I'm trying to make a circuit that has a load resistance of 2 Ohms, load current draw of 2A and voltage of 5V.

The circuit can only be fed with a 5V@500mAh battery.

Here is a diagram I have thought of: enter image description here

Please excuse me for any technical errors since I'm new to the field.

I would love to get some suggestions regarding whether this circuit is correct.

Thank you.

  • \$\begingroup\$ Not a technical error but a fundamental error: You want to supply 10 watt to a load, from a 2.5 watt supply. \$\endgroup\$ – pipe Dec 4 '18 at 22:18
  • 3
    \$\begingroup\$ It's a 2.5 Wh supply. It might supply 10 W for 10 or 15 minutes. \$\endgroup\$ – Transistor Dec 4 '18 at 22:24
  • \$\begingroup\$ Use a D44H11 (NPN) with emitter to ground, base to a \$150\:\Omega\$ resistor to your supply rail, and your \$2\:\Omega\$ load between the supply rail and the collector. The D44H11 can handle the load (and so much the better if it heats up a bit.) \$\endgroup\$ – jonk Dec 5 '18 at 7:20

I'm trying to make a circuit that has a load resistance of 2 ohms, load current draw of 2 A and voltage of 5V.

There is something not quite right with your description. I take it you mean that you have a 5 V supply and that you want to provide a constant 2 A through a 2 Ω load.

From Ohm's Law we can calculate the voltage drop across the resistor as \$ V = IR = 2 \times 2 = 4 \ \text V \$. That means that the transistor will need to drop 1 V from the 5 V supply.

Now you have a problem. You have selected a Darlington transistor. You will lose 0.7 V across each of its base-emitter junctions so that to get 4 V at the emitter you will need 5.4 V at the base. You will need an even higher voltage at the left end of the 430 Ω resistor. (Please number your components R1, R2, etc. to aid discussion.) In other words, your circuit will not work.


simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) NPN version. (b) PNP version.

There's more bad news. The TIP120 Darlington transistor will have a voltage drop of 2 V when in saturation (turned fully on). That means that you only have 3 V for your 2 Ω resistor so you will only get 1.5 A through it.

If you replace the Darlington with a regular transistor it will drop < 0.5 V in saturation so you will have 4.5 V or so for the load and get 2.25 A through your load.

  • \$\begingroup\$ Yeah, but then he has to supply something like .2 amps of base current. \$\endgroup\$ – WhatRoughBeast Dec 5 '18 at 1:02
  • \$\begingroup\$ Thank you for helping, is there any other more efficient way to deliver a 2A current @5V through a 2Ohm load while the max source current is 0.5A@5V? \$\endgroup\$ – Danaro Dec 5 '18 at 11:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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