OpAmp Question

âShow My Homewrok - Circuit x

1. Find V0 when $Va = 2V$, $Vb =2.25V$, $V c = 4V$ and $V d = 4.5 V$

2. If Va, Vb and Vd are held constant then what is the minimum value of Vc that shall not saturate the op amp?

3. If Va, Vb and Vd are held constant then what is the maximum value of Vc that shall not saturate the op amp?

Answers

The circuit presented in the figure is a summing amplifier on the negative and positive inputs.

Let Ra =20 k, Rb=18 k , Rc=30 k and Rd=20 k be the resistances in series with the inputs Va, Vb,

Vc and Vd, Rf the resistance from the output to the negative input of the amplifier (Rf =180 K)

and Rg the resistance from the positive input of the amplifier to the ground (Rg =20 K)

and $R_{minus} = Ra parallel Rb =20 K parallel 18 K = 9.47 K$

Va is seen at the output as

$Va_{out} = -Va*Rf/Rb$

Analog for Vb

$Vb_{out} = – Vb*Rf/Rb$

Vc is seen at the output as

$Vc_{out} =Vc*(Rg/(Rc+Rg))*(1+Rf/R_{minus})$

(if it were only the negative reaction and no resistors on the positive IN it would have been

$Vc_{out} = Vc*(1+Rf/R_{minus})$. Because of the negative reaction Vc gets divided by the branchformed by Rc and Rg and we get the result above)

Analog for Vd

$Vd_{out} =(Vd (Rg/(Rd+Rg))*(1+Rf/R_{minus})$

The total output voltage is

$Vo =-Rf*(Va/Ra+ Vb/Rb) + Vc(Rg/(Rc+Rg))*(1 +Rf/R_{minus}) +$

$+ Vd(Rg/(Rd+Rg))(1+Rf/R_{minus})$

the first 2 terms are the amplified voltages Va and Vb the 3rd term is the amplified voltage Vc and the last term is the amplified voltage Vd

$Vo = -Va*180/20 – Vb*180/18 + Vc*(20/50) *(1+180/9.47) +$

$+Vd*(20/40)*(1+180/9.47)$

$Vo =-9*Va -10*Vb +8*Vc +10*Vd$

$Vo = -9*2 -10*2.25 +8*4 +10*4.5 = 36.5 V$

The op amp is saturated

$Vo = +4 +8*Vc$

The minimum value for Vc is when Vo =0

$Vc =-4/8 = -0.5 V$

The maximum value for Vc is when Vo =40 V

$Vc =(40-4)/8 = 4.5 V$