LTSpice Question about Zener Diodes

[Puffie40]

Hi,
I'm trying to simulate a 48VDC P-MOSFET reverse polarity protection circuit in LTSpice, and I am having some trouble getting the Zener Diode section down to the right voltage.
The transistor I am using is a IRF9540N using the PSPICE library from Vishay.  I am using a 48V AC source to simulate the negative polarity in the simulation.

the trouble I am having is getting the gate voltage down to around 20-18VDC as specified by the datasheet.  Maybe it is because I forgot how Zener diodes work, but I have to add a second resistor in front of the diode to get the voltage below 20 volts, effectively having a voltage divider before the zener diode will work.  Otherwise, I am getting 30-32 volts at the gate.

Am I missing something?  The UFZ20B in this circuit has a breakdown voltage of 19.8V, so I sort of expected that would be the resulting voltage on the anode side of the zener diode.

[ledtester]

The resistor R3 will add to the gate-source voltage as any current which goes through the zener will also have to go through that resistor.

A quick simulation of just the zener shows that the zener current can reach in excess of 10mA which then adds another 6.8V to the gate-source voltage.

The use of R1 is sufficient to limit current through the zener, e.g.:



(from https://www.ti.com/lit/an/snva717/snva717.pdf )

[Konkedout]

It does not make sense to use a ~20V zener.  Generally there is no need to drive Vgs over 10V.  20V is the ABS MAX to avoid destroying the MOSFET.

I would choose a zener rated for 12, 13, 15, or 16V.  (with low zener current your actual Vgs might be a bit less than the rated zener voltage.  Also the zener in real hardware will have a voltage tolerance.)

[Puffie40]

Here is another attempt with just the Zener diode.  I selected a 11v zener diode, calculated the current so I had a margin under the zener's datasheet current of 20mA,  but I still have a VGS of 36VDC.

[Zero999]

Here is another attempt with just the Zener diode.  I selected a 11v zener diode, calculated the current so I had a margin under the zener's datasheet current of 20mA,  but I still have a VGS of 36VDC.

No you don't. You're misinterpreting the simulation results. You have a gate voltage of 36V.

Hopefully you can figure out what V_GS is. Note to others, please give the OP a chance at working it out for themself, before giving the answer.

EDIT: Here's a clue: you've plotted the gate voltage, relative to 0V, when you want the voltage relative to the source.

[Puffie40]

So you are saying Vgs on the datasheet is referring to the voltage difference between the Gate and source?  If I move the ground reference over to the output net, I measure net VGS at 11 volts, which is the Zener voltage of the KDZV11B.  Which makes sense.

I guess that would be my fallacy in this simulation, as a lot of example transistor circuits show the source pin connected to ground.

so therefore, a voltage of 35volts in relation to ground is to be expected at net VGS then,  Is that correct?

[Zero999]

So you are saying Vgs on the datasheet is referring to the voltage difference between the Gate and source?

Of course, the only voltages the data sheet can specify are relative to the gate, source and drain. The whole component can float at any voltage with respect to the circuit's 0V node.

If I move the ground reference over to the output net, I measure net VGS at 11 volts, which is the Zener voltage of the KDZV11B.  Which makes sense.

To be pedantic, the gate voltage is at -11V, with respect to the source.

I guess that would be my fallacy in this simulation, as a lot of example transistor circuits show the source pin connected to ground.

That can be changed. After running the simulation, right click a wire connected to the transistor's source and click "Mark Reference", a probe symbol will then appear on said wire, then any node you click on will plot the voltage relative to the source.

so therefore, a voltage of 35volts in relation to ground is to be expected at net VGS then,  Is that correct?

I'm not sure where you got that voltage from. If V_S = 48V and V_GS = -11V, then V_G should be 37V.

[Puffie40]

I'm not sure where you got that voltage from. If V_S = 48V and V_GS = -11V, then V_G should be 37V.

It might be in relation to the 10 amp load resistor I have in the circuit.  When I remove it, I do get closer to 37 volts.

[Zero999]

I'm not sure where you got that voltage from. If V_S = 48V and V_GS = -11V, then V_G should be 37V.

It might be in relation to the 10 amp load resistor I have in the circuit.  When I remove it, I do get closer to 37 volts.

Then it's because you're dropping around 2V between the source and drain, which is to be expected given the on resistance is 0.2 Ohm. You need to use a MOSFET with a lower on resistance, or you might as well use a plain diode.
https://www.irf.com/product-info/datasheets/data/irf9540.pdf