
#1
Feb2912, 04:13 PM

PF Gold
P: 2,551





#2
Feb2912, 04:17 PM

Mentor
P: 39,606

It's an underdefined problem as stated.
Are you assuming an "ideal" opamp? Or one that has input bias currents? And you should probably label the + and  inputs of the opamp, although that wouldn't change the answer to the problem. 



#3
Feb2912, 04:29 PM

HW Helper
P: 6,189

I believe you are right. :)
Btw, Vout does not "pass" through a resistor. A "current" passes through a resistor, and only if the voltages on both sides of the resistor are different. In this case those voltages are the same. However, I am assuming the opamp has its + and  inputs configured like a voltage follower. If they are reversed, the circuit is unstable and Vout will either go up to Vcc+ or go down to Vcc. Either way, there will no (significant) current flowing through the resistor since the opamp has a resistance that is near infinity. EDIT: 



#4
Feb2912, 04:41 PM

PF Gold
P: 2,551

Opamp true or falseSo, if it's an inverter though, the it's no longer a Voltage follower, so I can't use the same Vout = Vin law. That part I understand, though I wouldn't be sure what to make of it if the inputs were indeed reversed. 



#5
Feb2912, 05:03 PM

HW Helper
P: 6,189

To figure out what the circuit does, you need to apply the basic opamp law that says: $$V_{out} = (V_+  V_) A_{OL}$$ where ##A_{OL}## is the so called openloopgain which is typically about 1000000 and where ##V_{out}## is limited to the range Vcc to Vcc+. From there you can try and "solve" the circuit. After solving the circuit, you should consider what would happen if ##V_{in}## would rise just the tiniest fraction, say ##10^{10} V##. Either the circuit "pushes back" or it "diverges" to another solution where ##V_{out}## is either Vcc+ or Vcc. 



#6
Feb2912, 11:56 PM

PF Gold
P: 2,551

I'll make sure I have this printed for reference. Thanks a lot Klaas! :)




#7
Mar112, 02:49 AM

PF Gold
P: 2,551

I have another question for true and false.
For an ideal operative amplifier there's infinite resistance at the entrance and an infinite amplification in an open circuit I answered True, because I know the first sentence is true, but what is it about infinite amplification in an open circuit? I thought amplification is limited to the formulas ( Vout = ...) and the parameters of the opamp. I didn't think it can ever get "infinite". 



#8
Mar112, 04:26 AM

HW Helper
P: 6,189

The formula ##V_{out}=(V_+  V_) A_{OL}## gives the amplification. In a reallife opamp the amplification ##A_{OL}## is about 1000000. In an ideal opamp it is infinite. 



#9
Mar112, 05:58 PM

HW Helper
Thanks
PF Gold
P: 4,404

Zero bias current → zero current thru R, therefore zero voltage across R, therefore output =  input. And since + input =  input, output voltage = Vin. So it's a voltage follower. 



#11
Mar112, 06:15 PM

HW Helper
P: 6,189

Since I'm producing ideal opamps (virtually) I'm charging an infinite cost. ;) 


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