Can someone help me with finding v_out of these two circuits? And maybe a small explanation hah ty.

 

Welcome to AAC!

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

 

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.

 

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?

 

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.

 

Show us what you think they are.

0=-2mA*R+v_1-v_o

 

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?

 

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.

 

Because I guessed the voltage over the resistor is opposite of the voltage at inverting input.

You didn't need to guess anything.

Given this:

What is the voltage drop on the resistor? What does that make the output voltage?

 

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.

 

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?

 

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

 

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?

 

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.

 

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.

 

Okay ty! A tip on finding Vo at fig(c)?

 

A tip on finding Vo at fig(c)?

Show your attempt so I can see where you need help.

 

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?

 

Show your attempt so I can see where you need help.

Vo=1,1V, because (-) input has to be 2,2V right?

 

There are several ways to structure node equations. In your method, are currents leaving the node considered to be positive or negative?

Negative. But what way does the current flow in the upper loop? It has to come from 2mA, but I guess it would also travel back with the feedback?