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I've selected this MOSFET for a motor control application [picture attached]: IRF200P222 Datasheet: https://www.infineon.com/dgdl/Infineon-IRF200P222-DS-v01_02-EN.pdf?fileId=5546d4625b3ca4ec015b3e42ba4a0744

I'm going to use low side configuration. I've chosen the 1EDN751x gate driver which has two possible ways of using it. I attach them. What is the best way to use it? And how can I compute a proper gate resistor? The gate driver power supply is 12V. The dc link voltage is about 150VDC.

Datasheet: https://www.infineon.com/dgdl/Infineon-1EDN751x_1EDN851x_Rev%202.0-DS-v02_01-EN.pdf?fileId=5546d462576f34750157e176df0b3ca7

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  • \$\begingroup\$ You have a single control line, not a differential, and you have a single gate which should be driven both high and low. So why would you consider using the second configuration? Which fits your schematic? \$\endgroup\$ – WhatRoughBeast Jun 9 '18 at 15:33
  • \$\begingroup\$ Both of the configurations are single-channel driver. One of them just has sink and source separated and the other not. \$\endgroup\$ – Blue_Electronx Jun 9 '18 at 16:18
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The MOSFet you selected has a high \$R_{DS,on}\$ at \$V_{GS}=4V\$, so I'd go with the 'Z=8' type driver, definitely not the 'Z=7'.

According to fig 8 of the MOSFet datasheet you have to supply 100nC to the gate to switch it on. This also has to be removed during switch-off. Now the speed (current) with which you transfer this charge to and from the gate determines how fast the MOSFet will switch.

We don't know what your plan is, but let's assume the worst and push your driver to the values mentioned in table 12 of its datasheet: 4A source, 8A sink.

The source and sink impedances are 0.9 and 0.4 Ohm respectively and the gate resistance is 1.3 Ohm. Sourcing 12V DC in 2.2Ohm gives 6A, and sinking 5.5V gate 'plateau' (fig. 8) over 1.7Ohm gives 3.2A. During switch-on therefore an additional gate resistor is required in order to limit the current to 4A: 12/4=3Ohm total, or 3-2.2=0.8 additional gate resistance.

During switch off no resistor is required, so we can by-pass the additional gate resistor with a diode from gate to sink pin. This diode has to act fast, and have a low voltage, a Schottky diode is therefore advised.

I hope this answers your question.
P.S.: Until now your switching frequency, type of motor, inductance thereof, and values in the snubber circuit are unknown but important for a correct functioning, e.g.: not blowing up, of the contraption.

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  • \$\begingroup\$ Thank you for your useful answer. You're right about going with the Z=8. The problem is that it's not available in Infineon authorized distributors, maybe they are out of production for now. I chose a PWM frequency of 25 kHz (could be lower). The motor is PMDC brushed 4HP @125VDC. I haven't computed the snubber yet. \$\endgroup\$ – Blue_Electronx Jun 9 '18 at 18:07
  • \$\begingroup\$ The only option that I have for now is going with the 2EDN series which have two drivers. I guess I will have to let one of them useless. \$\endgroup\$ – Blue_Electronx Jun 9 '18 at 18:12
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    \$\begingroup\$ Using only one won't be much of a problem. \$\endgroup\$ – HarryH Jun 9 '18 at 18:18
  • \$\begingroup\$ One more question. If I take I^2*Rg, it would throw a very high power rating, but usually gate resistor power rating is low. I know it has to do something with the pulse peak charging current, but I'm kind of confused. \$\endgroup\$ – Blue_Electronx Jun 9 '18 at 22:22
  • \$\begingroup\$ Let's say the time it takes to switch on the MOSFet is \$\delta t_{switch on}\$. Divided by the switching period that gives you the relative time, \$\tau\$, that this current is flowing. Your dissipation is then \$\tau i² R_g\$. \$\endgroup\$ – HarryH Jun 10 '18 at 9:33

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