--- title: "Log-antilog circuit: Vo in terms of Vi and VR" subject: "Advanced Electronics" chapter: 5 key: "semester_4/advanced_electronics:log-antilog-circuit-vo-vi" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/log-antilog-circuit-vo-vi site: BCT Engineering --- # Log-antilog circuit: Vo in terms of Vi and VR Advanced Electronics · Chapter 5: Op-Amp Bipolar Transistor Logarithmic Amplifier - Asked in: 2074 Asoj, 4 marks; 2074 Chaitra, 7 marks; 2070 Asar, 4 marks (asked 3 times) - Video: [LOG & ANTILOG AMPLIFIER Explanation and solved Numericals-YouTube](https://www.youtube.com/watch?v=YU0y4CfR2rc) — Electronics Circuit Hub - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/log-antilog-circuit-vo-vi - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-5 ## Question For the circuit shown in figure below, find the relationship between V0 and Vi (find the expression for V0 in terms of Vin and VR; Q1 and Q2 are matched transistors). [Figure: Input V through R1 to the inverting input of op-amp A1 (non-inverting input grounded). Transistor Q1 is the feedback element of A1: base grounded, collector at the inverting input, emitter at the A1 output. The A1 output drives the emitter of matched transistor Q2, whose base is connected to the reference voltage VR. The collector of Q2 goes to the inverting input of op-amp A2 (non-inverting input grounded), which has feedback resistor R2 from its output Vo to its inverting input.] ## Short answer The first op-amp with Q1 forms a log amplifier giving VE = -VT ln(Vi/(R1*Is)). Q2 with base voltage VR has IC2 = (Vi/R1) exp(VR/VT). The second op-amp converts this current: Vo = (R2/R1) Vi exp(VR/VT). Matching cancels Is, yielding a voltage-controlled exponential gain amplifier. ## Answer The first stage (A1 with Q1) is a log amplifier. Q2 with A2 is an antilog stage whose base is lifted by VR. The overall circuit multiplies Vi by an exponential function of VR, a voltage-controlled gain. #### Circuit (as given) ```text Q1 (base GND) C +-----|<|-----+ E Vi-[R1]-+--(-)\ | GND-(+) A1>-----+---VE---+ | E VR ---- B --[Q2] | C R2 +----+---/\/\/--+ | | (-)\ A2 | GND-(+) >-------+--> Vo ``` #### Assumptions Q1 and Q2 are matched npn transistors (same Is). The op-amps are ideal, so both inverting inputs are at 0 V. The transistor law is: ```text IC = Is · exp(VBE / VT), so VBE = VT ln(IC / Is) ``` #### Stage 1 (log amplifier) The current Vi/R1 flows into the collector of Q1 (collector at virtual ground, base grounded): ```text IC1 = Vi / R1 VBE1 = VB1 − VE = 0 − VE VE = −VBE1 = −VT ln( Vi / (R1·Is) ) ``` #### Stage 2 (antilog amplifier) Q2 has its base at VR, its emitter at VE, and its collector at the virtual ground of A2: ```text VBE2 = VR − VE = VR + VT ln( Vi / (R1·Is) ) IC2 = Is · exp(VBE2 / VT) = Is · exp(VR/VT) · exp( ln( Vi/(R1·Is) ) ) = Is · exp(VR/VT) · Vi / (R1·Is) = (Vi / R1) · exp(VR / VT) ``` The collector current IC2 is drawn from A2's inverting node. It is supplied through R2 from the output, so: ```text Vo = IC2 · R2 ``` #### Result ```text Vo = (R2/R1) · Vi · exp(VR / VT) ``` - **Is cancels** because the transistors are matched. The output does not depend on the strongly temperature-dependent saturation current. - With VR = 0, Vo = (R2/R1)Vi, a plain linear amplifier. - The gain changes exponentially with VR: every 60 mV of VR (at 300 K, VT ≈ 26 mV) multiplies the gain by about 10. So the circuit is a **voltage-controlled exponential gain amplifier**. - VT = kT/q still appears, so a temperature-compensating resistor is needed in the VR path for accuracy. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the first stage (A1 with Q1)? A. A log amplifier B. An antilog amplifier C. A comparator D. A peak detector Answer: A. The transistor in feedback gives a logarithmic output. (Key point: Stage 1 function) 2. What is VE (A1 output) in terms of Vi? A. VT ln(Vi/(R1 Is)) B. -VT ln(Vi/(R1 Is)) C. Vi R1 D. -Vi Answer: B. With base at 0, VE = -VBE1 = -VT ln(IC1/Is) and IC1 = Vi/R1. (Key point: Emitter voltage) 3. Why does Is cancel in the final result? A. Q1 and Q2 are matched, so their Is values are equal B. Vi is large C. R2 removes it D. VT removes it Answer: A. IC2 = Is exp(VR/VT) Vi/(R1 Is). (Key point: Cancellation of Is) 4. What is the final Vo? A. (R2/R1) Vi exp(VR/VT) B. (R1/R2) Vi exp(-VR/VT) C. (R2/R1) ln Vi D. Vi VR Answer: A. Vo = IC2 R2 = (R2/R1) Vi exp(VR/VT), a voltage-controlled gain. 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