3
\$\begingroup\$

I found a couple of old bridge rectifiers. After reading some docs and tutorials about reverse polarity protection decided to give them a try. The problem is nobody offered a solution against the voltage drop (and the power loss) after the rectifier, which is usually mounted at the load side not at the supply side.

What is the way to avoid this drop - using higher voltage as input or additional circuit to overcome this at the load side?

Note that I chose this way of protection so the protected device will continue to work

\$\endgroup\$
0

4 Answers 4

Reset to default
6
\$\begingroup\$

If you want to avoid the drop, you need a different device that doesn't use as diodes for current control. Diodes have a drop, the best you can do is switch to a diode that has a lower voltage drop, usually 0.2 is as good as it gets.

Mosfets have low resistances, you can either deisgn your own active rectifier by matching mosfets, or buy an active rectifier IC. Another good option is to use a gate driver IC like the LT4350

enter image description here
Source: Storing the charge from a MOSFET Bridge Rectifier

\$\endgroup\$
5
  • 2
    \$\begingroup\$ Should we mention what happens if you place a capacitor between +OUT and -OUT to OP? But then again, it appears OP isn't using it as a full bridge rectifier. Just as a "make it work regardless of how I connect my DC power supply" \$\endgroup\$
    – Harry Svensson
    Commented Jun 19, 2019 at 21:07
  • \$\begingroup\$ Can you explain what will happen? \$\endgroup\$
    – 1000Gbps
    Commented Jun 19, 2019 at 21:35
  • 1
    \$\begingroup\$ Reverse polarity in most cases wouldn't need a filter cap. \$\endgroup\$
    – Voltage Spike ♦
    Commented Jun 19, 2019 at 21:47
  • 3
    \$\begingroup\$ @1000Gbps It's only a problem if your voltage source contains AC. See here if it's still not obvious. laptop2d's design will only work if your source is like a battery or some other fairly steady output, like a buck converter. If you use his design where the source is a transformer then you can watch as the mosfets burn up, assuming you have a capacitor between +OUT and -OUT. \$\endgroup\$
    – Harry Svensson
    Commented Jun 19, 2019 at 22:07
  • 1
    \$\begingroup\$ You may wish to add gate voltage (Vgs) limiting or comment on the need.. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jun 20, 2019 at 11:41
5
\$\begingroup\$

What about this simple solution from www.ti.com/lit/an/slva139/slva139.pdf

enter image description here

You should put a zener and a large resistor to protect your MOSFET if the load voltage is larger than Vgs max:

enter image description here

enter image description here

\$\endgroup\$
5
  • \$\begingroup\$ The idea is to always power-on the device even with reversed polarity \$\endgroup\$
    – 1000Gbps
    Commented Jun 19, 2019 at 21:34
  • 3
    \$\begingroup\$ Then please edit your question to reflect that -- reverse polarity operation is different from reverse polarity protection. \$\endgroup\$
    – TimWescott
    Commented Jun 20, 2019 at 15:23
  • \$\begingroup\$ Protection against reversed polarity doesn't mean that we should stop powering on the device when wrong polarity is used \$\endgroup\$
    – 1000Gbps
    Commented Jun 21, 2019 at 10:24
  • 2
    \$\begingroup\$ @1000Gbps Protection against reversed polarity doesn't mean that we should power on the device when wrong polarity is used either. - If I say that I'm going to the fridge to grab some milk, does this mean that I am also going to grab some juice? They are not mutually exclusive, meaning they have nothing to do with each other. I can grab milk without grabbing juice. Or grab milk and juice. - You can have protection against reversed polarity and operate during reversed polarity, or you can have protection against reversed polarity and not operate during reversed polarity. \$\endgroup\$
    – Harry Svensson
    Commented Jun 26, 2019 at 16:52
  • \$\begingroup\$ Yes, but we want to have the both - always power on and protected without using UPS \$\endgroup\$
    – 1000Gbps
    Commented Jun 26, 2019 at 17:05
2
\$\begingroup\$

If all you want to do is avoid damage you can use a single Schottkey diode in the power lead (as opposed to ground) and end up with less drop. The board won't work if power is applied in reverse, but it won't go up in smoke.

One place I worked used a fuse, and a diode on the other side of it that would crowbar the supply to one diode drop in reverse. If the board was connected backward the fuse would blow -- so technically it would be "broken", but it would be an easy fix. You need to take care if you do this -- you need to size the diode so that it doesn't get damaged before the fuse blows, and you need to make sure the system isn't going to be damaged by the brief short across the power leads.

\$\endgroup\$
-1
\$\begingroup\$

For reverse polarity protection there are several different circuits based on MOSFETs.

There are also dedicated ICs to control the MOSFETs appropriately.

Those solutions are often called "ideal diode" circuits. If you search for that term you will easily find everything from circuit diagrams to ICs ("ideal diode controllers") to complete modules you can buy.

\$\endgroup\$
3
  • \$\begingroup\$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review \$\endgroup\$
    – RoyC
    Commented Jun 21, 2019 at 6:48
  • \$\begingroup\$ @RoyC I disagree. The OP is asking for a diode-equivalent circuit. There are several different circuits and/or ICs which are commonly called "ideal diode". That search term will yield a lot of options; you just have to know that you want to search for "ideal diode" to find them. \$\endgroup\$
    – JimmyB
    Commented Jun 21, 2019 at 8:47
  • \$\begingroup\$ @RoyC Added a little more explanation. I still don't see the need to include a host of circuit diagrams for the different possible solutions when a simple search for the term given will yield all the available options from which the OP can then choose. \$\endgroup\$
    – JimmyB
    Commented Jun 21, 2019 at 8:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.