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Unknown voltage of a Op amp

Offshore again

Jun 24, 2016
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Hi,
Hoping for a prompt in the right direction regarding a unkown voltage on a Non Inverting Op Amp.
Reading through my notes i understand how the resistors R3 and R2 set up a potential divider which supply the Inverting input on the Amp.
And the fact that neither input is connected to Earth means the Virtual earth principle cannot be used.

My course notes show that:
V−=V+ because the op-amp is ideal and V+=Vin

So is this as simple as as saying V- =100mV ???
upload_2016-6-24_13-12-42.png

Any help would be greatly appreciated
 

Offshore again

Jun 24, 2016
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Thanks Harald,

I assume its the same course that you mention (its a distance learning course) but not sure how many or who the other people are.

The question threw me a bit as it seems to easy....so thanks for the re-assurance
 

LvW

Apr 12, 2014
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The question threw me a bit as it seems to easy....so thanks for the re-assurance

The principle behind this simple answer is as follows:
As long as the opamp is operated in its linear region (and negative feedback, as in your circuit, enlarges this region considerably) the input differential voltage of a real opamp is in the µV range (because of the very large open-loop gain Aol).
During calculation of voltages and currents within such a circuit this small differential voltage can be neglected if compared with all other voltages. That means: We can treat the opamp as an IDEAL unit with infinite gain and (consequerntly) zero input voltage.Hence V+ = V- (V+ - V-=0) and V+=0 for V-=0 (virtual ground principle).
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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one should also point out the voltage drop across R1 for a perfect op-amp would be...?
 

Offshore again

Jun 24, 2016
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In an ideal op amp if R1 is amp impedance then it should be near infinite resistance so therefore the full voltage would be dropped.....is that correct?
 

LvW

Apr 12, 2014
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In an ideal op amp if R1 is amp impedance then it should be near infinite resistance so therefore the full voltage would be dropped.....is that correct?

In your circuit, R1 is an external ohmic resistor of - perhaps - som kOhms.
 

Offshore again

Jun 24, 2016
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In your circuit, R1 is an external ohmic resistor of - perhaps - som kOhms.

So the voltage drop across the resistor would be proportional to its ohmic value.
If it were a variable resistor (POT) then it would adjust the gain......
Am i on the right lines or going down the wrong track??
 

LvW

Apr 12, 2014
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So you can find the answer to your questions (post#6 and 8) by yourself?
 

Offshore again

Jun 24, 2016
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So because there is no current (ideal OP-Amp) then the full 100mV is dropped across the resistor.
And with V- = V+ this is why the Unknown voltage V (in original post) is 100mV
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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So because there is no current (ideal OP-Amp) then the full 100mV is dropped across the resistor.
And with V- = V+ this is why the Unknown voltage V (in original post) is 100mV

Reconsider this in light of V = IR
 

Offshore again

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If there is no current then the Voltage drop must be zero across the resistor
V=IR If I is zero then the V must be zero
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Be careful with what you say. "Voltage" and "Voltage drop" are two different things.
 

Offshore again

Jun 24, 2016
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I have done a bit more digging and think I see where we are going.
Using Kirchoffs Voltage laws
Vin (100mv)-V1 (volt drop across resistor)= V+

So therefore volt drop across resistor is 100mV and V+ = 0V
 
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