--- title: "Design Vo = (Vx/Vy)^α·Vz with log-antilog" subject: "Advanced Electronics" chapter: 6 key: "semester_4/advanced_electronics:design-vx-vy-power-alpha" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/design-vx-vy-power-alpha site: BCT Engineering --- # Design Vo = (Vx/Vy)^α·Vz with log-antilog Advanced Electronics · Chapter 6: Log-Antilog Circuit Applications - Asked in: 2069 Asar, 7 marks; 2068 Chaitra, 7 marks (asked 2 times) - Video: [Analog Divider using log antilog amplifiers](https://www.youtube.com/watch?v=Ump575Pbe-0) — Electronics for You - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/design-vx-vy-power-alpha - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-6 ## Question Design a circuit that produces output voltage V0 = (Vx/Vy)^α · Vz using Log and Antilog amplifiers, where α > 1 and Vx, Vy and Vz are analog input voltages. [Hint: ln(x^a) = a ln x] ## Short answer Log amps give -K ln Vx, -K ln Vy, -K ln Vz. A difference amp gives K ln(Vx/Vy), a non-inverting amp scales by alpha, another difference amp adds ln Vz, and an antilog amp yields V0 = (Vx/Vy)^alpha Vz. Set Rf = (alpha - 1)R1; all inputs must be positive. ## Answer The idea is to work in the logarithmic domain: ```text ln V0 = α (ln Vx − ln Vy) + ln Vz V0 = exp[ α ln(Vx/Vy) + ln Vz ] = (Vx/Vy)^α · Vz ``` So the circuit must find logs, subtract, multiply by α, add, and take the antilog. **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 #### Block diagram ```text Vx->[LOG A1]--Vx'-->(-)[A4: diff amp] Vy->[LOG A2]--Vy'-->(+)[ gain 1 ]--Vd | [A5: non-inverting, gain α]<-+ | Ve v (+)[A6: diff amp]--Vs Vz->[LOG A3]--Vz'->(-)[ gain 1 ] | v [ANTILOG A7] --> V0 ``` #### Design steps 1. **Log amplifiers** (transistor in feedback, matched, all with the same K): ```text Vx' = −K ln Vx, Vy' = −K ln Vy, Vz' = −K ln Vz ``` 2. **Difference amplifier A4** (all four resistors R = 10 kΩ), with Vy' on the (+) input and Vx' on the (−) input: ```text Vd = Vy' − Vx' = K (ln Vx − ln Vy) = K ln(Vx/Vy) ``` 3. **Non-inverting amplifier A5** with gain α = 1 + Rf/R1: ```text Ve = α K ln(Vx/Vy) = K ln( (Vx/Vy)^α ) ``` Since α > 1, a non-inverting amplifier can give it directly: ```text Rf = (α − 1) R1 ``` For example, α = 2.5 with R1 = 10 kΩ gives Rf = 15 kΩ. Make Rf variable to set any α. 4. **Difference amplifier A6** (all resistors R), with Ve on the (+) input and Vz' on the (−) input: ```text Vs = Ve − Vz' = K ln( (Vx/Vy)^α ) + K ln Vz = K ln( (Vx/Vy)^α · Vz ) ``` 5. **Antilog amplifier A7:** ```text V0 = exp(Vs/K) = (Vx/Vy)^α · Vz ``` #### Check Take Vx = 4 V, Vy = 2 V, Vz = 1.5 V, α = 2. Then V0 = (2)² × 1.5 = 6 V. **Notes:** Vx, Vy and Vz must be positive. The log amps use matched transistors, so Is cancels, and temperature-compensating resistors, so the drift of VT is compensated. The output is scaled by the reference voltage (1 V here). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is ln V0 for V0 = (Vx/Vy)^alpha Vz? A. alpha ln Vx + ln Vy + ln Vz B. ln alpha + ln Vx - ln Vy C. (ln Vx - ln Vy)/alpha + ln Vz D. alpha(ln Vx - ln Vy) + ln Vz Answer: D. ln(a^b) = b ln a, so the power becomes a multiplication by alpha. (Key point: Log domain idea) 2. Which resistor relation gives a non-inverting gain alpha? A. Rf = R1/alpha B. Rf = (alpha + 1) R1 C. Rf = (alpha - 1) R1 D. Rf = alpha R1 Answer: C. Gain = 1 + Rf/R1 = alpha, so Rf = (alpha - 1) R1. (Key point: Gain for alpha) 3. For alpha = 2.5 and R1 = 10 kOhm, what is Rf? A. 4 kOhm B. 15 kOhm C. 25 kOhm D. 10 kOhm Answer: B. Rf = 1.5 x 10 k = 15 k. (Key point: Rf for example) 4. For Vx = 4 V, Vy = 2 V, Vz = 1.5 V, alpha = 2, what is V0? A. 6 V B. 3 V C. 12 V D. 8 V Answer: A. (4/2)^2 x 1.5 = 4 x 1.5 = 6 V. 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