# Why is output equal to non inverting input opamplifier

I wanted to make a simple current source. The circuit was build so that the laser would get a constant current.

I bought the opamplifier opa552 from china for 1 euro because it could handle large currents and voltages. But after some measurement I came to the conclusion that the output is equal to the non inverting input and that no current goes through the 3 ohm resistor. I have removed the capacitor in the circuit but that didn't help.

After some time the opamp started to smoke so now it is broken. Is my setup wrong or do you think the opamp pins of my opamp don't correspond with the datasheet of the opa552 or is is just a bad opamp? (all the cables are connected correctly and according to datasheets). Shall I buy new opamp and try it again?

I know I could also build a current source with other methods, by using transistors but I am just curious why this is not working.

By the way, my opamp needs 3 grounds, I have connected all three of them.

Describing the circuit: Voltage divider so that non inverting input=0.35 Resistance of 3 such that current is=0.35/3 A. THis current will also go through the laser. For testing I have replaced the laser with 4 resistors. Capacitors so that when turning on system laser will not get a too high voltage.

• You need to take a step back from the circuit and do some calculations. The current through the laser flows from + 12 V through "something" (in your circuit that's the opamp) then the laser an the 3 ohm resistor. Do a calculation on how much power each device will dissipate. My guess is that the opamp needs to dissipate too much power. Yes it can deliver 117 mA but can it do that while dropping 5 V ? You should really consider adding a power transistor to do the actual dissipation. Commented Mar 14, 2017 at 11:07
• "I bought the opamplifier opa552 from china for 1 euro" => Considering the real price of this OPA, good chance you got a fake. Also, circuit as drawn is wrong, as said by others. Commented Mar 14, 2017 at 11:21
• If the voltage as marked on the schematic is correct, the circuit is working. Whether it is dissipating too much power is another question. Commented Mar 14, 2017 at 12:01
• @dannyf If the voltage marked on the schematic is correct, something is very wrong. Putting 12V into a potential divider of 100R and 3300R won't give you 0.35V. Commented Mar 14, 2017 at 13:07
• Also, that input voltage divider is upside down - cannot possibly work as drawn.
– user16324
Commented Mar 14, 2017 at 13:18