
#1
Feb2712, 12:53 AM


#2
Feb2712, 11:31 AM

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Thanks
PF Gold
P: 4,408

As a start, you can either ignore R_{x} or the circuit is in saturated mode ... this is a consideration since the output current is defined to be 3 mA. The rest is just the usual KVL equations of an active op amp circuit (input voltages equal etc.).




#3
Feb2712, 11:35 AM

Mentor
P: 11,421

Rb and R1 form a voltage divider. What voltage are they dividing? What's the opamp going to try to do with that voltage? 



#4
Feb2712, 11:59 AM

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P: 6,189

What kind of opamp is that?
I'd say it looks a bit like an inverting amplifier, but it's not quite it.
KVL and KCL should tell what it does. 



#5
Feb2712, 01:07 PM

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PF Gold
P: 4,408

I hadn't noticed the output voltage is given. Since R1 is given, all components are defined implicitly.
You can solve this problem without writing any KVL equations. Hints: what must Ra be to satisfy Iz_min? What must Rb be to give a 12V output? And finally, what must Rx be to give 3 mA of current flowing into the op amp? 



#6
Feb2712, 01:24 PM

P: 1,784

There are a few opamps with open collector outputs. This could be one of them.




#7
Feb2712, 02:25 PM

P: 1,506

I would say it is a noninverting amplifier with voltage of 5V on the + input and voltage gain of R1/(R1+Rb).
Vout =+12V so Rb can be calculated.... and so on 



#8
Feb2712, 03:07 PM

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PF Gold
P: 4,408





#9
Feb2712, 04:13 PM

P: 1,506

gain of (R1+Rb)/R1 = 12/5 = 2.4




#10
Feb2912, 03:40 AM


#11
Feb2912, 07:00 AM

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P: 4,715

+Vs is the power supply (you've previously seen it as +Vcc) to power the OPAMP. There is no Vcc here, the negative power supply here is ground (you've previously seen OPAMP's negative supply as Vcc, but not with this arrangement here).




#12
Feb2912, 07:05 AM

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P: 4,715





#13
Feb2912, 07:13 AM

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#14
Feb2912, 01:12 PM

PF Gold
P: 2,551

3 2 I1 = 0 I1 = 1 mA There you go... 



#15
Feb2912, 03:38 PM

HW Helper
P: 6,189

The arrow indicates that the wire goes to a power supply. It does not indicate the direction of the current. The actual current flows away from the 25V power supply. 



#16
Feb2912, 04:36 PM

PF Gold
P: 2,551

OH! OH! Now I get it :) Or being my name. Telling me that I should take this exercise and try to solve it. He just put the "e" in front of the "k" by accident :) 



#17
Feb2912, 05:05 PM

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P: 6,189

OOOOOhhhhhhhhh!!!! AHHHHHHA!!!
Or take it! Yes, I think I understand now what he meant. ;) 



#18
Feb2912, 05:27 PM

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P: 4,715

Hint: you know V_{+} so determine V_. 


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