BYV143-40 .