# Op amp in circuit: find load current

1. Feb 25, 2013

### Color_of_Cyan

1. The problem statement, all variables and given/known data

http://imageshack.us/a/img27/7043/homeworktest2prob5.jpg [Broken]

Find the load current IL (in microamperes) in the circuit.

2. Relevant equations

V = IR

KCL, KVL,

voltage division, current division,

Op-Amp Output V = (Gain)(V+ - V-)

(Gain) = -Rf / Rin

3. The attempt at a solution

I'm not sure how to figure out the voltages at the nodes before the inputs to the op-amp

I think the gain is just -5k / 10k

I'm pretty sure all I would need to figure out is the input node voltage or whatever the Vin is for each input.

But I'm guessing:

V out = (5V - 3V)(-5kΩ/10kΩ) = -1V

IL = -1V/4000Ω

= -250μA

?

Last edited by a moderator: May 6, 2017
2. Feb 25, 2013

### CWatters

Perhaps check here...

http://www.ustudy.in/node/2939 [Broken]

Your circuit doesn't have an R3 so substitute R3=0 into the transfer equation.

Last edited by a moderator: May 6, 2017
3. Feb 25, 2013

### Staff: Mentor

Color_of_Cyan, do you know the basic characteristics/properties of an ideal op-amp? Can you describe them?

If you know these basic things and the rule about the potential difference between the op-amp's inputs when there's feedback from the output to the negative input then you will be able to derive the output voltage fairly easily from the diagram.

4. Feb 25, 2013

### Color_of_Cyan

CWatters, woops it did have a resistor there, I fixed it now though.

Thanks... although that's kind of weird. Is there any way to derive that? Or are you supposed to know for each type of op amp?

I still got V out = -1V with that by the way.

And the current through the 4k out is still -1A / 4000 ohms right?

gneill I really don't; my professor went through it pretty fast. I looked something up though and says that the + and - inside is connected by a resistor with a big value and the gain is supposed to be high.

The output is supposed to also be between here -5V to 5V.

That's pretty much all I know

5. Feb 25, 2013

### CWatters

If the gain of the opamp is very high then the +ve and -ve inputs must have very similar voltages or the output will saturate (be driven to the supply rail). For example if the opamp had a gain of 100,000 and the max output swing was say +/- 5V then the inputs cannot differ by more than 5/100,000 = 50uV or it will saturate.

In your circuit you can easily work out the voltage on the +ve input using the potential divider rule. Then using the above you know that the -ve input will be at a very similar voltage. Then you can look at the currents through the 5k and 10k to work out the output voltage.

Aside: On many opamp circuits (eg simple inverter) the +ve pin is 0V (or earth). Therefore if the amplifier is designed to operate in it's normal linear mode the -ve input must also be very close to 0V. This is known as a "virtual earth" because although it's not connected to 0V(earth) the -ve pin must be very close to it.

6. Feb 25, 2013

### CWatters

For example..

+ve pin...
3 * 3/9 = 1V

therefore -ve pin must also be 1V or very close to it!

Current through 10K resistor is

(Vin-1)/10K = 4/10k

Current through 5K resistor is

(1-Vout)/5K

These currents are the same as "none" goes in the -ve Pin (high impedance inputs).

Solve to give Vout = -1V