2

I’am using IRF520 mosfet according to this schema to shut down Arduino by low level on output to 10K ohms resistor

Schema of my device isenter image description here

The issue is that IRF520 took too much place

Is it possible to substitute IRF520 with smaller mosfet?

Additional info:

I measured average power consumption – it’s about 22 mA on 12V = 0.264W

IRF520 is for 70W max according to datasheets.

The base schema is

enter image description here

Improve this question
1
  • 1
    check out irlz44n much better option.
    – Ace
    Feb 11, 2020 at 10:16

1 Answer 1

Reset to default
1

For what you are trying to do, any transistor, bjt or MOSFET, would work. 2n2222 or 7000 for example. Irf5x0 is actually a poor choice here. It only worked due to the low load being switched.

Improve this answer
8
  • I going try 2N7000 - futurlec.com/Transistors/2N7000.shtml, since 200mA should be enought for my schema. What do you think about resistors? Should I change them or I can left 10K and 120K?
    – Stepan Novikov
    Aug 8, 2017 at 11:05
  • 1
    That transistor has a very high R_DSON. You should be looking for sub-ohm resistances.
    – Majenko
    Aug 8, 2017 at 12:04
  • @Majenko I'm not very good with mosfets, should I increase resistance of R1 or R2 for 2N7000?
    – Stepan Novikov
    Aug 8, 2017 at 16:42
  • 1
    No. Choose a different mosfet with a lower resistance.
    – Majenko
    Aug 8, 2017 at 16:47
  • No need to increase r1 r2. If anything they are already too big. 2n7000 or 2n7002 if you want even smaller devices is more than adequate for your application. Its rds is 6ohm max and if you look at the id vs. Vds chart it is even smaller for a Typical 2n7000. So people arguing for its high rds are reading the spec without understanding it.
    – dannyf
    Aug 8, 2017 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.