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I'm trying to speck out a heat sink for a 3 phase rectifier. The rectifier consists of 3 individual rectifier blocks. I need a heat sink for each block.

The rectifier is for a generator and all the specs I have to work with is that the DC output will be 468VDC at 112ADC.

I will use 3 individual heat sinks. How do I calculate the power dissipated per individual block?

enter image description here

Individual block.

schematic

The best i've been able to come up with is that at 112A a diode will drop about 2v which will give a power dissapation of 224W but that seems too high because that is peak conduction and will only spend a small amount of time there.

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    \$\begingroup\$ 224W is not all that high when your system is allegedly delivering over 50kW - its less than 0.5% \$\endgroup\$
    – brhans
    Jul 18, 2016 at 15:22
  • \$\begingroup\$ For a quick on-line calculator, Semikron has a tool : semikron.com/service-support/semisel-simulation.html \$\endgroup\$
    – Marla
    Jul 18, 2016 at 15:34
  • \$\begingroup\$ Your positive leg diodes will share the current between them so the power will be \$ I_F \times V_F \$ where \$ V_F \$ is the average forward voltage of the diode that is on (and there's always one on). Don't forget to multiply by two for the diodes in the negative leg. (It's not clear from your post whether you have done that in the 2 V drop allowance.) \$\endgroup\$
    – Transistor
    Jul 18, 2016 at 15:38
  • \$\begingroup\$ @transistor I only accounted for a single diode in the 2V drop. \$\endgroup\$
    – vini_i
    Jul 18, 2016 at 17:12

3 Answers 3

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schematic

simulate this circuit – Schematic created using CircuitLab

I ran a simulation.

enter image description here

The simulation implies that no two diodes in the same block (indicated by the dashed boxes) conduct at the same time. Instead they take turns conducting the full DC current in an almost 50/50 raster. This would further imply that at any time a block would not dissipate more than the losses of a single diode.

In this way each block behaves as though it were dissipating 224 W continuously.

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With single phase, the calculation is a bit iffy, because the reservoir capacitor makes the rectifiers draw very high peak currents.

With three phase like you have, the situation is much easier, as with no reservoir capacitor, the current is ripply rather than very peaky, and is actually calculable.

The total dissipation will be evenly spread out between the three diode sets. The total will be a little more than 224W, as the RMS value of the current (the heating effect) is still higher than the average value, but with 3 phase only slightly higher. I guess 10 to 20% should be plenty of tolerance, though if somebody wants to do the sums properly and comment, I'll change the figure to the correct one.

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As posted in the comment to your OP, your positive leg diodes will share the current between them so the power will be IF×VF where VF is the average forward voltage of the diode that is on.

... a diode will drop about 2v which will give a power dissipation of 224 W but that seems too high because that is peak conduction and will only spend a small amount of time there.

You haven't shown any capacitance in your load and very often it isn't required in a poly-phase rectifier because one of the phases is always "up". This means that each diode conducts pretty much a constant current with constant voltage drop (your 2 V) for 120° of the mains cycle.

enter image description here

_Figure 1. The APTDF400AK120G I/V curve shows that at somewhat over 100 A that \$ V_F \$ is about 2 V as you stated and decreases with temperature to a minimum of 1.25 V. Your choice of 2 V is safe.

So power for the positive-side diodes will be \$ V_FI = 2 \times 112 = 224~W \$ and the same for the negative side diodes giving 448 W total.

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    \$\begingroup\$ As the diodes warm up the foward volts will drop to maybe 1.5V which will reduce your expected power losses .Using lots of diodes in paralell will reduce the current per diode and hence terminal voltage and the total thermal resistance .This will reduce heatsink needs and may be cost effective these days. \$\endgroup\$
    – Autistic
    Jul 18, 2016 at 21:13

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