# OP-AMP question?

Discussion in 'Homework Help' started by error404, Apr 25, 2010.

1. ### error404 Thread Starter New Member

Apr 25, 2010
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How can i find Vo voltage and Io current? This question is very complicated for me. Please help me.

2. ### Wendy Moderator

Mar 24, 2008
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The op amp will likely go to full voltage on either power supply rail, so it depends on the component being used. Very few op amps anywhere near the power supply rails.

You don't have a component for the op amp, you don't have power supply definitions.

In short, as stated the question is unanswerable.

3. ### retched AAC Fanatic!

Dec 5, 2009
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Being a homework question, it is either ideal or a 341 .

4. ### hgmjr Retired Moderator

Jan 28, 2005
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Show us how you think the solution to the problem should be set up.

hgmjr

5. ### mik3 Senior Member

Feb 4, 2008
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69
Find V- (inverting input voltage with respect to ground) and V+ (non inverting input voltage with respect to ground).

Then equate V- and V+ (ideal op amp) and you should find Vo with respect to Vin.

6. ### hgmjr Retired Moderator

Jan 28, 2005
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218
Using the hint Mik3, see if you can take a stab at solving the problem and post it here.

hgmjr

7. ### mik3 Senior Member

Feb 4, 2008
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69
V+=(Vo*12k)/(12k+8k)=0.6*Vo , voltage divider rule

current through 4k and 16k is: assume current flows from Vo to 6V

I=(Vo-6)/(4+16)k=(Vo-6)/20k

thus

V-=6+I*4k=6+4k(Vo-6)/20k=6+(Vo-6)*0.2=6+0.2*Vo-1.2=0.2*Vo+4.8

assuming V-=V+

0.6*Vo=0.2*Vo+4.8

thus

0.4*Vo=4.8

thus

Vo=12V

Io=12V/10k=1.2mA

8. ### hgmjr Retired Moderator

Jan 28, 2005
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218
Millman's Theorem Solution.......

$\small V_-=V_+$

$\frac{\frac{6}{4K}+{\frac{V_o}{16K}}}{\frac{1}{4K}+{\frac{1}{16K}}}=\frac{\frac{V_o}{8K}}{\frac{1}{8K}+{\frac{1}{12K}}}$

$\frac{6(16K)+4K(V_o)}{16K+4K}=\frac{12K(V_o)}{12K+8K}$

$\frac{6(16K)+4K(V_o)}{20K}=\frac{12K(V_o)}{20K}$

$\normalsize {6(16K)+4K(V_o)}={12K(V_o)}$

$\normalsize {8K(V_o)}={6(16K)}$

$\large {V_o}={\frac{6(16K)}{8K}}$

$\large {V_o}={6(2)}$

$\large {V_o}=12V$

****************************************

$V_+=\frac{12K(12)}{12K+8K}$

$V_+=\frac{12K(12)}{20K}$

$V_+=7.2V$

****************************************

$I_o=\frac{V_o-V_-}{16K}+\frac{V_o-V_+}{8K}+\frac{V_o}{10K}$

$I_o=\frac{12-7.2}{16K}+\frac{12-7.2}{8K}+\frac{12}{10K}$

$I_o=\frac{12-7.2}{16K}+\frac{12-7.2}{8K}+\frac{12}{10K}$

$I_o=0.0003+0.0006+0.0012=0.0021A$

hgmjr

9. ### mik3 Senior Member

Feb 4, 2008
4,846
69
Hgmjr,

You are right about Io. I thought Io is the current through the 10k resistor only.

Apr 25, 2010
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