The IRF520 datasheet does not specify a maximum Rds(on) for a Vgs of 5V. So there is no guarantee that any IRF520 you buy will even turn on at 5V. Some may some may not some may be somewhere inbetween.
"Arduino friendly" I guess they can say that without fearing any legal implications.
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Yes, it will. This can be gleaned from the datasheet which gives the characteristic curve for a number of Vgs, including 5V.
So what is the Maximum Rds(on) at a Vgs of 5V
By determining the current and using Ohm's law on the working point in the MOSFET's curve. Since the current and the working point are interrelated, you'll either have to approximate or iterate towards the definitive value. The easiest way, however, is to make a Spice simulation using a decent model of the MOSFET, which tends to be reasonably close. As you probably know, MOSFET curves are kind of steep and as a result, slight variances can have big effects, so it's usually impossible to theoretically determine the exact Rds in the device's linear region.
So you don't know
You try to help, so do I. If you feel what I say is incomplete or inaccurate, I'd appreciate a straightforward rectification or response. There's no reason to be disingenuous.
The graph on the datasheet seems to show a drain-to-source current of 4A when using a 5V gate signal, am I reading this wrong? Can you explain where your info comes from?
I asked a simple question
No, that's correct, but keep in mind that the graph is plotted with Vgs and Vds being applied directly to the device, no additional components, and the Vds supply is a perfect power source that'll supply whatever current the MOSFET happens to draw. It's also measured under pulse conditions (the Infineon datasheet mentions a 20us pulse), not sustained. These are different conditions than a real-world use case where you'd have this part acting as a low-side switch for some kind of load. You can infer a few things from the chart, but you need to be aware of a few limitations to it.
hmm ok, I'm new to this so will have to do more reading about it, thanks though.
I tried having the Vgs constantly on and measured the output with a multimeter and it was showing 12V, but this does not hold true once I connect something to the output. Does that point to a particular issue?
You'd have to show the schematic of the setup you're doing the measurements on. But yes, you can expect differences depending on what kind of load is being switched.
However, from a practical viewpoint, the IRF520 is just not an ideal choice for switching larger loads from a 5V logic pin (and certainly not 3.3V) and you're better off choosing a different part. If you want to get on with your project, this is what I'd suggest.
For learning/academic purposes, it can be instructive to study how the IRF520 performs, but keep in mind it may run hot during your experiments. You can burn yourself on it and it may melt plastics nearby.
The setup is basically exactly as illustrated previously in this thread (see eg. post #13) just swapping the LED for a multimeter. I have a small piezo buzzer that I connected which makes the output voltage drop to approx 3V...
I understand what you mean and will probably buy a different module, but I'm just frustrated because the existing one works perfectly (it doesn't behave as described above and continues to output 12V) it's just these new ones I bought that don't - so I'm hoping to work out what the problem is before I go shopping again.
See post #16
guaranteed to work with 5V drive signal
yep I've bookmarked that one, thanks!
See post #21
Just don't buy any module that has uses the IRF520
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