Answer to Question #124908 in Electric Circuits for NNNNN

Question #124908

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 ?)


1
Expert's answer
2020-07-03T10:05:08-0400

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\n\\begin{aligned}\n\\small i_1 &= \\small \\frac{V_1-V_p}{R_1} = \\frac{V_1 -0}{R =} = \\frac{V_1}{R}\\\\\n\\small i_2 &= \\small \\frac{V_2}{R\/3} = \\frac{3V_2}{R} \\\\\n\\small i_3 &= \\small \\frac{V_3}{R\/2} = \\frac{2V_3}{R}\\\\\n\\small i_4 &= \\small \\frac{2V_4}{R}\\\\\n\\small i_5 &= \\small \\frac{4V_5}{R}\n\\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\n\\begin{aligned}\n\\small i &= \\small \\frac{V_p-V_o}{R}= \\frac{0-V_o}{R}=\\frac{-V_o}{R} \\cdots\\cdots(2)\n\\end{aligned}"

  • By (1) = (2),

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


1). Only V1

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

2) Only V2

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

3) Only V3

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

4) Only V4

  • Same results as "only V3"


5) Only V5

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


6) When all voltages equal to Vi

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


Note: double arrows are to show the alternating current.



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS