Answer to Question #124908 in Electric Circuits for NNNNN

Question

Design an adder circuit using op-amp to satisfy following equation:

-Vo = V1 + 3 V2 + 2 V3 + 2V4 + 4V5

The supply voltage of the op-amp is +/- 13V and output saturation voltage is +/- 12 V.

The circuit should always operate in its linear range. What are the minimum- maximum voltage values of V1, V2, V3, V4 and V5 when only one of them is connected to the input at a given time (i.e. other input voltages are zero)? What is the value of maximum voltage Vi if V1 = V2 = V3 = V4 = V5 = Vi ?)

Answer 1

Explanations & Calculations

• Refer to the figure attached

• Feed back resistor is (which connects the inverting terminal to the output) is also the same value R.
• $V_p$ is zero as the non inverting input is grounded.
• So considering all the inputs, currents through each could be found as,

$\qquad\qquad \begin{aligned} \small i_1 &= \small \frac{V_1-V_p}{R_1} = \frac{V_1 -0}{R =} = \frac{V_1}{R}\\ \small i_2 &= \small \frac{V_2}{R/3} = \frac{3V_2}{R} \\ \small i_3 &= \small \frac{V_3}{R/2} = \frac{2V_3}{R}\\ \small i_4 &= \small \frac{2V_4}{R}\\ \small i_5 &= \small \frac{4V_5}{R} \end{aligned}$

• Now all these currents pass through the feed back resistor —R— to the output circuit. Therefor, $i = i_1+i_2+i_3+i_4+i_5 \cdots\cdots(1)$

• Considering the feedback resistor (V = iR),

$\qquad\qquad \begin{aligned} \small i &= \small \frac{V_p-V_o}{R}= \frac{0-V_o}{R}=\frac{-V_o}{R} \cdots\cdots(2) \end{aligned}$

• By (1) = (2),

$\qquad\qquad \begin{aligned} \small \frac{-V_o}{R}&= \small \frac{V_1}{R} + \frac{3V_2}{R} +\frac{2V_3}{R} + \frac{2V_4}{R} +\frac{4V_5}{R}\\ \small \bold{-V_o} &= \small \bold{V_1 +3V_2 +2V_3 +2V_4 +4V_5} \end{aligned}$

1). Only V₁

$\qquad\qquad \begin{aligned} \small V_1 &= \small -V_o \cdots(\text{gain = 1})\\ \small V_{1-min} & =\small -(12V) = -12V\\ \small V_{1-max} & =\small -(-12V) = 12V \end{aligned}$

2) Only V₂

$\qquad\qquad \begin{aligned} \small V_2 &= \small \frac{-V_o}{3}\cdots(\text{gain = 3})\\ \small V_{2-min}&= \small \frac{-(12)}{3} = -4V\\ \small V_{2-max}&= \small \frac{-(-12)}{3} = 4V \end{aligned}$

3) Only V₃

$\qquad\qquad \begin{aligned} \small V_3 &= \small \frac{-V_o}{2} \\ \small V_{3-min} &= \small \frac{-(12)}{2} = -6V\\ \small V_{3-min} &= \small \frac{-(-12)}{2} = 6V \end{aligned}$

4) Only V₄

• Same results as "only V₃"

5) Only V₅

$\qquad\qquad \begin{aligned} \small V_5 &= \small \frac{-V_o}{4}\\ \small V_{5-min}&= \small \frac{-(12)}{4} = -3V\\ \small V_{5-max} &= \small \frac{-(-12)}{4}= 3V \end{aligned}$

6) When all voltages equal to V_i

$\qquad\qquad \begin{aligned} \small -V_o &=\small 12V_i\\ \small V_{i-max} &= \small \frac{-V_{o-min}}{12}\\ &= \small \frac{-(-12V)}{12}\\ &= \small \bold{1V} \end{aligned}$

Note: double arrows are to show the alternating current.