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# Op-amp circuit

Discussion in 'General Electronics Chat' started by Dritech, Apr 6, 2012.

1. ### Dritech Thread Starter Distinguished Member

Sep 21, 2011
826
6
Hi all,

I found the attached circuit of an op-amp. I built the circuit and it worked, but i have difficulties understanding how the circuit really operates. Can someone explain to me how it works and what is the use of each resistor in the circuit??

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2. ### panic mode AAC Fanatic!

Oct 10, 2011
1,773
538
R1 and R2 are voltage divider forming reference point Vg.
Vg=5V*R2/(R1+R2)
Vg=0.054V

Approximately, because current through RG is neglected for now.

Bypass capacitors are used to eliminate any noise by providing low impedance paths to DC ground (capacitors act as open circuit for DC but low resistance for fast changing signals like noise).

Vin is test voltage used as input to the circuit. Note: opamps are assumed to have no current flowing in or out of inputs and have same voltage on both inputs. this means that we have Vin at noninverting input (+) but also, same voltage is at inverting input (-).

RG is input resistor for inverting input. It is used as sensing resistor whisg defines current from Vg towards input (-).
Since no current (ideally) goes in or out of input of the opamp, the whole current that comes through RG continues through RF to output. Ratio of RF and RG defines gain (amplification) of opamp.

situation at opamp inputs (any voltage difference) causes opamp output to change. this change is rapid and brings circuit in balance mentioned earlier - it does everything it can to ensure that voltage difference between inputs is zero.
in this case (we assumed that current through RG was from left to right) output would try go to negative. there just isn't much room for that since Vg is tiny.
hence normal operation for this circuit is when Vin is greater than Vg, current would flow thrrough RG (and RF) from right to left. and the output would be at some level greater than Vin.

in fact we can calculate this as
(Vo-Vin)/RF=(Vin-Vg)/RG
then we get
Vo-Vin=(Vin-Vg)*(RF/RG)
Vo=Vin(1+RF/RG)-Vg*RF/RG
you already know all the values (RF, RG, Vg) so you get nice linear relationship between Vo and Vin.

gain is (1+RF/RG)=(1+9)=10

so output voltage is

Vo=10*Vin-0.486

since the circuit is powered from 5V, Vo cannot be bigger than 5V, therefore maximum Vin is 0.55V:

5=10*Vin-0.5
5.5=10*Vin
Vin=5.5/10=0.55V

similarly, output cannot go below zero

0=10*Vin-0.5
0.5=10*Vin
Vin=0.5/10=0.05V

So Vin must be in range 0.5...0.55V

RL is just a load resistor.

Last edited: Apr 6, 2012
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3. ### Dritech Thread Starter Distinguished Member

Sep 21, 2011
826
6
Thanks for explaining in details. Really appreciated and what is the use of RL in the circuit?

4. ### crutschow Expert

Mar 14, 2008
23,116
6,854
RL helps pull the output voltage down to 0V when the input is 0V. The outputs of many op amps operated from a single supply do not go completely to 0V without such a pull-down resistor.

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