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I'm designing a circuit for automotive use, where I control various kinds of inductive loads (motors, solenoids, etc) up to 20A (Máx). In the design i'm looking into using 2 SMD N channel MOSFET in parallel (Vishay-Siliconix/SQJ142EP-T1_GE3) to control these loads with a 100Hz PWM signal and in parallalel with the load i'm using a SMD flyback diode to reduce voltage spikes (STMicroelectronics/FERD20H100SB-TR).

Since I'm only using SMD components i did not want to use any kind of heatsinks to save space in the PCB, but i'm having problems to find out if this diode will need one. How would I calculate that that?

P.S. The ambient temperature this PCB will be subject will be about 65°C - 70°C (149°F - 158°F).

P.S.S. Here is a picture of my design schematics enter image description here (Schematic)

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  • \$\begingroup\$ Draw a schematic and add all power supply rails and load current information. Don't try and describe the circuit in words; use a schematic. That is why EEs draw schematics. \$\endgroup\$
    – Andy aka
    Mar 19, 2021 at 18:00
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    \$\begingroup\$ Just an unrelated comment: your mosfet gates are basically connected directly to the car's positive supply. Any excursions above Vgs(max) - typically 20V - may kill your mosfets. Such spikes are fairly likely in a car; automotive environment is harsh electrically. Also, why the first mosfet stage, and why the choice for optocouplers? \$\endgroup\$
    – marcelm
    Mar 19, 2021 at 18:50
  • \$\begingroup\$ I'm aware of the voltage spikes, i'm still looking into a protection circuit for that (maybe add a TVS in the gate), in fact I'm open to ideas. The optocouplers are used for protection, because the system will be assembled and disassembled a lot, if something is not connected properly it's way easier fo fix a damaged optocoupler than a mosfet and it also protects my control circuits. The first mosfet stage is just an easy way I found to invert the signal to the 2nd stage (since I can't change the 2nd stage mosfets). \$\endgroup\$ Mar 19, 2021 at 20:56
  • \$\begingroup\$ Fair enough. Note that with the 10k gate resistors, the mosfets will be fairly slow to switch. That may or may not be what you want. \$\endgroup\$
    – marcelm
    Mar 19, 2021 at 21:31

3 Answers 3

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Essentially the diode will need to dissipate the inductive energy stored in the load, at 100 times per second. You’ll need to establish the inductance of the load and can then determine the power rating for the diode. If you don’t know what the load will be then you have a problem. You might imagine that a motor could dump quite a significant amount of energy if it’s operating as a generator and being actively driven.

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  • \$\begingroup\$ That's the problem, I don't know which load will be connected, I only know it's inductive. If I consider the voltage drop at the diode (shown in the datasheet) and multiplied by the expected max current, would it be a good approximation of the power dissipation? \$\endgroup\$ Mar 19, 2021 at 21:01
  • \$\begingroup\$ The diode current is unlikely to exceed to motor current, but since the switching frequency is low I’d be inclined to rate the diode for continuous operation at that current. For a relatively small diode a 50ms on-time isn’t far from DC in terms of instantaneous heating of the die. \$\endgroup\$
    – Frog
    Mar 20, 2021 at 4:26
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Driving an inductive load at 100Hz, sounds like you are attempting to control the amount of current flowing through it.

If you are not letting the current decay to 0A, then what you have is an average current in the inductor, and thus also in the diode. The diode current will be inversely proportional to the PWM duty cycle of the inductor. eg. At 60% duty cycle, the diode will be on for 40% of the time. The diode current will follow the \$\frac{R}{L}\$ decay of the inductor.

Once you know your average current in the inductor and your duty cycle, then it's plug and chug: \$Power = Diode{_p}{_w}{_m} * Iavg *Vfd\$

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Time ago I failed to pass the UL tests of an AC/DC flyback power supply due to the heating of two ultra fast 3A SMD diodes in parallel at the output.

The required output current was 1A so I put two diodes in parallel.

At 40 Celsius degrees ambient temperature the Mosfet's controller was so hot that went in protection and shutdown by itself.

Two years later I redesigned the power supply using Mosfet synchronous rectification on the high side of the output and I passed all the UL 60950 tests.

I passed also all the EMC tests because the synchronous Mosfet was on the high side.

I started from an NXP's demo board of a PSU that showed good EMC performance. I adapted it to my needs and I had success. It was a 92.8 % efficient PSU able to deliver 50W at 12 V.

The switching Mosfet was an Infineon: 650 V, SMD, 285 mOhm

My suggestion is:

If you have heating problems, use synchronous rectification at the output and push the efficiency above 90%.

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