# Help with op-amp circuits

#### sebholm

Joined Dec 18, 2020
20
Can someone help me with finding v_out of these two circuits? And maybe a small explanation hah ty.

#### dl324

Joined Mar 30, 2015
14,321
Welcome to AAC!

Post your attempts at solutions so we can see where you're having problems.

#### RBR1317

Joined Nov 13, 2010
690
The method that is used to solve op-amp problems like this is to write the node equation for the (-) op-amp input, then set the (-) node voltage equal to the (+) input voltage. Because an ideal op-amp operating in its linear range with negative feedback works to keep both input terminals at the same voltage. Also, an ideal op-amp has no current flowing into the input terminals.

#### sebholm

Joined Dec 18, 2020
20
In figure (b) I have thought of doing kvl in upper loop. The (-) input has 3,3 V. The current into resistor in feedback is also 2mA. But I can´t figure out if the elements in the kvl loop have plus or minus in front.
If I use kcl, is the node equation for figure(b): (3,3V-v_o)/R+2mA=0?

#### dl324

Joined Mar 30, 2015
14,321
But I can´t figure out if the elements in the kvl loop have plus or minus in front.
Show us what you think they are.

#### sebholm

Joined Dec 18, 2020
20

#### dl324

Joined Mar 30, 2015
14,321
0=-2mA*R+v_1-v_o
Why did you treat the current source as -2mA? That isn't how I'd interpret it. Are we to assume that v_1 is the voltage at the inverting input? What is that voltage?

#### sebholm

Joined Dec 18, 2020
20
Because I guessed the voltage over the resistor is opposite of the voltage at inverting input. Yes sorry, v_1 is the inverting input with 3,3V.

#### sebholm

Joined Dec 18, 2020
20
What is the voltage drop on the resistor? What does that make the output voltage?
Drop over resistor must be (2mA*3kohm)=6V and with output I dont know. I think my teacher´s answer confused me.

#### dl324

Joined Mar 30, 2015
14,321
Drop over resistor must be (2mA*3kohm)=6V and with output I dont know. I think my teacher´s answer confused me.
If you have 3.3V on one end of the resistor and a drop of 6V across the resistor, what is 3.3V - 6V?

#### sebholm

Joined Dec 18, 2020
20
Hah -2.7V. But then why is this kvl not correct:
0=-2mA*R+v_1-v_o
v_o=-6+3.3

#### dl324

Joined Mar 30, 2015
14,321
Hah -2.7V. But then why is this kvl not correct:
0=-2mA*R+v_1-v_o
v_o=-6+3.3
Looks correct to me.

3.3V - 6V - Vo = 0
Vo = 3.3V - 6V = -2.7V

Why did you bother with a loop equation when you knew you had 3.3V on one end of the resistor and a 6V drop across it?

Last edited:

#### sebholm

Joined Dec 18, 2020
20
Just my teacher´s answer, and could not figure out why in the kvl, you should minus Vo and plus Vin. I´m pretty new to this.

#### dl324

Joined Mar 30, 2015
14,321
Just my teacher´s answer, and could not figure out why in the kvl, you should minus Vo and plus Vin. I´m pretty new to this.
If that's the way your teacher wrote it (without spaces), shame on him/her.

It should have been 0V = 3.3V - (2mA)(3k) - Vo. This makes it clear that you're using the standard convention for the current direction and that it's causing a voltage drop across the 3k resistor.

When you write loop equations, you need to be consistent. We usually add voltage sources and subtract voltage drops.

When I first read your sloppy equation, I thought you had the current direction wrong. Inserting spaces and using parenthesis makes intention clear.

• atferrari

#### sebholm

Joined Dec 18, 2020
20
Okay ty! A tip on finding Vo at fig(c)?

#### dl324

Joined Mar 30, 2015
14,321
A tip on finding Vo at fig(c)?
Show your attempt so I can see where you need help.

#### RBR1317

Joined Nov 13, 2010
690
If I use kcl, is the node equation for figure(b): (3,3V-v_o)/R+2mA=0?
There are several ways to structure node equations. In your method, are currents leaving the node considered to be positive or negative?

#### sebholm

Joined Dec 18, 2020
20
Show your attempt so I can see where you need help.
Vo=1,1V, because (-) input has to be 2,2V right?