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What is happening here?

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Mario87:
Hi all, I have a circuit as shown below using an IRF520 MOSFET and the pulse is controlled by an arduino.



Now when I use a 1K resistor, or even a 220 ohm resistor in place of R1 I get the following output, which is exactly what I am looking for....



Basically, the 5v rail is pulsed down to GND for 4038us, allowed to go high for 577us, then pulsed back down and repeats. Perfect.....

However, the circuit I will be using will eventually be drawing circa 2A on every pulse, so for a test I hooked up a 2.2ohm, 50w power resistor (V=IR -> 5=2.2I -> I=2.27A).

Problem is, when I do that I get the following output where it is only pulling the 5v rail down to circa 3.3v before being allowed to go high again, there are also very large overshoots when going high and going low. Only for a few ns, but overshoots none-the-less. The transistor also becomes very hot, pretty quick. Can anyone explain this to me, as I am a bit confused by it. Screenshot below....



MOSFET datasheet here... https://www.vishay.com/docs/91017/91017.pdf

isometrik:
If you look at the Vds vs Id vs Vgs characteristics curves, you will notice that the resistance of the IRF520 is about 0.4 ohm at Vgs = 5V.


iMo:
You need a driver which creates more than 5V at the gate..
When the mosfet is not fully opened, it works in a linear region and heats up.

madires:
Or use a logic level MOSFET. The spikes are generated by the power resistor since it's wirewound and therefore inductive.

Mario87:

--- Quote from: isometrik on September 29, 2019, 08:44:28 pm ---If you look at the Vds vs Id vs Vgs characteristics curves, you will notice that the resistance of the IRF520 is about 0.4 ohm at Vgs = 5V.
--- End quote ---

Ok, so I take it that because the resistance is approx 0.4 ohm then I am effectively creating a voltage divider with the 2.2ohm resistor, so I will never be able to pull the voltage down to 0v, as some of it will always be across Vgs (approx 0.6v by my calcs). However, this obviously becomes a negligible voltage at higher resistances of R. Is that right?


--- Quote from: imo on September 29, 2019, 08:48:51 pm ---You need a driver which creates more than 5V at the gate..
When the mosfet is not fully opened, it works in a linear region and heats up.
--- End quote ---

Ok, I have tested and you appear to be correct. When looking at the graphs on the datasheet (I hate logarithmic scales), it appears that at 5v Vgs & 5v Vds the drain current is circa 1.4-1.58A

However I cannot find anything in the datasheet that states in order to drive a Vds of 5v I need to have a Vgs geater than 5v...so what is happening here?? I am still a bit confused due to lack of information in the datasheet


--- Quote from: madires on September 29, 2019, 08:56:26 pm ---Or use a logic level MOSFET. The spikes are generated by the power resistor since it's wirewound and therefore inductive.

--- End quote ---

I think you are correct, logic level MOSFET would be best. I did actually suspect inductance in the wirewound resistor causing the spikes, so I added a flyback diode to try and eliminate them, but it didn't actually do anything.

Adding a 1k resistor to the gate however, did remove the spikes when the gate powered off (voltage output went to 5v).

Any further explanations on the above?

It is clear you guys are correct, but due to lack of info on the datasheet I am still a bit perplexed as to why it reacts as it does.

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