Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

24V-27A DC motor drive, thermal considerations

Status
Not open for further replies.

alexxx

Advanced Member level 4
Joined
Apr 17, 2011
Messages
1,013
Helped
273
Reputation
552
Reaction score
270
Trophy points
1,383
Location
Greece
Activity points
7,936
I need to drive a 24V/27A DC motor (650W). This motor must operate even when mains power is off, so a 24V lead-acid battery will be used. The trasnformer will be 1KW. My concern is the power dissipation on bridges, diodes and mosfets when the motor is moving. The following components seem sufficient for the job.

https://export.farnell.com/multicom...sp=All&searchView=table&rpsku=rel3:KBPC5004FP
https://export.farnell.com/infineon/ipd50p04p4l11atma1/mosfet-p-ch-40v-50a-to-252-3/dp/2443434
https://export.farnell.com/on-semic...meResp=All&searchView=table&iscrfnonsku=false

But the problem is the total power dissipation. For example the diode will have a voltage drop of 0.3V, at 25A this will be 7.5W. The mosfet will dissipate something less than 7W and the bridge more than 20W!

1. I think that a good idea is to mount the bridge on the transformer, for the best possible cooling. On the other hand maybe I could use a DC power supply at 1KW, to avoid the use of the bridge. But could a powerful motor be driven from such PSUs?
2. About the schottky diode and the mosfet, if I use convencional components with a heatsink on them, would this be enough for 7-8W?
3. Can anyone suggest a thermal protection for the motor? A simple NTC (ie at 40A) would be OK? Or something else must be used for such high dc motor currents?

Thanks in advance.
 

Hi,

1)
You ask about mounting on a transformer, but we don't see the transformer.
Usually a transformer is no heatsink. If you can use it this way depends on the thermal situation of the transformer. It's own power dissipation, it's temperature, the ambient temperature and the thermal resistance of the transformer to the ambient.
If you consider to use the iron core..then you should know that the thermal resistance of iron is way worse than aluminum.
Avoid conductive material directely touching the iron core to avoid eddy currents.

2)
You need to take care about waveform. Thus we can't verify your calculations.
For power dissipation calculation use the
* RMS current for the Mosfet
* average current for the diode
We don't see a schematic...so don't know if the schottky diode is necessaray at all....what's the use of it?

3)
We don't have enough information. Can you be sure the torque is always within specified limits? Can you be sure the heat spreading is within specified limits? If you can answer "yes" on both, then you could use a current measurement technique. A motor protection fuse with thermal trigger. A current measurement...I2t trigger...
If you are unsure: use a thermofuse directely at the motor.

Klaus
 
  • Like
Reactions: alexxx

    alexxx

    Points: 2
    Helpful Answer Positive Rating
But the problem is the total power dissipation. For example the diode will have a voltage drop of 0.3V, at 25A this will be 7.5W.
1. I think that a good idea is to mount the bridge on the transformer, for the best possible cooling.

Thanks in advance.

Absolutely not. A transformer generates significant self heating, due to copper and core losses. Use a separate heatsink.

Also... you cannot use a 30 volt Schottky with a 28 volt output (this voltage required for full battery charging).
How come, you may ask?
Schottkys at close to its rated voltage have very large reverse currents (which rapidly increases with temperature, see figures 3 and 4) which lead to early failure due to thermal runaway.
And that is before taking into account the sum of the Back-EMF and the Peak of the sinewave voltage, which you have not considered.

I'm afraid that you will have to use a normal rectifier diode.
In which case........ Its voltage will be even higher than 0.7 volt at its rated current..........actually a good rule of thumb (validated both by my own measurements and datasheet values), is to assume a diode Vf close to 1.0 volt.
 
  • Like
Reactions: alexxx

    alexxx

    Points: 2
    Helpful Answer Positive Rating
Thank you both for the detailed information. I have not yet come to a full schematic, but since I take care of the thermal issues, it won't be a problem. I have a quick diagram, you can find it attached. D1 is needed to separate the motor voltage (transformer powered or battery powered). S1 (mosfet) is also acquired by the application to cut the battery from the motor. About calculations I didn't analyse them for quickness reasons. The bridge datasheet has a graph: current Vs voltage forward drop. For the mosfet I multiplied 25A * 25A * 0.01Ohm (P=I*I*Rds on).


schmitt trigger said:
I'm afraid that you will have to use a normal rectifier diode
Very good point about the 30V it is very close to 28V. But why not use a schottky in general? Because for each 100mV voltage drop at 25A, I get 2.5W of extra power dissipation on the diode! The below schottky seems sufficient, why not use it?
https://www.tme.eu/en/details/mbr40250tg/tht-schottky-diodes/on-semiconductor/

Another solution is an industrial purpose commercial AC-DC converter of 1KW. But this will only save me from the transformer, the bridge and the very big input capacitance. The schottky and the mosfet must still be part of the power circuit.

IMG_20170419_114833.jpg
 

Very good point about the 30V it is very close to 28V. But why not use a schottky in general? Because for each 100mV voltage drop at 25A, I get 2.5W of extra power dissipation on the diode! The below schottky seems sufficient, why not use it?
https://www.tme.eu/en/details/mbr40250tg/tht-schottky-diodes/on-semiconductor/
For the simple reason that, on Schottky diodes, as the Vrrm increases, so does its Vf.

Did you check the forward voltage drop of the proposed diode? It is 0.86 volt, almost the same as a regular diode.

Granted, you will still save some power, but by far not as much as your original calculations had shown.......but by all means, go ahead and use it. It still saves some power.

There are ways to build a full-bridge Mosfet synchronous rectifier. That would provide the lowest losses, by far. Let me see if I can find the book.
 
  • Like
Reactions: alexxx

    alexxx

    Points: 2
    Helpful Answer Positive Rating
Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top