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As far as I understand, diode bridges are mainly used to convert AC into DC, but one can also use them just to ensure an expected DC output polarity for arbitrary DC input polarity. I have some small energy devices (3V-5V, <1A) that require an expected polarity and I want to safely connect them to a power source that is likely to be used with different polarity. How do I find the right type of diode bridge and what disadvantages exist when using it? Given a safe span of input current, does the diode bridge just act like a simple resistor? If so, how high is its virtual resistance, so how much energy would I loose compared to ensuring right polarity by other means?

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    \$\begingroup\$ If you don't care about guaranteed operation and only want to protect the downstream devices, you could add a fuse in series with the small device and a diode between input and return (on the unit side of the fuse) - if the polarity is correct, the loss is minimal (just the resistance of the fuse), if the polarity is reversed, the diode conducts heavily and opens the fuse, keeping your device safe. \$\endgroup\$ – Adam Lawrence Oct 19 '12 at 14:41
  • \$\begingroup\$ @Madmanguruman: Thanks, that'll be the best solution. With the right fuse I also get overcurrent-protection for free. \$\endgroup\$ – Jakob Nov 3 '12 at 14:49
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The main problem with a diode bridge is the fact that you always have two diodes in series with your circuit, and this creates a voltage drop of about 1.4 V between the power source and the load.

The power loss is simply this voltage drop multiplied by the load current.

It also means that you cannot connect the negative side of the load, which you might ordinarily consider to be "ground", to any external ground, which might be connected to either side of the power source.

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    \$\begingroup\$ For a lower voltage drop/power loss, one could use Schottky diodes.. \$\endgroup\$ – m.Alin Oct 17 '12 at 11:23
  • \$\begingroup\$ Thanks for pointing to the main problem. I wonder how the 1.4V is calculated and how it differs with Schottky diodes. What are limitations of the latter? \$\endgroup\$ – Jakob Oct 17 '12 at 12:01
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    \$\begingroup\$ He's saying 1.4V because the assumed forward voltage of a standard diode (Vfw in a datasheet) is 0.7V. So since it passes through two of them you get 1.4V drop. Schottky diodes have a lower forward voltage drop. Really though Vfw is a function of how much current you are drawing. So if with very little current is used your voltage drop would be less, however we all talk about 0.7V as the usual drop for a diode. Mostly for convenience sake. \$\endgroup\$ – Some Hardware Guy Oct 17 '12 at 12:28
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    \$\begingroup\$ The 1.4V comes from the nominal 0.7V forward voltage drop of two diodes in series. Of course the forward voltage drop seen in a real circuit will vary depending upon the amount of forward current through the diodes. An application using Schottky diodes can expect to see a lower forward voltage drop. There can be quite a range of forward voltage drops seen with the Schottky diodes dependent upon component selection and forward current. For mA level applications you can find these diodes with Vf as low as 0.2V and a high current version in the Amp range may be as high as 0.6V or more. \$\endgroup\$ – Michael Karas Oct 17 '12 at 12:31
  • \$\begingroup\$ Ok, so now I know what (Schottky diode bridge) and which parameter (voltage drop Vf) to look for to calculate the expected efficiency. Thanks a lot! \$\endgroup\$ – Jakob Oct 17 '12 at 14:57

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