--- title: "Square root circuit and √((V1V2)²+(V3V4)²)" subject: "Advanced Electronics" chapter: 6 key: "semester_4/advanced_electronics:square-root-vector-magnitude" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/square-root-vector-magnitude site: BCT Engineering --- # Square root circuit and √((V1V2)²+(V3V4)²) Advanced Electronics · Chapter 6: Log-Antilog Circuit Applications - Asked in: 2072 Chaitra, 3+4 marks; 2071 Shrawan, 3+4 marks (asked 2 times) - Video: [Circuit Design Numerical, Log Anti-Log Amp || 2072 Chaitra || See question in Description](https://www.youtube.com/watch?v=9ETRV6-OuAM) — I Love Electronics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/square-root-vector-magnitude - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-6 ## Question Construct a circuit that generates square root of an input voltage and derive its input-output relationship. Draw a circuit that produces the output voltage V0 = √((V1·V2)² + (V3·V4)²). ## Short answer Use a log amp, an inverting stage of gain -1/2, and an antilog amp so V0 = exp(0.5 ln Vi) = sqrt(Vi). For V0 = sqrt((V1V2)^2 + (V3V4)^2), form each square with log-antilog stages, sum them, then pass the sum through a square-root block. Inputs must be positive. ## Answer #### Square-root circuit Taking a square root means multiplying the logarithm by ½: √Vi = exp(½ ln Vi). **Building blocks used** (all voltages in volts, Vref = 1 V, K = scale factor such as VT or 1 V/decade): - Log amp (L): output = −K ln(Vin), for Vin > 0 - Antilog amp (E): output = exp(Vin/K) - Inverting amp of gain −m: output = −m × input ```text R/2 +---/\/\/---+ | | Vi -> [LOG] --V1'--[R]---+--(-)\ | >---+--> V2' GND--(+)/ V2' -> [ANTILOG] -> Vo ``` **Derivation:** 1. Log amp: V1' = −K ln Vi 2. Inverting amplifier with gain −(R/2)/R = −½: ```text V2' = −½ × (−K ln Vi) = (K/2) ln Vi ``` 3. Antilog amp: ```text Vo = exp(V2'/K) = exp(½ ln Vi) = Vi^(1/2) ``` **Result:** Vo = √Vi. With a general reference, Vo = √(Vref·Vi). For example, Vi = 9 V gives Vo = 3 V (Vref = 1 V). The input must be positive. #### Circuit for V0 = √((V1·V2)² + (V3·V4)²) Plan: form (V1V2)² and (V3V4)² with log–antilog stages, add them, then take the square root. ```text V1->[LOG]-+ +->[summer, gain -2]->[ANTILOG]->(V1V2)^2 -+ V2->[LOG]-+ | [SUMMER] <----+ V3->[LOG]-+ ^ | +->[summer, gain -2]->[ANTILOG]+ | S V4->[LOG]-+ (V3V4)^2 v S -> [LOG] -> [gain -1/2] -> [ANTILOG] -> V0 ``` **Step by step:** 1. Log amps give −K ln V1, −K ln V2, −K ln V3 and −K ln V4. 2. **Inverting summer with gain 2** (input resistors R, feedback 2R): ```text Va = −2(−K ln V1 − K ln V2) = 2K ln(V1V2) ``` The antilog gives exp(Va/K) = **(V1V2)²**. In the same way the second branch gives **(V3V4)²**. 3. A **non-inverting summer** (or an inverting summer followed by an inverter) gives: ```text S = (V1V2)² + (V3V4)² ``` 4. The **square-root block** (log, gain −½, antilog) gives: ```text V0 = √S = √( (V1V2)² + (V3V4)² ) ``` All inputs must be positive (or first passed through precision rectifiers, since the result depends only on squares). The circuit uses 8 log or antilog amps (5 log, 3 antilog) plus the summing and scaling op-amps. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is a square root obtained with log-antilog blocks? A. Add the log to itself B. Take the antilog and then log C. Multiply the log by 1/2 and take the antilog D. Multiply the log by 2 and take the antilog Answer: C. sqrt(Vi) = exp(1/2 ln Vi). (Key point: Square root idea) 2. What gain does the inverting stage have in the square-root circuit? A. -1 B. -1/2, using feedback R/2 with input R C. -2, using feedback 2R D. +1 Answer: B. Gain = -(R/2)/R = -1/2. (Key point: Gain stage) 3. How is (V1 V2)^2 formed? A. Sum the logs of V1 and V2 with a gain of 2, then antilog B. Subtract the logs and take the antilog C. Take the antilog and square the result D. Multiply V1 and V2 using a resistor Answer: A. 2K ln(V1 V2) in antilog gives (V1 V2)^2. (Key point: Squaring branch) 4. After forming (V1V2)^2 and (V3V4)^2, what completes V0? A. Subtract and take log B. Multiply them C. Take the antilog twice D. Add them and pass the sum through a log, gain -1/2 and antilog Answer: D. The square-root block gives V0 = sqrt(S). 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