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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.

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  • \$\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
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    \$\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
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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.

schematic

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.

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  • \$\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

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