Question

Op-Amp Voltage Adder / Summing Amplifier R13 =R R14=Rf 2k<TEXT> 1k BAT7 <TEXT> 9V <TEXT> POT A RV5 R10 =R1 R1=R2=R3=R=1K Rf=2K Vout=Va+Vb+Vc=1.43V <TEXT> 309 РОТ В RVO 1k <TEXT> <TEXT> R11=R21 R11=R2A NU6 POTC RATA <TEXT> R12=R31 Vout 1-46 9V <TEXT> HRV7 741 + 1k <TEXT> <TEXT +0.09 <TEXT +0.67 Volts +0.67 Volts +1.44 Voits Volts = Vout=(1+(Rf/R))*((Va/R1)+(Vb/R2)+(Vc/R3)) ((1/R1)+(1/R2)+(1/R3)) E VIDEOS

Can you please give the hand calculation of this circuit with details when its Non-Inverting and Inverting

Answer 1

Ans.

Given below is a simplified diagram of the same above circuit ,excluding the voltage sources and potentiometers (I have drawn the opamp inverted please note that):

1K 2K Vb E M / Vout R3 1K

When there is a feedback in an opamp the voltages in + (Non inverting pin) and - (Inverting pin) will be equal.

Thus V1=V2 :

As no current goes into the Non- inveting and inverting pins we can write node equation at V1 and V2 to find each voltage and equate it thus:

Node equation at V1:

$\frac{V_{1}-V_{a}}{R1}+\frac{V_{1}-V_{b}}{R2}+\frac{V_{1}-V_{c}}{R3}=0\\\\ \Rightarrow V_{1}=\frac{\frac{V_{a}}{R1}+\frac{V_{b}}{R2}+\frac{V_{c}}{R3}}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}}\rightarrow Eqn1$

Node equation at V2:

$\\\frac{V_{2}-0}{R}+\frac{V_{2}-V_{out}}{R_{f}}=0\\\\ V_{2}=\frac{\frac{V_{out}}{R_{f}}}{\frac{1}{R}+\frac{1}{R_{f}}}\\ \\\Rightarrow V_{2}=\frac{V_{out}}{1+\frac{R_{f}}{R}} \rightarrow Eqn2$

Equating Eqn 1 and 2 we get :

$\frac{V_{out}}{1+\frac{R_{f}}{R}}=\frac{\frac{V_{a}}{R1}+\frac{V_{b}}{R2}+\frac{V_{c}}{R3}}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}}\\ \Rightarrow V_{out}= (1+\frac{R_{f}}{R})(\frac{\frac{V_{a}}{R1}+\frac{V_{b}}{R2}+\frac{V_{c}}{R3}}{\frac{1}{R1}+\frac{1}{R2}+\frac{1}{R3}})\rightarrow Eqn3$

Thus we get the output equation. now put R1=R2=R3=1K=R, Rf=2K in Eqn 3 we get:

$V_{out}=V_{a}+V_{b}+V_{c} \rightarrow Eqn4$

Now to find Va ,Vb and Vc look into the initial figure ,we can fing Va by using Ohms law the only part of Va is given below :

9V 1K*(100-50)%=500 0hm Va 1K*50%=500 Ohm --

Thus Va is : $V_{a}=9-\frac{500*18}{500+500} = 0V$

in same way we can find Vb and Vc .

Circuit of Vb is :

9V OV 1K*(100-54)%=460 Ohm Vb 9V 1K*54%=540 Ohm

Thus Vb is : $V_{b}=9-\frac{540*18}{460+540} = 0.72V$

Circuit of Vc is :

4* 9V 1K*(100-54)%=460 Ohm Vc 9V 1K*54%=540 Ohm

Thus Vc is : $V_{c}=9-\frac{540*18}{460+540} = 0.72V$

Putting Va , Vb ,and Vc in Eqn 4 we get :

If any doubts or errors please comment.