# Current Source using Op Amp

Op amps can be wired to set the amount of  current through a load instead of the load setting the current.

You wire the load between the output and inverting (-) input. Then put a resistance from the (-) input to ground. A desired voltage is applied to the non inverting (+) input.

The 2 main determining factors of what the op amp output current will be is the voltage at the (+) input and the resistance between the (-) input and ground.

The output voltage will be the combined voltage at the (+) input and the voltage that the load needs when it passes that much current. The output makes sure that the voltage across the (-) to ground resistor stays at 5.1V in the example circuit. With a 1,000Ω being used there, that is 5.1V/1,000Ω = 0.0051A (5.1mA) of current that you can expect the output to provide. Don’t forget though that there is a load between the output and the (-) input. The output will need to step up the voltage to what the load will be dropping at that much current.

If the load drops about 2V (such as a red LED) then the output will have to be about 7.1V to make sure that 7.1V-2V=5.1V reaches the (-) input.

The main limits of this circuit is the voltage the output can provide (often about 1V less than the supply voltage), and how much current it can provide. Which should be around 20mA.

Assorted integrated circuits (ICs) kit. Included is the LM358. It is an Affiliate link ad that supports this channel.