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Reverse Polarity Protection on High Current P-Mosfet Circuit

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Z

Zaprooder

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Hi,

Thanks for reading this. Although reverse polarity protection is easily findable on searching the forum and on Google etc.. The specifics of my project makes it hard to apply most of the common results from such searches to what I want to do. In short, the circuit has a high power switching P-Channel Mosfet directly between VCC, load and Ground. There's amicrocontroller and stuff elsewhere on the circuit, but I thought a simple diode on the input to the 3.3V regulator would sort that side of things out. As for the Mosfet itself, my thinking was that when connected in reverse, the internal diode on the Mosfet would start to conduct and ruin my day. So my thinking was, if I placed a SC Diode and 100mA PTC across the load, it would serve 2 functions, it would dampen back EMF from the load (which is slightly inductive) and also when the battery is put in backwards the diode would start to conduct, the diode in the Mosfet would conduct and the current would exceed 100mA causing the PTC to trip. I was unsure if any back emf might cause a trip, but I didn't think that was likely.
**broken link removed**

I would very much appreciate any advice, pointers on mistakes I've made etc..

Thanks.
Z
 

First of all, I think those PTC fuses take a relatively long time to trip; your MOSFET could be a distant memory before the current is interrupted. You don't say what your current and voltages are...

I don't know what an "SC diode" is. Schottky? (That's what your symbol looks like).

Is there any reason you can't tolerate a diode in the MOSFET path, just like you've done for the regulator?
 

    V

    Points: 2
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Sorry, my mistake. The VCC is at 12V and the load is as you can see 1.2 ohms, although this can range from 0.5 ohms to 4.0 ohms. So the current is very high, so a diode in the mosfet path is not really an option, it'd have to be the size of a dustbin. Shame about the trip times of those PTC fuses, at least I know that now, thanks.

I think my only option is to use a second p-channel mosfet in a reverse polarity protection configuration.
 


    V

    Points: 2
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Well...... I never even knew such things were available. That would solve the whole lot. One of those on the power supply should be enough. Plus I can take out the regulator diode too, as it wouldn't be needed anymore.
You have made me happy, thanks :)

EDIT: It's pulsed, PWM source to control the average power to the load.

I could just do this:
**broken link removed**
 
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Just keep in mind, if you've got a 0.5 ohm load, that's about 24 amps. The drop across the diode is about 0.3V==> 7.2 Watts, which is not insignificant.
 

It's unlikely that a 0.5 ohm load would be driven at 100% duty cycle so the thermals would be averaged I imagine, I would think there's a thermal lag on any component before it heats up enough to cause damage? The average output power would typically be around 30-40 watts, although some crazy users may be up in the 70-80 watts area. With a 0.5 ohm load @ 40 watts, if my maths is correct, means I need an RMS voltage of
V²/R = P thus V = SQR(RP)
Around 4.47V
Meaning a duty cycle of 4.47² / 12² = 0.139 (13.9%) so the average current is 0.139 * (12/0.5) = about 3.33 Amps.
So should be fine I hope. If my maths is ok that is ;)

EDIT: My maths sucks, ignore all that should be nearly 9 amps.
 
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If you are trying to deliver 80 watts from a 12 volt supply, that's an AVERAGE current of 6.7 Amps (80/12--assuming 100% efficiency).

6.7 x 0.3 (Diode drop) is a more reasonable 2 Watts dissipated by the diode (still not insignificant. You might need a heat sink or a different package)

You didn't have to do all that other math, but you get extra credit. ;-)
 

    V

    Points: 2
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Yeah, lol. I am still thinking in micro controller code, as the controller has to do all that math to get the correct duty cycle. Also, the power is only applied to the load for a short time, maybe 4 or 5 seconds every couple of minutes or so at most. It's not constant, so it should be fine.
Thanks barry, I've got some thinking to do, some data sheets to read and some calculator bashing.

Cheers
Z
 

If you had a FET where the body terminal was separate from
the source, you could (and people do) diode-block the body.
See this in products like HV / high-Z-OV-protected analog
switches.

But this would be abnormal in the discrete power device world.

A second, series device with some sort of reverse polarity
detection and drive, might be the answer. Depending on reaction
time required the "sensing" drive might be as simple as a resistor
to ground (energizing the PMOSFET when incoming supply is up)
that gets overridden by a switch to VLOAD when it is not (i.e.
VLOAD>VIN). That might just be a PNP with E=VLOAD, C=gate
of blocking FET, w/ resistor, B=VIN (w/ some appropriate limiting
resistance).
 

    V

    Points: 2
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Hi,

Just a comment on the schematic.
With switching i reccomend to put a capacitor from FET's drain to gnd. This prevents from ripple in the line to your voltage regulator.

If it is possible, then try to use " low side switching" by connecting the load to vcc and drain, and source to gnd.
The function stays the same, but the gate driver gets more simple. Less parts, less space.

Klaus
 

    V

    Points: 2
    Helpful Answer Positive Rating
Hi,

Just a comment on the schematic.
With switching i reccomend to put a capacitor from FET's drain to gnd. This prevents from ripple in the line to your voltage regulator.

If it is possible, then try to use " low side switching" by connecting the load to vcc and drain, and source to gnd.
The function stays the same, but the gate driver gets more simple. Less parts, less space.

Klaus

Hi Klaus, thanks for replying. The schematic I posted is somewhat simplified, there are indeed capacitors to deal with ripples. I am using a mosfet gate driver IC because of the high frequency of the PWM, which is only 1 component so that's not really too much of a problem. I cannot unfortunately use low side switching for two reasons. First is that the load has a common earth (the power supply negative is directly connected to one side of the load). The other is that the micro controller has to detect the resistance of the load. This is far far easier to do when the load is on the ground rail and disconnected from VCC.
 

Hi,

ok i see,

i just thought, that driving high side FETs usually need the bootstrap circuit...

common ground.. OK. Then i´d expect more stray inductivity in the lines to load. D3 must be next to Q2 to catch those voltage peaks.

Measuring load resitance:
This is far far easier to do when the load is on the ground rail
Really? You don´t use the current and voltage with the power pwm?

Klaus
 

I do, however detecting the load resistance during driving would be too late. The user can select a power level to drive the load, in order to do that the micro controller has to know the load resistance and level of VCC (it's a battery) before operation in order to figure out what duty cycle to use. It's also required so the user cannot operate the load if the load's resistance is too low etc.. the display also has to display the load resistance and also display the predicted voltage/power/current before the load is fired.

The load could be made common to VCC I suppose. It would require some work on the enclosure as the battery negative is on the enclosure shielding as is the load. But then I get nightmares on how to detect the load resistance as it's on a variable rail. If I reverse the point of view so VCC becomes 0V and ground becomes -VCC then I'd have to be generating negative test currents to test the load, it just seems simpler to me all round if I use high side switching.
 

You might want to look at "hot swap controller" type driver
products. They may or may not exist for your voltage*current
requirements but they are somewhat popular for USB and
automotive applications. Maybe you can even find an integrated
load switch product that rolls it all up (see comments about
the usefulness of body node access, which can be had by the
IC designer).
 

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