1
\$\begingroup\$

I have a 12V AC power supply that I want to convert to 12V DC using a bridge rectifier. I have LED lights that will draw up to 200 watts.

By my calculations, that should be about 17 amps (200W/12V = 16.667A). So would any bridge rectifier rated over 12V and over 17A work?

I bought the following thinking it would be more than enough but it started smoking when I connected it. I did check that it was connected to the proper terminals per specs, and verified that I was getting 12VAC from the power supply. http://www.amazon.com/KBPC50-10-1000V-Housing-Bridge-Rectifier/dp/B00GYTWCP2

\$\endgroup\$
3
\$\begingroup\$

Did you try reading the datasheet for the device? There, you'll see the forward voltage drop of diodes inside of the rectifier. In general, you need to find and read the datasheet for every electrical component you're using.

Drop times current is (sort of, more or less) power that needs to be dissipated. We also have maximum operating temperature. So armed with that information, a cooling system needs to be devised to work with the rectifier.

So if we look at real numbers, we have maximum drop of 1,2 V per element. At one time, we have two diodes conducting, so that's 2.4 V times 17 A giving us a bit less than 40 W in the rectifier alone.

Next, we have operating temperature of the semiconductor junction. It's from -65 to +150 degrees Celsius.

So what we have next is the procedure how to get the temperature increase above ambient for the unit. So let's look first at the thermal resistance between the junction and the case. It's 1.5 K/W. One kelvin and one degree Celsius are of the same magnitude, so we can get the temperature increase this way.

So at 40 W, our junction will be at temperature 60 K higher than the temperature of the case. Looking back at operating temperatures, this gives us maximum case temperature of 90 degrees Celsius.

Now we need to estimate the thermal resistance of the heatsink we need to select for our use. Let's say that our ambient temperature is 25 C. This gives us maximum temperature difference of 90-25=65 C between the heatsink and the surrounding air. So now we do the backward calculation of what we've had in the previous step. We already have the power of 40 W and 65 C, so we divide them and get thermal resistance of 1,625 K/W for the heatsink. Lower would be better, to give you more room to work with higher ambient temperatures and to run the rectifier cooler.

We now know what sort of heatsink we need. Some of the basic tutorials for heatsinks are available here and here. They are a bit more "academic". This one from Sparkfun sees a bit more "applied". Using those resources, you "just" need to find a suitable heatsink for the rectifier.

In real life, you'll probably need a big hunk of metal with lots of ribs and a big fan blowing air on it to just keep the rectifier cool. In my personal opinion, it would be a better idea to look for alternative sources of DC current that would be a bit more efficient.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks for the helpful information. I have a followup question regarding the 12V power supply. It converts from 120V AC input to 12V AC output. If my 200W LED lights are drawing 17 amps on 12V, does that mean the power supply is drawing 1.7 amps from the 120V side of the power supply? \$\endgroup\$ – confusedGFCI Feb 20 '16 at 22:14
  • \$\begingroup\$ @confusedGFCI Glad to be of assistance. For your question, the answer would be sort of. Usually, we use the Power_in=Power_out for transformers, but in reality that's not the case, since the transformer would introduce some losses of its own. I myself would do my calculations with transformer using around 2 A on the high-voltage side, as a sort of safety margin. \$\endgroup\$ – AndrejaKo Feb 20 '16 at 22:44
1
\$\begingroup\$

If you have some old scrapped atx power supplies, you may salvage 40Amps 3 pins dual schottky diodes in to247 packages.

Get 4 of these, wire each of the dual diodes as a single diode by shorting out two outer pins. Now built your bridge rectifier using these to247.

These packages are easy to mount on heatsinks with very low chip-to-sink thermal resistance.

Very cheap and super reliable high power rectifier. This is what I have done for my 12 volt 30Amp battery charger PSU.

\$\endgroup\$
0
\$\begingroup\$

Did you mount that bridge rectifier to a suitable heatsink? The metal case is meant to provide a good thermal connection to a heatsink.

Any bridge rectifier carrying over 5 Amps will dissipate significant power, and must be mounted on a suitable heatsink to remove the resulting heat. In your case, the bridge will probably drop 2 volts - at 17 amps, that's 34 watts.

\$\endgroup\$
1
  • \$\begingroup\$ Paralelling two of them (both mounted to a heat sink) could help spread that heat out some, but this is more or less a hack. \$\endgroup\$ – rdtsc Feb 18 '16 at 20:32
0
\$\begingroup\$

To answer your question: Yes any diode bridge with the rating you posted would work, with the caveat of the voltage drop on a diode.

Explanation for understanding: We don't have perfect diodes in the world. So diodes have a real voltage drop which means they dissipate some power. Your diode bridge started smoking because it got too hot from the heat dissipated by the diodes. At best the diodes would drop 0.7V. Using that as a general basis you'd be generating around 0.7V*17A*2 (two diodes on at any given time) leads to about 24W of heat dissipated. There are only really two options then.

  1. Heat sink the diode bridge.
  2. Use a MOSFET bridge chip. These use mosfets and monitor the voltage input and output to turn the mosfets on and off to act like diodes. An example chip is Linear Devices' LT4320.

Which ever you chose, remember to consider all thermal parameters in the design as 17A is a lot of current and heat loss is proportional to the current squared for MOSFET on resistance. Or alternatively, you can just use the P=I*V equation again and measure the voltage drop across the MOSFETs (or diodes) to find heat loss.

\$\endgroup\$
2
  • \$\begingroup\$ Heat loss is proportional to \$I^2\$ for a resistor, not for a diode. \$\endgroup\$ – Spehro Pefhany Feb 18 '16 at 20:52
  • \$\begingroup\$ @SpehroPefhany en.wikipedia.org/wiki/Joule_heating Plus I'll adjust my answer since I was commenting more on the mosfets, which I hadn't done a calculation for, than the diodes which I calculated using V*I. \$\endgroup\$ – Dave Feb 18 '16 at 20:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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