changing the input of an inverting op-amp

Hi i have attached the figure of an inverting op-amp.

op-amp one of the aspect is
"The output does what it takes to make sure the two input voltages are the same".

I can understand how the output voltage Vout expression is brought about. Its quite simple and straightforward. I found this example online
Let's take an example; let R1 = 10k, R2 = 100k, Vin = 0.2v.
A = -100k / 10k = 10 * 0.2v so the output voltage is -2v.

I understand this calculation. We have determined the output voltage to be -2V. But from here on can anyone explain me with simple equation how the feed back works such that the 0.2V input voltage becomes 0 (since one of the input is grounded) and since op-amp tries to match both the inputs how is 0.2V reduced to 0V. I can understand only till finding the output voltage but not how this -2V is fed back such that the input is reduced to 0 matching the gnd. Can anyone kindly explain this part. Thank you!!!

dear for this type of configuration, you consider ground at joining point of two resisters (at the input of Amp).
then you calculate the output of the amplifier.
May I got your point, if not, ask again.

Thank you so much!!! I get it now. So itseems like more than the opamp the feedback resistor is mainly responsible for matching the inputs. But then again i though opamp amplifies the difference of the inputs. So op-amp gain is 4. so regardless of the gain if the difference is 0 then wer is the amplification? Won't the output also be 0. I thought the more the difference then more output (amplified) voltage. Unless if the inputs are not exactly matched and there exists a very minute difference.

preethi19 said:

Thank you so much!!! I get it now. So itseems like more than the opamp the feedback resistor is mainly responsible for matching the inputs. But then again i though opamp amplifies the difference of the inputs. So op-amp gain is 4. so regardless of the gain if the difference is 0 then wer is the amplification? Won't the output also be 0. I thought the more the difference then more output (amplified) voltage. Unless if the inputs are not exactly matched and there exists a very minute difference.

Click to expand...

The difference is NOT zero! However, this differential voltage is so small that we can neglect the error if we ASSUME it would be zero.
In reality it is in the µV range (Vdiff=Vout/Aol with Aol=open-loop gain).

Hi,

So op-amp gain is 4.

Click to expand...

A typical OPAMP gain is in the range of millions.
Therefore if you see 1V of output voltage .. then (calaculated back) this means 1uV of input voltage difference. About nothing. This is why we assume it to be zero.

Dont mix OPAMP gain (= open loop gain) with OPAMP circuit gain (= closed loop gain).

--> In the example of post#1:
* OPAMP gain is in the range of millions
* But OPAMP circuit gain is -10

Klaus

The opamp has an extremely high gain but the negative feedback reduces the circuit gain to a useable amount. Therefore the difference voltage between the inputs is extremely low.

Use Ohm's Law to calculate the output voltage since an opamp has an extremely low input current:
1) Input resistor to inverting opamp= 10k, feedback resistor= 100k.
2) Input voltage to the input resistor= +0.2V.
3) Current in the 10k resistor= 0.2V/10k= 20uA.
4) Current in feedback resistor is also 20uA so the output voltage= 20uA x 100k= -2V.

Can you pls explain the difference between OPAMP gain (open loop) and OPAMP circuit gain (closed loop). I understand op-amp open loop gain being in range of millions. But why can't a closed loop gain have that range too??? Also gain be a negative value like -10???

Open loop gain is the opamp without any negative feedback. The gain is maybe a million, the noise and distortion are very high and the opamp is amplifying its own input offset voltage. A frequency compensation capacitor inside reduces frequencies above about 10Hz. An opamp is never used open loop but a similar comparator is and it does not have a frequency compensation capacitor.

Closed loop gain is with negative feedback that reduces the gain to a useable amount, reduces noise and distortion and reduces its amplification of its input offset voltage. The high frequency response is extended.

It is usually wrong to say the gain is -10 but an opamp can attenuate (reduce) a signal and have negative gain. An opamp can be inverting so that the phase of its output is negative when compared to its input or it can be non-inverting then the input and output phases are the same (positive phase).

Thank you all for the help so far!!! I have one last question. Say if the non-inverting terminal is given 5V and not grounded and inverting terminal=0.2V, R1=10k and R2=100K. Can anyone pls let me know the expression for Vout. thank you!!!

preethi19 said:

Thank you all for the help so far!!! I have one last question. Say if the non-inverting terminal is given 5V and not grounded and inverting terminal=0.2V, R1=10k and R2=100K. Can anyone pls let me know the expression for Vout. thank you!!!

Click to expand...

In this case, you have two input voltages and you can apply the superposition rule:

Vout1=Vin1*(-R2/R1)
Vout2=Vin2*(1+R2/R1)
Vout=Vout1+Vout2.

Of course, this applies only as long as the opamp output is below the limits set by the supply voltages.

Example circuit:
Vout1=0.2*(-10)=-2V and Vout2=5*(1+10)=55V.
Vout=+53V.
That means: The output will be at the pos. limit set by Vcc.

Hi,

usually in an amplifier circuit the non-inverting voltage is known somehow.
In your case it is 5V

An OPAMP (when poroperly working) regulates the output in a way that the inverting input is the same as the non-invcerting input.
Therefore you can consider the inverting input = non-inverting input = 5V.

Now look at R1.
At one side you have 0.2V and t the other there is 5V.
--> The voltage across R1 = 5V-0.2V = 4.8V
--> the current through R1 is: I = U/R = 4.8V / 10k = 480uA

The current into the inverting input is considered to be zero, therefore the current thorugh R2 is the same as through R1.
--> I_R2 = I_R1 = 480uA
--> The voltage across R2 is: U = R * I = 100k x 480uA = 48V

Now you know the voltage at one side of R2 (= inverting-input-voltage) = 5V, and the voltage across R2 = 48V, therefore the voltage at the other side of R2 (=OPAMP output) = 5V +48V = 53V.

****
Generally:

U_OUT = U_+in + R2 * ( U_+in - U_in) / R1

here:
= 5V + 100k * (5.0V - 0.2V) / 10k
..
= 53V

Klaus

Oh got it!!! So for a 5V input at both the +ve and -ve input terminals we obtain output of 53V. Then can someone let me know what is the expression of the gain of the opamp in this case?

53/5=10.6 ---> What is the gain expression. Becoz by superposition rule if i add
(-Rf/Rin) + (1+Rf/Rin) i don't get 10.6 so i can see this is the wrong way to bring about the derivation of the gain. Can you please clear this?? Thanks a lot again!!!

Hi,

No. It seems you still mix open_loop_gain with closed_loop_gain.

Indeed one input is at 5V and the other is very, very close to 5V. Therefore we consider it to be 5V.
(Theoretically it is 4.999947V ... at an open_loop_gain of 1million).
But when you consider the input_offset_range to be +/-2mV ... it is somewhere inbetween 4.997947V and 5.001947V.
So nobody cares about the 53uV when the offset is +/-2mV = 4000uVpp.
We simply consider it to be 5V.)

If both inputs were exactely at 5V, then the differential input voltage is 0V. And that is what cares: The difference between +In and -In.

****
When +In is connected to 5V, then then input voltage and the output voltage of the OPAMP_circuit refer to those 5V.
--> U_In = 0.2V - 5V = -4.8V
--> U_out = 53V - 5V = 48V
OPAMP_circuit_gain = U_out/U_in = 48V/(-4.8V) = -10
OPAMP_circuit_gain = - R_fb / R_in = - 100k / 10k = -10

Klaus