Working on a method to have an adjustable flash on an LED neon strip.
Have attached the schematic for the test circuit, plan to use a 7805 to power the 555 but testing with a 9v battery I get 4 volts on the 555 output.
The issue is the gate has a maximum of 4v on the gate.
thinking of a voltage divider or Zener diode on the gate to avoid applying too much voltage to the gate.
I show 2 LEDs but D3 is just for testing. D2 will be a 12v LED neon strip. the 9v will be 12 v.
and yes the breadboard circuit works as showen
You need to link datasheets, part numbers and why no heatsink? That schematic is not helpful.
To switch 2A and only dissipate 0.2 W (not too hot) you need a 0.1V drop (2A*0.1V=0.2W) which means Ron = 1V/2A = 0.5 ohm
The IRF520 is 0.27 ohm max at Vgs=10 but more than 2x this at 150'C so a heatsink is needed.
Here's an update to follow if you like the 30:1 frequency range You can make the IRF520 work if you follow my advice. Use 12V for gate voltage add small heatsink. There are about 1,000 logic level FETs to choose from.
The Vgs(th) is the gate voltage that turns it OFF, you need the gate voltage that turns it ON which is usually 10V for a common old Mosfet. A moderm "logic level" Mosfet fully turns on with a 5V gate voltage.
Technically Vt or Vgs(th) or gate threshold voltage is defined by the range of voltages due to mfg tolerances that result in current of 250µA with Vgs=Vds. So iif Vt was say 2.5V then RdsOn= 2.5V/0.25 mA = 10 kohm. . not quite off or on.
1. OK with 1 to 2V actually with 12V gate drive you can use 2 to 4V too. ok?
2. I gave 0.2 ohm as a limit for the example using the IRF FET getting hot (TO220 without a heatsink and not an ideal target for a FET resistance. ok?
3. But there are tons of different FETs in catalogs, so why not get one to handle 5A easily. or RdsOn of <= 25 mohms. 5^2 * 0.025 = 63 mW , cool ')