Help starting a different Op Amp Problem

[Asphyxiated]

Homework Statement

An original picture can be seen http://images3a.snapfish.com/232323232%7Ffp733%3A%3A%3Enu%3D52%3A%3A%3E379%3E256%3EWSNRCG%3D335%3B%3B7%3A542347nu0mrj".

Homework Equations

The Attempt at a Solution

I am not really sure how to start, how would I find out what the voltage is at the two input terminals of this op amp? Typically they would just equal the power source but the power is being divided but the resistors are not in series or parallel so how do I get going here?

 

[Interesting]

If you make a loop around the + terminal, it will give you the voltage. Then you can figure out the gain (R2/R1). Then you can figure out the output voltage, and then get io.

 

[Asphyxiated]

Hey, could you possibly draw on the pic i posted and show me what you mean by 'make a loop around the + terminal'? I missed this lecture in class so I really don't know what I am doing. If i could see an example it would help more but if you will just show me the way it would be very much appreciated.

 

[Interesting]

http://imgur.com/vF77U

Make a loop is the same as saying sum all the voltages in the loop. This is a simply voltage divider, and it will give you the voltage that's present at the + terminal. This is equal to the voltage at the - terminal. The gain of you op amp is a division of two resistors, and there's a sign associated with it. Your book should help you there.

 

[Asphyxiated]

Hey, sorry this took so long to post but I had a project that needed to get done for school. Hopefully you'll respond! Anyway,

Summing the voltages in the loop you get:

5000i+10000i=3\rightarrow15000i=3 \rightarrow i=\frac {3}{15000} =.0002 A=.2 mA=200 \mu A

voltage drop across the 5k resistor:

5000 \Omega * .0002A= 1v

so voltage at the terminal is 2v, which is correct according the answer sheet so then to find Vo we can use the fact that the current through the 2k resistor is the same as the current through the 8k resistor like so:

i_{1}=i_{2}

\frac {V_{s}-V_{1}}{R_{1}} = \frac {V_{1}-V_{0}}{R_{f}}

now solve this for Vo:

R_{f}(V_{s}-V_{1})=R_{1}(V_{1}-V_{0})

\frac {R_{f}}{R_{1}} (V_{s}-V_{1})= (V_{1}-V_{0})

-V_{0} = \frac {R_{f}}{R_{1}} (V_{s}-V_{1}) - V_{1}

V_{0} = -\frac {R_{f}}{R_{1}} (V_{s}-V_{1}) + V_{1}

V_{0} = -\frac {8}{2} (1) + 2 = -4+2 = -2v

this is also correct. Now I need to find Io and if we apply KCL to the node where Io points we should get:

i_{0} = \frac {V_{0}}{8000}+\frac {V_{0}}{4000} = \frac {-2}{8000}+\frac {-2}{4000} = -.00075 A = -.75mA=-750 \mu A

the answer is suppose to be -1mA which you could round to from -.75mA but since the other answers came out exactly this seems wrong to me. Can I get a little more help?

 

[Interesting]

In your io equation, it's not Vo/8k. You have a voltage missing. Vo/8K assumes that the voltage over your 8K resistor is Vo, but there's a voltage that isn't ground connected to your 8K resistor.

 

[Asphyxiated]

great,

i_{0} = \frac {V_{0}-V_{1}}{8000} + \frac {V_{0}}{4000} = -.001 a = -1 mA

thanks greatly!