How to turn on/off a 5 V, 10 A relay from 5 V, 0.2 mA input signal

Question

I am working on a project where I have to switch on/off a 5 V, 10 A relay to turn on/off a 230 VAC LED light using 3.3 V, 0.2 mA of input voltage and current respectively from the microcontroller.

I know that the current supply is not enough to trip the relay. Please provide a solution so that I can provide proper current and voltage to the relay module.

Note: I tried to trip the relays using transistor and MOSFET. Nothing seems to work.

^{simulate this circuit – Schematic created using CircuitLab}

The relay coil requires 5 V, 10 A. I have used an IRFZ44N MOSFET and a TIP122 transistor.

Rated coil power is 0.36 W (48 VDC, 0.51 W)

Answer 1

The relay coil requires 72mA (360mW/5). That's easily accomplished with a BJT such as 2N4401.

^{simulate this circuit – Schematic created using CircuitLab}

You could also use a small MOSFET such as AO3400 (SOT-23 package).

IRFZ is inappropriate because it takes too much drive voltage, and the TIP122 Darlington is inappropriate because it drops too much voltage.