- #1

peasngravy

- 72

- 6

- Homework Statement
- FIGURE 1 shows two amplifier circuits, each of which has one unknown

voltage (shown highlighted). Select, from TABLE A, the most

appropriate value for the unknown voltages for each circuit. Assume that

the op-amps are ideal and that the magnitude of their output voltages is

less than their maximum peak output voltage swing (VOM).

- Relevant Equations
- V+ = V1/(R+10R)

(V2 - V-)/R = -(Vo - V-)/10R

Hi - I have made an attempt at both of these circuits, but I am not 100% sure I am correct. Would it be possible to check if I have this right, and if not, please give me some pointers as to where I have gone wrong? Thanks

Given data:

Vo = 5V

I have called the voltage feeding into the non-inverting input V1 = 0.5V

voltage feeding into the inverting input = V2

(V2 - V-)/R= -(Vo - V-)/10R

V2 - V- =(-Vo + V-)/10

10(V2 - V-)-V- = -Vo

10V2 - 11V- = -Vo

Plug in values

(10*0.5) - 11V- = -5

-11V- = -10

V- =10/11 = 0.909V

V+ = V- = 0.909V

V+ =V1/(10+1)

V1 = V+ * 11

V1 = 0.909*11 = 10V

V+ =Vo/(R+R)

V+=12/2=6V

V+ = V- =6V

Due to short from inverting input to output, output voltage (V) is 6V

**First op-amp**Given data:

Vo = 5V

I have called the voltage feeding into the non-inverting input V1 = 0.5V

voltage feeding into the inverting input = V2

*For non inverting input V+*= V1/(1+10)*For inverting input V-*:(V2 - V-)/R= -(Vo - V-)/10R

V2 - V- =(-Vo + V-)/10

10(V2 - V-)-V- = -Vo

10V2 - 11V- = -Vo

Plug in values

(10*0.5) - 11V- = -5

-11V- = -10

V- =10/11 = 0.909V

*As it is an ideal op-amp:*V+ = V- = 0.909V

*Therefore:*V+ =V1/(10+1)

V1 = V+ * 11

V1 = 0.909*11 = 10V

**Second op-amp***At non-inverting input*:V+ =Vo/(R+R)

V+=12/2=6V

*As it is an ideal op-amp*:V+ = V- =6V

Due to short from inverting input to output, output voltage (V) is 6V