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I have too much heat in a rectifier so there is a considerable efficiency loss. I've been searching about it and I've found it could be because forward voltage or even because high frequency AC input.

The context:

  • There is a small 3-phase generator providing 30Vac, 60A (max) and 1700Hz. I use this rectifier: http://ixapps.ixys.com/DataSheet/VUO52-16NO1.pdf
  • I use a 2200uF Cap for the ripple in the outside. After this I connect a DC/DC converter to set the DC voltage.

My question is if 1700Hz could be responsible of considerable losses for this normal kind of rectifier.

I see that it will be too much current for this rectifier so I am using far from the full power in the generator and still heats too much. I want to find another rectifier or even buy 6 diodes, and I don't know if in my case I must search for fast diodes or directly for bridge rectifiers. If the most important will be lower forward voltage or if fast recovery is needed also. This diode could be enough?: VS-150EBU02

Thanks in advance for any kind of help

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  • \$\begingroup\$ Is your DC/DC converter a switching converter? What's its output voltage? I'm thinking along the lines of voltage conversion first, then rectification. \$\endgroup\$ – Michael Gorsich Jun 2 '17 at 14:40
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    \$\begingroup\$ How much heat is too much? 60A at two diode drops (say 2V) is about 120W worst-case. Are you seeing more than this? Less? \$\endgroup\$ – marcelm Jun 2 '17 at 14:50
  • \$\begingroup\$ Michael, I think so. The dc-dc converter is a non-isolated buck- boost regulator, which employs synchronous rectification \$\endgroup\$ – Víctor Jun 2 '17 at 15:20
  • \$\begingroup\$ Marcelm, I didn't measure the heating energy, but it seems like too much because i fitted a big heatsink (around 100x70x50mm) and it is burning my hand in one minute. \$\endgroup\$ – Víctor Jun 2 '17 at 15:24
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The reverse recovery of your diodes does influence the power dissipation in your application, but in this case is extremely unlikely to be the major cause of power/efficiency loss. You could start by comparing a fast rectifier to you current rectifier. The VUO52-16NO1 and a much faster VUE130-12NO7 are a good comparison.

While you gain lower switching losses in the VUE130- the power loss is actually dominated by the difference in Vf in your application (which I assume is a car type alternator). The VUO52- has a Vf of only 1.4 V for a total bridge loss of about 160 W at 60 A, and the VUE130- has a Vf of 2.7 V for a total loss of about 320 W at 60 A. The reverse recovery losses for the VUO52- is unlikely to be more than about 16 W. Because you have a large output filter capacitor the diodes go into reverse recovery just after the peak of each phase voltage, but If has already dropped to close to zero dropping the stored charge considerably. Yes, there will be a reverse current flow, but insignificant compared to If max. This might be worthwhile reading for you.

You could use fast and low Vf Schottky diodes to reduce the voltage losses. For example the APT60S20B or the VS-100BGQ100 (more appropriate because it can be screwed down) would more than half your power loss.

The best way to reduce both switching and forward power loss would be to use a half or full synchronous rectifier of course. More complex electronics, but power losses in the rectifier can be reduced to just a few watts with fully synchronous.

SMART and Synch Rectifiers

If you look up SMART rectifier controls (such as this) you will see ways to use FETs as ideal rectifiers. There are lots of these controllers but be careful that you understand that some of these devices only operate at low frequencies, so may not be suitable.

One simple way to implement synchronous rectification is to sense the current in the lower diode of each pair and use a FET on the upper diode. This is simple and reduces your power loss in half. You could implement this with the 3 phase rectifier you already have.

Then you can get creative and build your own microprocessor solution. However if you aren't at that level you could consider using an open source ESC controller as a synchronous rectifier ....here is a demo of that on YouTube. Here the user is simply using the ESC32 as-is, but you could make it automatically track speed by altering the firmware. Again, depends on you capabilities.

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  • \$\begingroup\$ Thanks Jack. The generator is 3-phase permanent magnets RC motor. I have low voltage and high current, so yes, bridge rectifier has big losses because the forward voltage in the diodes. Ok, perhaps with VS-100BGQ100 both losses can be reduced: 0'82 Vf (at 100A) and 45ns reverse recovery time. On the other hand I though that synchronous rectifiers are still in development. I don't know if I can find some comercial one for my features. I see that for example Texas Instruments has some controller for doing it with MOSFET, but Im not sure if it's a down-to-earth for me. \$\endgroup\$ – Víctor Jun 3 '17 at 12:24
  • \$\begingroup\$ The reverse recovery losses are insignificant, you will not change your total power consumption by more than a percent. Choose any Schottky diode with low Vf and ignore the Trr. The VS-100BGQ100 I recommended seems your best bet, which gets you to about 100W losses at 60 A. Synch rectification is well established and understood, but it is usually complex. I have seen some very creative ways to implement....I'll look for references and post to the answer. \$\endgroup\$ – Jack Creasey Jun 3 '17 at 16:45
  • \$\begingroup\$ Jack, I have searching about the issues you wrote. I will start trying with VS-100BGQ100 schottky diodes, as I must do the system now and I can manage it. In parallel I am going to continue searching about synchronous rectification and also about ESC, for improve efficiency in the system in the next step. Thanks for your help! \$\endgroup\$ – Víctor Jun 6 '17 at 10:58
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My question is if 1700Hz could be responsible of considerable losses for this normal kind of rectifier

Absolutely, if you look at a standard 1N400x diode it has a reverse recovery time of 30 us - this means that when the diode goes from forward conduction to supposedly reverse blocking, it actually takes 30 us: -

enter image description here

Some 1N400x manufacturers are better than others of course and some data sheets don't even specify it.

So, it's acting as a short for 30 us every half-cycle of the power frequency. Given that 1700 Hz has a cycle time of 588 us, a 1N400x rectifier would be wanting to short the supply (straight after the zero crossing point) for 10% of the half-cycle time. So, 10% of 180 degrees is 18 degrees and your sin wave will have reached 30.9 % of its peak negative voltage value and of course, the diode will still be trying to short out that supply.

Your rectifier does not quote reverse recovery time but it does say it's intended for 50/60 Hz applications and this usually means it has a poor reverse recovery time and I would bet it's in the tens of microseconds just like the 1N400x diode.

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  • \$\begingroup\$ Thanks por the explanation Andy. If the losses increase by 10% it means I should use fast diodes for sure! Yes, the datasheet of the rectifier doesn't talk about reverse recovery time. For the moment I don't find bridge rectifiers with this information, so I will check fast diodes. \$\endgroup\$ – Víctor Jun 2 '17 at 15:32
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    \$\begingroup\$ The losses don't increase by 10%, the losses increase massively compared to using a diode with RR time less than 1 us. You have to remember that at 10% into the half cycle waveform, the voltage is sin(18 degrees) = 30.9% of the peak voltage and this, is still largely being shorted out. At 1 us RR time, the voltage is about 1% of Vpeak and much less of a problem if shorted. \$\endgroup\$ – Andy aka Jun 2 '17 at 15:43
  • \$\begingroup\$ Ok Andy. I understand and it convinces me to try to solve it with fast diodes. If I solve the problem I will back to you. Thanks a lot \$\endgroup\$ – Víctor Jun 2 '17 at 16:07

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