Help with op-amp circuits

RBR1317

Joined Nov 13, 2010
591
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.
 

Thread Starter

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?
 

Thread Starter

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.
 

dl324

Joined Mar 30, 2015
12,224
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?
 
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Thread Starter

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
12,224
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.
 

Thread Starter

sebholm

Joined Dec 18, 2020
20
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?
 
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