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How is it possible to modify this reverse polarity protection circuit, to switch off when more than 5.5V will be applied. I was thinking about voltage divider, and zener diodes, or voltage comparator linked to gate of MOSFET, but I'm not sure which way is better one. I have DDS generator which requires 5V , 100mA and I'd like to avoid using voltage regulator due to voltage drop. So I could power it up from 5V and in the same way it will switch off if reverse polarity connected or more than 5V was provided.

enter image description here

Thank you

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One way to do it is as shown below.

D1 provides reverse voltage protection and U2 is used to generate the shutdown signal for Q1.

In use, VIN is set to zero volts, R1 is set to maximum resistance, and the load (R5) is connected.

Then, VIN is increased until VOUT rises to the desired trip point voltage.

R1 is now decreased until VOUT falls abruptly to zero volts and, finally, VIN is rocked back and forth around the trip point while monitoring VOUT, just to make sure there are no problems.

The plot was made using LTspice, setting R1 to 12.7k, R2 to 1k, and varying VIN between -6 and +6 volts.

The image can be viewed full-screen by left-clicking the image, and the LTspice circuit list follows the plot just in case you want to play with the circuit.

enter image description here

Version 4
SHEET 1 1252 680
WIRE 192 0 160 0
WIRE 224 0 192 0
WIRE 336 0 288 0
WIRE 496 0 336 0
WIRE 608 0 496 0
WIRE 768 0 704 0
WIRE 800 0 768 0
WIRE 336 96 336 0
WIRE 496 128 496 0
WIRE 528 128 496 128
WIRE 656 128 656 64
WIRE 656 128 608 128
WIRE 656 176 656 128
WIRE 800 176 800 0
WIRE 160 208 160 0
WIRE 496 240 496 128
WIRE 336 288 336 176
WIRE 416 288 336 288
WIRE 656 304 656 256
WIRE 656 304 576 304
WIRE 336 336 336 288
WIRE 160 464 160 288
WIRE 336 464 336 416
WIRE 336 464 160 464
WIRE 496 464 496 416
WIRE 496 464 336 464
WIRE 800 464 800 256
WIRE 800 464 496 464
WIRE 160 528 160 464
FLAG 160 528 0
FLAG 192 0 VIN
FLAG 768 0 VOUT
SYMBOL res 784 160 R0
SYMATTR InstName R5
SYMATTR Value 50
SYMBOL voltage 160 192 R0
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value PULSE(-6 6 0 1 1 .1)
SYMBOL res 320 80 R0
WINDOW 0 -54 33 Left 2
WINDOW 3 -71 65 Left 2
SYMATTR InstName R1
SYMATTR Value 12.7k
SYMBOL res 320 320 R0
WINDOW 0 -49 39 Left 2
WINDOW 3 -57 67 Left 2
SYMATTR InstName R2
SYMATTR Value 1.0k
SYMBOL schottky 224 16 R270
WINDOW 0 32 32 VTop 2
WINDOW 3 0 32 VBottom 2
SYMATTR InstName D1
SYMATTR Value 1N5817
SYMATTR Description Diode
SYMATTR Type diode
SYMBOL pnp 704 64 M270
WINDOW 0 50 62 VLeft 2
WINDOW 3 77 90 VLeft 2
SYMATTR InstName Q1
SYMATTR Value 2N2907
SYMBOL res 640 160 R0
SYMATTR InstName R4
SYMATTR Value 1k
SYMBOL res 624 112 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R3
SYMATTR Value 10k
SYMBOL Comparators\\LT6703-3 496 304 R0
SYMATTR InstName U2
TEXT 184 488 Left 2 !.tran 2.1
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  • \$\begingroup\$ Is there any reason why you replaced the PMOS shown in the original diagram with a Schottky (D1) for reverse voltage protection? I imagine for high-current applications, D1 would dissipate a lot of heat (and generate a voltage drop). \$\endgroup\$ – tangrs Oct 5 '16 at 22:49
  • \$\begingroup\$ @ tangrs I replaced diode because my question was how to create polarity protection and over-voltage block, but all the sudden people started to focus not on the question how to implement task, but on this diode. So I replaced it. \$\endgroup\$ – Dmitrij Kirikil Oct 6 '16 at 9:27
  • \$\begingroup\$ @tangrs: The reason is that I couldn't find any PMOS in my LTspice library that worked, so I went for the jellybean 2N/PN2907 and 1N5817 in order to provide a working concept. If you can come up with a working PMOS solution I'd be delighted to see it. \$\endgroup\$ – EM Fields Oct 6 '16 at 11:05
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You can not. The body diode will conduct. You need another transistor for that.

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  • 3
    \$\begingroup\$ @DmitrijKirikil he's not talking about a fly-back diode. He's talking about the intrinsic diode in the body of the MOSFET. \$\endgroup\$ – tangrs Oct 5 '16 at 8:57
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    \$\begingroup\$ @DmitrijKirikil you get a body diode whether you want it or not. \$\endgroup\$ – Steve G Oct 5 '16 at 9:01
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    \$\begingroup\$ @DmitrijKirikil we know it works. The reason it works is because of the body diode (it's even mentioned in the video) as Wossname mentioned. You will need a separate over-voltage protection circuit connected after the MOSFET if you want to achieve what you want. \$\endgroup\$ – tangrs Oct 5 '16 at 9:12
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    \$\begingroup\$ @DmitrijKirikil It won't work because even if you managed to switch off the MOSFET, the internal body diode will still conduct. Have a look at the schematic in the video you linked. Notice how there is a diode going from drain to source? That diode will conduct regardless of whether the MOSFET is on or off. \$\endgroup\$ – tangrs Oct 5 '16 at 9:20
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    \$\begingroup\$ You can easily try this out with your existing circuit. Connect the gate directly to the source, this will turn off the mosfet. Now check if this has turned off the voltage to the load. It will not, because the body diode still conducts. The load will get 4.3V because the body diode drops 0.7V. \$\endgroup\$ – Steve G Oct 5 '16 at 9:23

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