--- title: "Matched-transistor log amp and temperature effects" subject: "Advanced Electronics" chapter: 5 key: "semester_4/advanced_electronics:log-amp-matched-temperature" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/log-amp-matched-temperature site: BCT Engineering --- # Matched-transistor log amp and temperature effects Advanced Electronics · Chapter 5: Op-Amp Bipolar Transistor Logarithmic Amplifier - Asked in: 2073 Shrawan, 7 marks - Video: [3.5.2 | Compensated Log Amplifier | V H Mankar |](https://www.youtube.com/watch?v=nBws5jUTeKI) — Vijay Mankar - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/log-amp-matched-temperature - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-5 ## Question Explain the operation of log amplifier using matched transistor. Explain how the effect of temperature can be minimized in this circuit. ## Short answer Matched transistors cancel Is, so output is Vo = K VT ln(Vin/Vref). To minimize temperature drift due to VT, the gain K is made temperature-dependent by using a resistor with +0.33%/°C temperature coefficient, keeping K VT constant. The reference current sets the zero-crossing point. ## Answer A log amplifier produces Vo ∝ ln(Vin). The basic one-transistor log amp gives Vo = −VT ln(Vin/(R1·Is)). This has two temperature-dependent terms: Is (which rises very steeply with temperature) and VT = kT/q. The matched-transistor log amp removes Is and a temperature-sensitive resistor removes the VT effect. #### Circuit ```text Vin-[R1]-+-(-)\ Q1 (C at -, E at out, B GND) | A1 >---Vo1---[R3]--+--[R4]--+ GND-----(+)/ | | (-)\ | A3>---+--> Vo (+)/ Vref-[R2]-+-(-)\ Q2 | | A2 >--Vo2--[R3]----+--[R4]--GND GND------(+)/ Q1, Q2: matched pair on one chip (same temperature) R3: temperature-sensitive resistor (RTC) ``` #### Operation 1. **A1 with Q1 (log of input).** The current Vin/R1 flows through Q1, and A1's output settles at −VBE1: ```text Vo1 = −VT ln( Vin / (R1·Is) ) ``` 2. **A2 with Q2 (log of reference).** It is driven by a fixed current Vref/R2: ```text Vo2 = −VT ln( Vref / (R2·Is) ) ``` 3. **A3 subtracts the two outputs** with gain K = R4/R3: ```text Vo = K (Vo2 − Vo1) = K·VT·ln( Vin·R2 / (Vref·R1) ) ``` With R1 = R2: **Vo = K·VT·ln(Vin/Vref)**. The output follows the logarithm of the input ratio. It is about 60 mV per decade per unit K at room temperature. #### How the temperature effect is minimized 1. **Is is removed by matching.** Q1 and Q2 are made on the same chip (a matched pair), so they have the same Is and the same temperature. In the subtraction, ln(1/Is) − ln(1/Is) = 0, so the large drift of Is disappears. 2. **VT is compensated by a temperature-sensitive gain.** The output is still proportional to VT = kT/q, which rises by about 0.33 %/°C at 25 °C. So the gain K is made to fall at the same rate. R3 (or RTC in a non-inverting gain stage, K = 1 + R2/RTC) is a resistor with a positive temperature coefficient of about +0.33 %/°C. As T rises, RTC rises and K falls, so K·VT stays constant. 3. Using a reference current (Vref/R2) also fixes the zero point of the log curve (Vo = 0 at Vin = Vref). 4. Low-bias-current, low-offset op-amps reduce errors at small input currents. The result is a log amplifier that is accurate over several decades of input with little drift. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which two quantities make the basic one-transistor log amp drift with temperature? A. The op-amp gain and bias current B. Is and VT = kT/q C. R1 and Vin D. Vref and Rf Answer: B. Vo = -VT ln(Vin/(R1 Is)) contains both Is and VT, which depend on temperature. (Key point: Two temperature terms) 2. How does the matched pair remove the effect of Is? A. Q1 and Q2 on one chip have equal Is and temperature, so ln(1/Is) - ln(1/Is) = 0 in the subtraction B. Is is made zero by cooling the chip C. A reference current makes Is constant D. A large Rf swamps Is Answer: A. A3 subtracts Vo1 and Vo2, and since both have the same Is it drops out. (Key point: How Is is removed) 3. How is the VT effect minimised? A. Use a negative-temperature-coefficient resistor so the gain rises with T B. Operate at a fixed Vin C. Use a larger Vref D. Use a positive-temperature-coefficient resistor so the gain K falls at the same rate as VT rises, keeping K VT constant Answer: D. VT rises about 0.33 %/degC, and an RTC with a positive coefficient reduces K by the same amount. (Key point: Compensating VT) 4. What is the role of the reference current Vref/R2? A. It increases the temperature drift B. It sets the polarity of the input C. It makes the output linear D. It fixes the zero point of the log curve, with Vo = 0 at Vin = Vref Answer: D. With R1 = R2, Vo = K VT ln(Vin/Vref), which is zero at Vin = Vref. (Key point: Reference current role) ## Related in Advanced Electronics - [Matched-transistor antilog amp and temperature](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aantilog-matched-temperature) — Advanced Electronics, chapter 5 - [Log amplifier transfer function, matched transistors](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Alog-amp-matched-transfer) — Advanced Electronics, chapter 5, asked 2× - [Output of temperature-compensated log circuit](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Atemperature-compensated-log-circuit) — Advanced Electronics, chapter 5 - [Output of log-ratio amplifier with RT compensation](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Alog-ratio-temperature-compensated) — Advanced Electronics, chapter 5 - [Log-antilog circuit: Vo in terms of Vi and VR](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Alog-antilog-circuit-vo-vi) — Advanced Electronics, chapter 5, asked 3× - [Antilog amplifier with matched transistors](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aantilog-matched-transistor-transfer) — Advanced Electronics, chapter 5, asked 2× ## Related in other subjects - [Op-amp properties and summing amplifier](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-summer-v1-2v2-3v3) — Basic Electronics Engineering, chapter 4 - [Summing amplifier using op-amp](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Asumming-amplifier-op-amp) — Fundamental of Electrical and Electronics Engineering, chapter 6