EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Iali on March 30, 2021, 05:22:48 pm
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Hi,
I have a simple dual D4184 mosfet pwm driver board that I’m testing with a 3Mohm dummy load trying to understand how a perfect square PWM signal translates on the power side of this circuit. I’ve attached a picture of what I’m seeing on my oscilloscope. The yellow signal is the driving PWM and the blue one is the output across the dummy load. To me it appears that the reason the power output has a trailing end is because the mosfets have capacitance due to the nature of how they are constructed. Is my understanding correct and is there anyway to correct this behaviour?
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There is no schematic, so it is unclear. Anyway, if the output is two mosfets in half bridge (aka push-pull) configuration, then this is not normal. It looks like maybe the low side mosfet is not turning on, so the capacitance disharges very slowly through resistor. Yet, the trace shows very linear voltage drop, it is not a typical RC exponential curve, so I don't know. Schematic needed. By the way, 3 megaohm resistor is a very high resistance. It is more like a dummy leak, then a dummy load :)
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Thanks for the response Manul. Here is the link to the board I’m testing
mosfet driver (https://www.amazon.ca/dp/B089KF8NNL/ref=cm_sw_r_cp_api_glc_i_XAAJWTXQANMJN72YG147?_encoding=UTF8&psc=1)
Not sure how it’s configured but I’m guessing not H-bridge as far as I know. I measured the output signal across the Vout +/- and I can see on the board that Vout+ is wired directly to Vin+, so looks like the mosfet is switching the ground side which I believe is a typical configuration. I’m assuming the mosfets are wired in parallel to increase power capacity.
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Yes, by looking the image of the board there are two mosfets in parallel switching on the low side. Your measurement also explains why the scope trace looked inverted (fast rise, slow fall). It happened, because you measured in reference to +DC, not ground. It is ok, but more common is to measure in reference to circuit ground.
So, if we think about mosfet switching on the low side, we can say that it will produce fast falling edge when it turns on (because on resistance is very small). When it turns off, the voltage at its drain will rise, because it is pulled up by your dummy load. The output capacitance of two D4184 mosfets at low applied voltages can reach 2nF (look capacitance graph in the datasheet). It is quite a lot. Your resistor is also huge. If you would use smaller resistor it would look more rectangular, because it would pull up a lot stronger. Try 1K resistor.
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Thanks for this explanation Manul. You are right and I'm not sure why I didn't reference ground instead of V+ but I'll definitely try that next and also use a smaller resistor and see if the results are more square. Much appreciated!