1
\$\begingroup\$

Situation: I'm trying to implement a reverse polarity circuit using a p-channel mosfet and a voltage divider using a resistor + schottsky diode. More information here if you need a refresher.

The key point here is that the MOSFET is reversed. The current flows from Drain to Source, first through the internal diode but after turning on through the normal MOSFET path.

I've been reading various datasheets of various MOSFET's and they all seem to indicate that the maximum Vdrain-source voltage is negative. If I took this at face value it means that a MOSFET can't conduct in both directions, which I know to be false. Example.

Can I ignore this negative number and assume it's a maximum rating both ways or is there more at play here?

Edit: the actual circuit taken from Hackaday

\$\endgroup\$
1
\$\begingroup\$

FETs ideally don't care about the direction of the current. You simply have to invert your conventions and switch Source and Drain in your equations. Everything works exactly the same way.

The only reason that power FETs do care is because there is a Source-Body connection put in place during device fabrication. This creates a parasitic diode that conducts when the voltage is reversed. However, if your drain-source voltage is below the diode voltage, the device will still behave as a FET.

If you have access to the fourth, body, terminal. You can bias it externally so that the device always operates as a FET.

\$\endgroup\$
0
\$\begingroup\$

If you apply a positive drain-source voltage for P-channel FET the internal diode will start to conduct. Thus the positive voltage rating is irrelevant. The voltage over the diode depends on the current as it does for any diode.

The Vds voltage is as written, from drain to source. Meaning that at -10V, the drain is at 10 V lower voltage than source.

\$\endgroup\$
  • \$\begingroup\$ When this circuit starts conducting over the parallel diode, it turns on the FET itself. At this point there is a positive voltage over the Drain-Source, which most FETs I've looked at aren't rated for in the datasheets. The FET will conduct more current than the diode because of its low resistance. Will it blow up? \$\endgroup\$ – Tryphon Nov 20 '18 at 0:31
  • \$\begingroup\$ No it will not blow up. It is perfectly fine for current to flow either way through the channel of the MOSFET when it is on. The Rds and current limits are the same regardless of current direction (assuming FET is on). The positive voltage you are worried about is less than the maximum allowable diode forward voltage, so you know the voltage is OK (won't hurt anything). \$\endgroup\$ – mkeith Nov 20 '18 at 5:21
  • \$\begingroup\$ Actually, current through the diode will not turn on the FET. Just the diode conducts. In this situation V_ds is 0.7 V, the threshold of a diode. If you turn on the FET by applying negative V_gs, you can see that V_ds drops. \$\endgroup\$ – TemeV Nov 20 '18 at 6:46
0
\$\begingroup\$

Seems like you are on the right track. Current can definitely flow both ways through the channel of a power MOSFET when it is on. Not much voltage will be developed between drain and source this case (unless the current is very high). It is just Id*Rds(on).

When the P-channel MOSFET is off, and current is flowing from drain to source, you just need to respect the ratings of the body diode (which should be listed in the datasheet).

You can't use the same MOSFET for reverse-polarity and over-voltage protection, generally. If the FET is oriented so that it can cut off over-voltage current, then it will be "pointing the wrong way" to cut off reverse current. Over-voltage current and reverse voltage current flow in opposite directions, so two MOSFET's will be needed. [edited to make the point more clearly, hopefully!]

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.