--- title: "Matched-transistor antilog amp and temperature" subject: "Advanced Electronics" chapter: 5 key: "semester_4/advanced_electronics:antilog-matched-temperature" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/antilog-matched-temperature site: BCT Engineering --- # Matched-transistor antilog amp and temperature Advanced Electronics · Chapter 5: Op-Amp Bipolar Transistor Logarithmic Amplifier - Asked in: 2072 Chaitra, 7 marks - Video: [Temperature Compensated Antilog Amplifier](https://www.youtube.com/watch?v=MecXjeWutEM) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/antilog-matched-temperature - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-5 ## Question Draw the circuit diagram for matched transistor antilog amplifier and hence derive the input and output voltage relationship. Explain how the effect of temperature can be minimized. ## Short answer A matched-transistor antilog amplifier cancels Is, giving Vo = (Rf·Vref/R1)·exp[(RTC/(R2+RTC))·Vin/VT]. Temperature effect is minimized because the matched pair cancels Is, and a positive-temperature-coefficient resistor RTC compensates VT by rising with temperature to keep the exponent constant. ## Answer An antilog (exponential) amplifier produces an output proportional to the exponential of its input. Using two matched transistors with a reference current cancels the saturation current Is. A temperature-sensitive resistor then compensates VT. #### Circuit ```text Vref-[R1]--+--(-)\ |C A1>------+ Q1 GND-(+)/ | B=GND \E | +-----node E--+ |E Q2 \C Rf Vin-[R2]-+-B +---------+--------/\/\/----+ | | | [RTC] (-)\ | | A2>-------------+--> Vo GND GND-(+)/ ``` A1 forces Q1's collector current to Iref = Vref/R1 by driving the common emitter node E. The base of Q2 gets Vb = Vin·RTC/(R2 + RTC). #### Input–output relation Using VBE = VT ln(IC/Is): **Q1:** ```text IC1 = Vref/R1, base at 0 V VE = −VT ln( Vref/(R1·Is) ) ``` **Q2:** ```text VBE2 = Vb − VE IC2 = Is · exp(VBE2/VT) = Is · exp(Vb/VT) · Vref/(R1·Is) = (Vref/R1) · exp(Vb/VT) ``` **A2 (current-to-voltage):** ```text Vo = Rf · IC2 ``` **Result:** ```text Vo = (Rf·Vref/R1) · exp[ (RTC/(R2 + RTC)) · Vin / VT ] ``` For example, with K = RTC/(R2 + RTC) and 2.303VT/K = 1 V: ```text Vo = (Rf·Vref/R1) · 10^(Vin) ``` #### Minimizing the effect of temperature 1. **Matched pair.** Q1 and Q2 are made on the same chip and kept at the same temperature. Their Is values are equal and cancel, as shown above. Is is the most temperature-sensitive term, roughly doubling every 10 °C. 2. **Compensating VT = kT/q.** The exponent is K·Vin/VT, and VT rises about 0.33 %/°C. RTC is a resistor with a positive temperature coefficient (about +0.3 %/°C, with R2 ≫ RTC). Then K ≈ RTC/R2 rises with T at the same rate as VT, so K/VT, and therefore the exponent, stays constant. 3. A stable reference Vref and low-bias-current op-amps keep the scale factor Rf·Vref/R1 constant. With these measures the antilog output stays accurate over a wide temperature range. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What allows Is to cancel in the matched antilog amplifier? A. A stable Vref B. A large Rf C. Using a low-bias-current op-amp D. Q1 and Q2 are on the same chip with equal Is and temperature Answer: D. VE from Q1 contains ln(1/Is), which cancels Is in IC2. (Key point: Why Is cancels) 2. What is the role of RTC in the divider? A. It converts current to voltage B. It cancels Is C. Its positive temperature coefficient makes K rise with T at the same rate as VT, keeping K/VT constant D. It sets the reference current Answer: C. The exponent is K Vin/VT, and K = RTC/(R2+RTC) is made to track VT. (Key point: Role of RTC) 3. What kind of temperature coefficient must RTC have in the antilog amp? A. Any, as long as it is matched B. Positive, about +0.3 %/degC C. Negative, about -0.3 %/degC D. Zero Answer: B. VT rises with T, so K must also rise to keep K/VT constant. (Key point: Sign of coefficient) 4. Which is the antilog output relation? A. Vo = -VT ln(Vin) B. Vo = (Rf Vref/R1) exp[(RTC/(R2+RTC)) Vin/VT] C. Vo = (R1 Vref/Rf) ln(Vin/VT) D. Vo = Rf Is exp(Vin) Answer: B. IC2 = (Vref/R1) exp(Vb/VT) and Vo = Rf IC2. 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