Show us what you think they are.But I can´t figure out if the elements in the kvl loop have plus or minus in front.
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?0=-2mA*R+v_1-v_o
You didn't need to guess anything.Because I guessed the voltage over the resistor is opposite of the voltage at inverting input.
Drop over resistor must be (2mA*3kohm)=6V and with output I dont know. I think my teacher´s answer confused me.What is the voltage drop on the resistor? What does that make the output voltage?
If you have 3.3V on one end of the resistor and a drop of 6V across the resistor, what is 3.3V - 6V?Drop over resistor must be (2mA*3kohm)=6V and with output I dont know. I think my teacher´s answer confused me.
Looks correct to me.Hah -2.7V. But then why is this kvl not correct:
If that's the way your teacher wrote it (without spaces), shame on him/her.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.
Show your attempt so I can see where you need help.A tip on finding Vo at fig(c)?
There are several ways to structure node equations. In your method, are currents leaving the node considered to be positive or negative?If I use kcl, is the node equation for figure(b): (3,3V-v_o)/R+2mA=0?
Vo=1,1V, because (-) input has to be 2,2V right?Show your attempt so I can see where you need help.
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?There are several ways to structure node equations. In your method, are currents leaving the node considered to be positive or negative?
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