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Old 2nd November 2008, 05:40 AM   #1
xyz9915 is offline xyz9915  Pakistan
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Default Driving Eight IRFP064 with SG3525 directly

Hi, all
I want to drive eight IRFP064 with SG3525 directly (in each side). Please explain that is there any formula for calculating the number of MOSFET's?
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Old 2nd November 2008, 06:22 PM   #2
Eva is offline Eva  Spain
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The gate resistors (or ferrite beads) have to be sized to avoid exceeding the peak and average current capabilities of the driver IC. Measuring the voltage drop across the gate resistor (or a small series resistor) may help. When too many MOSFET are used, keeping the driver within its ratings usually results in too slow switching.

I have seen some commercial designs that blow SG3525A from time to time because the IC is abused by driving too hard too many MOSFET. Sometimes this may be fixed by adding a separate driver like IR4427 (or better).

Anyway, IRFP064 is almost an obsolete device, take a look a IRF2907Z for example, gate charge is the same but Rds-on is half, body diode is much faster and output capacitance is lower too.
I use to feel like the small child in The Emperor's New Clothes tale
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Old 3rd November 2008, 02:55 AM   #3
xyz9915 is offline xyz9915  Pakistan
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In many circuits (available on the net) I noted that, many transistors are used in parallel. Here is an example is attached. So it is requested to define a simple rule/formula for paralleling MOSFETS for SG3525 etc
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File Type: pdf h-bridge power stage.pdf (65.6 KB, 235 views)
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Old 3rd November 2008, 03:27 AM   #4
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But the circuit attached has buffers ie Q3 & Q4 and a a pair of totem poles also.
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