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I have a barrel jack powered design that I'm making, but I would like to prevent damage in the case of reverse polarity (sleeve is positive instead of negative) and overvoltage (plugging in a 12V adapter into a device intended for 3.3V). My design will only consume 500 mA worst case, but to clarify, the wall wart I've chosen can provide up to 1.5 A of current at 3.3V.

I'm using a P-channel MOSFET in addition to a PTC fuse and zener diode. Input Power Jack Design

Will this work? Will the zener diode blow in a high overvoltage case due to a slow PTC fuse? Is my selection for a MOSFET correct?

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    \$\begingroup\$ BC2721-ND is a Digikey part number, not the manufacturer part number. The manufacturer part number is PTCCL07H411DBE \$\endgroup\$
    – DKNguyen
    Jun 24, 2019 at 17:22
  • \$\begingroup\$ @DKNguyen Thanks for noticing, I fixed it. \$\endgroup\$
    – dylanweber
    Jun 24, 2019 at 17:28
  • \$\begingroup\$ DKN no matter.... it is still searchable, yet dylan could have included the datasheet. \$\endgroup\$
    – Tony Stewart EE75
    Jun 24, 2019 at 18:08

3 Answers 3

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The simplest way to implement reverse polarity protection is to put a diode in series with the PTC fuse. If the voltage on pin 1 is negative with respect to ground, the diode will no conduct, and the circuitry downstream won't see the negative voltage. Make sure the reverse voltage rating of the diode is large enough to handle any negative voltages you might expect to see on the input.

During normal operation, the diode will drop some voltage, but you can find low forward voltage diodes, Schottkys for instance, that can get down to the 300mV range. If you can handle the voltage drop, this is probably the easiest solution for reverse voltage protection.

As for overvoltage protection, as Sunnyskyguy said, changing the input voltage to a higher voltage (5-12V), and regulating it down to 3.3V is probably your best bet. I know you said you don't want to add in a separate power supply, but any overvoltage protection circuitry is probably going to be as complicated as adding an additional voltage regulator. There are plenty of other benefits to using an on board regulated supply, rather than relying on a wall wart to generate your 3.3V rail. You can have a large input voltage range, you are less sensitive to series voltage drops (from the PTC, from a diode, from the MOSFET, ...), the input current is less (assuming a switching regulator), etc.

Are you locked in to using a 3.3V wall wart, or is it possible to use a higher input voltage and step it down on board?

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  • \$\begingroup\$ Since you wanted feedback on the circuit you provided, I don't think it will do what you intended very well. Q2 looks like it doesn't do very much - it will turn on and stay on as long as the input is greater than Vg_thr. I assume the other part of the circuit is supposed to short Q3 to GND during an over voltage event, and blow the fuse. That will happen when Vin > Vbe(Q3) + Vz(D2), both of which are not well defined, but is approx. 0.7V + 3.6V = 4.3V. In addition, your circuit will experience still experience overvoltage for a brief period of time before the fuse blows. \$\endgroup\$
    – newothegreat
    Jun 24, 2019 at 19:35
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I think your Zener is has a 50/50 chance of frying since the PTC has a room temperature resistance of 3 ohms which means the initial current will be about (12V-3.6V)/3R = 2.8A which is way beyond what the Zener is listed as being able to handle. It would be more certain if pulse characteristics were provided for the Zener but they are not so we I can only compare it against the steady state ratings.

Your PMOS is fine because the important thing here is the maximum Vgs which is 20V and you only ever plan to apply 12V max.

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  • \$\begingroup\$ $V_{gs}$ in this instance cannot exceed the zener voltage, so I don't think I need to worry about maximum $V_{gs}$ \$\endgroup\$
    – dylanweber
    Jun 24, 2019 at 17:29
  • \$\begingroup\$ I suppose that's true. \$\endgroup\$
    – DKNguyen
    Jun 24, 2019 at 17:30
  • \$\begingroup\$ I changed the schematic to accommodate for the power across the zener diode. Will this work better? \$\endgroup\$
    – dylanweber
    Jun 24, 2019 at 17:50
  • \$\begingroup\$ That seems like the right track but there are a few issues. The last graph on page 3 shows the trip time. 410mA-615mA is the trip current and at x2 it takes 10sec to trip. At 4x it takes 1sec. 8x it is 0.3sec. There needs to be series base resistor or else the B-E will fry. And that BJT still neither handle a lower current for the longer trip time nor the higher currents for a shorter trip time. \$\endgroup\$
    – DKNguyen
    Jun 24, 2019 at 19:29
  • \$\begingroup\$ Also note that if Vin is too low (but still high enough to be dangerous), a 0.7V drop may not be maintainable across B-E to turn the BJT on since not enough current goes through R3. So maybe reduce the zener breakdown by 0.7V to accomodate. This isn't without its own issues since the series base resistor voltage drop gets added with Vzener and Vbe and will increase the voltage at which the clamp turns on but not as badly as before (the minimum Vin where clamping occurs is that which has enough V be leftover to appear across Rbase to push enough base current to fully turn on the BJT. \$\endgroup\$
    – DKNguyen
    Jun 24, 2019 at 19:42
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No linear design with/without reverse protection will be efficient with linear power dumped =Pd=(12V-3.3)*0.5A =3.85W with 3.3V x 0.5A =Pout =1.67W which what Zener must dump when no other load.

It is better to use a source at 5V or a 12V to 3.3V DC-DC converter. Your concept could be made to work inefficiently, but why waste heat, needing heatsinks and bigger current/power dissipating parts when a better choice is possible.

i.e. a FET to dump almost 4W with full load with heatsink

  • a shunt voltage regulator to dump 1.85W with no load with heatsink.
    • vs a 3 terminal DC-DC regulator to replace everything except J2 and Polyfuse for reverse protection or a diode.
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  • \$\begingroup\$ I changed the schematic to accommodate for the power across the zener diode. Will this work better, or should I pick a 5W zener diode instead? \$\endgroup\$
    – dylanweber
    Jun 24, 2019 at 17:51
  • \$\begingroup\$ Neither will work better than what I answered. \$\endgroup\$
    – Tony Stewart EE75
    Jun 24, 2019 at 18:06
  • \$\begingroup\$ Consider this choice that matches my solution digikey.com/product-detail/en/murata-power-solutions-inc/… \$\endgroup\$
    – Tony Stewart EE75
    Jun 24, 2019 at 18:09
  • \$\begingroup\$ I'm not looking to integrate a power supply into my product, only to implement overvoltage and reverse voltage protection. That device will not help me if there is more than 36V applied to the input, and won't help me if reverse voltage is applied. A 5V power adapter wouldn't work with my product if I used this, too. \$\endgroup\$
    – dylanweber
    Jun 24, 2019 at 18:16
  • \$\begingroup\$ @dylanweber a 5V supply reduces the losses in your 0.5A 3.3V requirement or a integrated DC–DC regulator is also more efficient. But your question is about OVP, rev. Input and yet causes OTP issues due to inefficient linear operation. So my suggestion works with a diode or PTC but that does not suit your unstated requirements Like this DC–DC 3 terminal part. Yet it fails your unstated criteria if >36V , DC–DC BUY solution and your design fails the OTP issues demanding heatsink for FET and BJT. Then just insert a 5W resistor in front of any suitable LDO \$\endgroup\$
    – Tony Stewart EE75
    Jun 24, 2019 at 21:05

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