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I have made a bench power supply that can erogate 0-30 volt and 0-2 A. Now, how can i protect my bench power supply from reversed voltage applied to it's output?

Typical p or n mosfet circuit dosent work in this circumstance-

I am thinking to create a separate circuit with a separate floating ground to sense a reverse voltage applied to the output of the power supply, but i am looking for an easier way to do it!

A diode, in antiparallel, and a fuse would do the job, but i am looking for something more reliable!

Any suggestion?

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  • \$\begingroup\$ Is "my bench power supply" the new supply that you made, or is it what you're using to provide the input power to your new-made supply circuit? \$\endgroup\$ – The Photon Mar 13 '18 at 20:54
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    \$\begingroup\$ A big Diode across the output and a Fuse is one way. If the output is connected to a reverse voltage source the fuse will blow but it should protect the rest of the supply. \$\endgroup\$ – Warren Hill Mar 13 '18 at 20:55
  • \$\begingroup\$ The bench power supply should also have a current limit functionality, but we take that as an axiom (otherwise it's not called bench power supply). \$\endgroup\$ – Laszlo Valko Mar 13 '18 at 20:57
  • \$\begingroup\$ Along with reverse voltage protection, what about over voltage from an external source? Back-EMF, etc. \$\endgroup\$ – Sparky256 Mar 13 '18 at 21:00
  • \$\begingroup\$ @Sparky256 i have setted some protection to prevent damage from external overvoltage applied to the output of the bench power supply. The only protection missing is for the reverse voltage! Laszlo Valko : yes i have designed protection for overcurrent! Warren Hill Hi! Your solution will work! But... I dont like it very much :) do you know other methods? The Photon yes, i am referring alway to the same bench power supply ( designed by me ) thank you all for the answer!!!!!! \$\endgroup\$ – Lorenzo Rosin Mar 13 '18 at 21:19
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I think your floating ground approach may be a nice simple way, as it lends itself to being an external add-on to any power supply, and especially battery chargers.

This arrangement doesn't draw any current from the battery except when there is a negative input < -0.6V. So 2 or 3 tiny lithium coin cells, and it will run for ever.

schematic

simulate this circuit – Schematic created using CircuitLab

When the output goes below the negative lead by 0.6V, Q1 turns on, and turns M1 off. R1, R2 values affect how abrupt the switch off is.

R2 will set the load current when off.

Fet should have very low R on, as it is outside the feedback loop.

The battery voltage needed depends on VGS.sat of the FET.

I have not tested this. (It looks like a great use for old PC motherboards, which have power fets, and lithium coil cells and holders)

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This is how I retrofitted an old bench supply that I need to tolerate reversed batteries connected to the output (without forcing current through them or having fuses/circuit breakers pop).

A fet switch optocoupler controlled the drive to the output stage.

schematic

simulate this circuit – Schematic created using CircuitLab

Normally D1 is energised, and the output stage gets drive. When the output goes below 0, D2,D3 take the current from opto D1, disconnecting the drive to the output stage. D2 lights during reverse protection. D3 protects D2 from 30V- LED's don't have high reverse voltage rating. D4 can be 0-2 diodes to set the voltage around 0, where it cuts off.

You need to watch for other components that might be affected. C1 is the usual electrolytic capacitor across the output. Adding C2 makes it bipolar, so -ve does not destroy it. However electros are not like diodes. They do not conduct reversed. On my supply C1 is a 63V electro. A 12V battery can be put reversed across the power supply. The 63V electro does not mind -12V :I did not need C2.

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  • \$\begingroup\$ You, i like you! Thank you very much for sharing your circuit. I will put an addictional transformer and i will design a circuit similar to yours, with some amp op, to detect even small reverse voltage. Thank you veru much! \$\endgroup\$ – Lorenzo Rosin Mar 14 '18 at 18:57
  • \$\begingroup\$ There is no particular need to have a super hard cutoff at 0. The power supply is current limited, and if it delivers 1A at 0V, it will deliver 1A at -1V. You just have to ensure that the feedback input and drive to output stage remain in linear range when output is negative. In the case above, the output voltage can go down to -1.2V (Q1+Q2) before the feedback amplifier output hits 0V. So ensuring cutoff by -1V is sufficient. \$\endgroup\$ – Henry Crun Mar 14 '18 at 19:22
  • \$\begingroup\$ The actual use case I know of is connecting a reversed battery. There are two functions: a) prevent damage to the psu b) if the battery is left reversed, prevent it being charged backwards. \$\endgroup\$ – Henry Crun Mar 14 '18 at 19:24
  • \$\begingroup\$ I have added in my design a component that can't tollerate voltage below -0.3V, so i will design a super aggressive cut off circuit :-) . This power supply bench will be use to play with battery, big inductor, voltage booster and other dangerous circuit for a non-protected bench power supply. Some times it will be used for some evaluation board. Thanks again my friend! \$\endgroup\$ – Lorenzo Rosin Mar 14 '18 at 19:34
  • \$\begingroup\$ What component can't survive -0.3V? That sounds like something is lacking a series resistor for protection. Remember a psu should never have an IC pin (e.g. opamp input) directly connected to the output \$\endgroup\$ – Henry Crun Mar 14 '18 at 19:37

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