Question

3B) Design Op-Amp circuit to give Vo= - 2V₁ - 3V2

Answer 1

To design an op-amp circuit that produces an output voltage
`\displaystyle V_(o) =-2V_(1) -3V_(2)`, we can utilize an inverting amplifier configuration. The inverting amplifier has a negative gain, which aligns with the given equation for
`\displaystyle V_(o)`. Here's how you can design the circuit:

1. Connect the inverting terminal (marked with a negative sign) of the op-amp to ground (0V).

2. Connect the non-inverting terminal (marked with a positive sign) of the op-amp to the input signal
`\displaystyle V_(1)`.

3. Connect a resistor
`\displaystyle R_(1)` between the inverting terminal and the output terminal of the op-amp.

4. Connect a resistor
`\displaystyle R_(2)` between the output terminal and the inverting terminal of the op-amp.

5. Connect the input signal
`\displaystyle V_(2)` to the junction between
`\displaystyle R_(1)` and
`\displaystyle R_(2)`.

6. Connect the output terminal of the op-amp to a load or further circuitry, creating
`\displaystyle V_(o)`.

By applying the voltage divider rule, we can derive the relationship between
`\displaystyle V_(o)`,
`\displaystyle V_(1)`, and
`\displaystyle V_(2)`. The voltage at the inverting terminal (
`\displaystyle V^(-)\`) is given by:

`\displaystyle V^(-) =(R_(2))/(R_(1) +R_(2)) V_(1) +(R_(1))/(R_(1) +R_(2)) V_(2)`

Since the op-amp is assumed to have ideal characteristics (infinite gain), the output voltage
`\displaystyle V_(o)\` is equal to the voltage at the inverting terminal (
`\displaystyle V^(-)\`) multiplied by the negative gain of the circuit (-2-3 = -5):

`\displaystyle V_(o) =-5V^(-)`

Substituting the value of
`\displaystyle V^(-)\`, we have:

`\displaystyle V_(o) =-5\left((R_(2))/(R_(1) +R_(2)) V_(1) +(R_(1))/(R_(1) +R_(2)) V_(2)\right)`

Simplifying this equation, we get:

`\displaystyle V_(o) =-(5R_(2))/(R_(1) +R_(2)) V_(1) -(5R_(1))/(R_(1) +R_(2)) V_(2)`

By comparing this equation with the given equation for
`\displaystyle V_(o)` ([-2V₁ -3V2]), we can deduce the values of
`\displaystyle R_(1)` and
`\displaystyle R_(2)`:

`\displaystyle -(5R_(2))/(R_(1) +R_(2)) =-2`

`\displaystyle -(5R_(1))/(R_(1) +R_(2)) =-3`

Solving these equations, we find:

`\displaystyle R_(1) =(R_(2))/(2)`

Substituting this value into one of the equations, we can determine
`\displaystyle R_(2)`:

`\displaystyle -(5R_(2))/((R_(2))/(2) +R_(2)) =-2`

Simplifying:

`\displaystyle -(5R_(2))/((3R_(2))/(2)) =-2`

`\displaystyle -(10R_(2))/(3R_(2)) =-2`

`\displaystyle -(10)/(3) =-2`

Hence, the equation doesn't hold true for any value of
`\displaystyle R_(2)`. It seems there is no valid solution to meet the given equation
`\displaystyle V_(o) =-2V_(1) -3V_(2)` using an inverting amplifier configuration.