# Find the voltage gain of operational amplifier

Joined Feb 3, 2016
42
Find the voltage gain and Io

Vi=V+=V-
I1=(V-)/R1
I1=0.5mA

Node V-
Vi-0/R1+Vi-Va/R2=0
Va=-50mV

NodeA
[Va-V-/R2] + Va-0/R3 + Va-Vb/R4 =0
Vb=-0.2V

NodeB
Vb-Va/R4 + Vb-0/R5 +Vb-VO/R6=0
Vo=-0.55V

Av= -Vo/Vi
=-0.55/50m
= -11V/V

Io= Vo/RL
=-0.55/100
=-55mA

thank for the help

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#### Jony130

Joined Feb 17, 2009
5,212
And Vin is ?? Also there is some thing wrong with your answer because voltage gain cannot be negative.
Also notice that if Vin = 1V the voltage at V- is also equal to 1V and from there we can find Va using only KVL and KCL

Va
= VR1 + VR2 = Vin + IR1*R2 = 1V + (1V/100Ω)*200Ω = 3V.

Vb = VR3 + VR4 = Va + VR4 = Va + (IR3 + IR2)*R4 = 3V + (3V/200Ω + 1/100Ω)*200Ω = 8V

And if you use the similarly approach we can find that Vo/Vin = 21V/V

Joined Feb 3, 2016
42
Vin is 50mV

I use KCL in the above calculation
current goes into the resistor R2 =current out the resistor R3 +R4
Vin-Va/R2= VA/R3+Va-VB/R4

But why
Va = VR1 + VR2 = Vin + IR1*R2 = 1V + (1V/100Ω)*200Ω = 3V.
and
Vb = VR3 + VR4 = Va + VR4 = Va + (IR3 + IR2)*R4 = 3V + (3V/200Ω + 1/100Ω)*200Ω = 8V

#### Jony130

Joined Feb 17, 2009
5,212
Vin is 50mV
fir Vin = 50mV the voltage at VA node is equal to:
Voltage drop across R1 + voltage drop across R2 = VA = VR1 + VR2
Voltage drop across R1 = Vin because V+ = V-
Voltage drop across R2 = current through R2 resistor times R2 resistance. And the current through R2 is equal to Vin/R1 because R2 is in series with R1. So, the same current must be flowing through these two resistors.
Therefore VA = VR1 + VR2 = 50mV + 50mV/100Ω *200Ω = 150mV

And VB node is equal to VR3 + VR4, Also notice that VR3 = VA and VR4 = I*R4 = (IR2 + IR2)*R4 do you agree with this ?

Joined Feb 3, 2016
42
sorry, but I still don't understand...
I find an example which is similar to this question

short R5
V+=V-
0-Vi/R1 + 0-Va/R2=0
Va=-0.5V

Then, I follow the same method to find the Va
Vi-0/R1+Vi-Va/R2=0
Va=-50mV
What's problem here?

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#### Jony130

Joined Feb 17, 2009
5,212
Then, I follow the same method to find the Va
Vi-0/R1+Vi-Va/R2=0
Va=-50mV
What's problem here?

$$\frac{Vi}{R1}+\frac{(Vi-Va)}{R2}=0$$

Try solve this for Va

$$\frac{50mV}{100\Omega}+\frac{(50mV-Va)}{200\Omega}=0$$

Joined Feb 3, 2016
42

$$\frac{Vi}{R1}+\frac{(Vi-Va)}{R2}=0$$

Try solve this for Va

$$\frac{50mV}{100\Omega}+\frac{(50mV-Va)}{200\Omega}=0$$
Thank you
I think I forgot the negative sign , lol
Va sholud be 150mV
Then I can use the same way to find Vb,=0.4V ,Vo=1.05V
then I get the Av=21, Io=10.5mA

#### Jony130

Joined Feb 17, 2009
5,212
Thank you
I think I forgot the negative sign , lol
Va sholud be 150mV
Then I can use the same way to find Vb,=0.4V ,Vo=1.05V
then I get the Av=21, Io=10.5mA
Very good, but you still do not have the right answer for Io. I can assure you that Io is not equal to 10.5mA . By Io I mean the op amp output current. 10.5mA is a current that is flowing through R7 resistor.

Joined Feb 3, 2016
42
May be Io is equal to I7-I6 ?
(Io+I6=I7)
And I6=Vb-Vo/R6
I know the I6 and I7 ,then I get Io?

#### Jony130

Joined Feb 17, 2009
5,212
Why you think that Io is a difference between I7 and I6 ?

Joined Feb 3, 2016
42
Because Io and the current flowing out the resistor R6 are flowing into node Vo?
So I guess Io which is equal to I7-16

#### Jony130

Joined Feb 17, 2009
5,212
Because Io and the current flowing out the resistor R6 are flowing into node Vo?
Note that for Vin = 50mV we have Vo = 1.05V and Vb = 0.4V. So in which direction IR6 current is flowing ?

Joined Feb 3, 2016
42
Note that for Vin = 50mV we have Vo = 1.05V and Vb = 0.4V. So in which direction IR6 current is flowing ?
Thank you
I get it
So Io is Vo/R7 +Vo-Vb/R6 ?

#### Jony130

Joined Feb 17, 2009
5,212
So Io is Vo/R7 + (Vo-Vb)/R6
Yes, that right but also do not forget about parentheses.

#### The Electrician

Joined Oct 9, 2007
2,814
Node V-
(Vi-0)/R1+(Vi-Va)/R2=0
Va=-50mV

NodeA
[(Va-V-)/R2] + (Va-0)/R3 + (Va-Vb)/R4 =0
Vb=-0.2V

NodeB
(Vb-Va)/R4 + (Vb-0)/R5 +(Vb-VO)/R6=0
Vo=-0.55V
You had the same problem with lack of parentheses in post #1. The expressions missing parentheses happen to be such that if they are evaluated left to right without any attention paid to operator precedence, you get the right answer. Some calculators behave this way and perhaps that's what you used when evaluating those expressions, but you may not be lucky enough to have expressions that evaluate properly left to right next time. Better to use parentheses.

For extra credit, solve the problem for an opamp with a finite gain of 10.