--- title: "Antilog amplifier with matched transistors" subject: "Advanced Electronics" chapter: 5 key: "semester_4/advanced_electronics:antilog-matched-transistor-transfer" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/antilog-matched-transistor-transfer site: BCT Engineering --- # Antilog amplifier with matched transistors Advanced Electronics · Chapter 5: Op-Amp Bipolar Transistor Logarithmic Amplifier - Asked in: 2073 Chaitra, 5 marks; 2069 Asar, 6 marks (asked 2 times) - 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-transistor-transfer - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-5 ## Question Derive the transfer function (output voltage relationship) of antilog amplifier using two matched transistors and two op-amps. ## Short answer Matched transistors cancel Is. Q1 sets emitter voltage VE = -VT ln(Vref/(R1*Is)). Q2 base receives Vb = K Vin, producing IC2 = (Vref/R1) exp(Vb/VT). The output op-amp gives Vo = (Rf*Vref/R1) exp(K Vin/VT), with K = R3/(R2+R3). The scale depends on VT remaining. ## Answer An antilog (exponential) amplifier gives an output proportional to the exponential of its input. With two matched transistors and two op-amps, the saturation current Is cancels out and a reference current sets the scale. #### Circuit ```text Vref-[R1]--+--(-)\ |C A1>------+ Q1 GND-(+)/ | B=GND \E | +-----node E--+ |E Q2 \C Rf Vin-[R2]-+-B +---------+--------/\/\/----+ | | | [R3] (-)\ | | A2>-------------+--> Vo GND GND-(+)/ ``` Q1 and Q2 are a matched pair with their emitters tied at node E. A1 forces Q1 to carry Iref = Vref/R1. The base of Q2 receives Vb = Vin·R3/(R2 + R3) (a divider used for scaling and temperature compensation). #### Derivation The transistor law is IC = Is·exp(VBE/VT). **Q1** (base at 0 V, collector at virtual ground): ```text IC1 = Vref/R1, VBE1 = 0 − VE VE = −VT ln( Vref / (R1·Is) ) ``` **Q2** (base at Vb, emitter at VE): ```text VBE2 = Vb − VE = Vb + VT ln( Vref/(R1·Is) ) IC2 = Is · exp(VBE2/VT) = Is · exp(Vb/VT) · Vref/(R1·Is) = (Vref/R1) · exp(Vb/VT) ``` **A2** works as a current-to-voltage converter. IC2 flows from the output through Rf: ```text Vo = Rf · IC2 = (Rf·Vref/R1) · exp(Vb/VT) ``` With Vb = K·Vin, where K = R3/(R2 + R3): ```text Vo = (Rf·Vref/R1) · exp( K·Vin / VT ) = (Rf·Vref/R1) · 10^( K·Vin / (2.303 VT) ) ``` #### Points to note - Is (strongly temperature dependent) cancels because Q1 and Q2 are matched. - VT = kT/q remains. Making R3 (or R2) a temperature-sensitive resistor keeps K/VT constant. - At 300 K, 2.303VT ≈ 60 mV. With K = 1, the output rises ten times for every 60 mV increase of Vin. - The output is always of one polarity, set by Vref. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What do Q1 and A1 do in the matched-transistor antilog amplifier? A. They amplify Vin before it reaches Q2 B. They convert the output current of Q2 into a voltage C. They remove the DC offset of A2 D. They force Q1 to carry the reference current Vref/R1, which sets the emitter node voltage VE Answer: D. A1 makes Q1 carry Iref = Vref/R1, and this fixes VE = -VT ln(Vref/(R1 Is)), which is later used by Q2. (Key point: Role of Q1 and A1) 2. Why does the saturation current Is not appear in the final output Vo? A. Rf is chosen to be equal to Is B. The divider R2-R3 removes Is C. Q1 and Q2 are matched, so the Is in VE from Q1 cancels the Is in the IC2 expression for Q2 D. Is is very small and can be neglected Answer: C. IC2 = Is exp(Vb/VT) x Vref/(R1 Is); the two Is terms cancel only if the transistors are matched. (Key point: Why Is cancels) 3. What is the output of the antilog amplifier? A. Vo = Rf Is exp(Vin/VT) B. Vo = (Rf Vref / R1) exp(K Vin / VT), with K = R3/(R2+R3) C. Vo = -VT ln(Vin/(R1 Is)) D. Vo = (Rf Vref / R1) ln(K Vin / VT) Answer: B. Vo = Rf x IC2 = (Rf Vref/R1) exp(Vb/VT) with Vb = K Vin. (Key point: Transfer function) 4. What is the job of A2 and Rf? A. Provide the temperature compensation of Is B. Convert the collector current IC2 of Q2 into the output voltage Vo = Rf IC2 C. Generate the reference current D. Subtract the logs of two inputs Answer: B. A2 is a current-to-voltage converter, so Vo = Rf x IC2. (Key point: Role of A2) ## Related in Advanced Electronics - [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× - [Matched-transistor antilog amp and temperature](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aantilog-matched-temperature) — Advanced Electronics, chapter 5 - [Output voltage of two op-amp IA](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Atwo-opamp-ia-output) — Advanced Electronics, chapter 4 - [Output of three log amps with summer](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Athree-log-amps-summing) — Advanced Electronics, chapter 6 - [IA features; 3 op-amp IA output voltage](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aia-features-3opamp-output) — Advanced Electronics, chapter 4 - [Output offset voltage of 741 inverting amplifier](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aoutput-offset-741-inverting) — Advanced Electronics, chapter 2 ## Related in other subjects - [Transfer function of op-amp with Rs ∥ C feedback](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-opamp-rs-c-r1-r2) — Control Systems, chapter 2 - [Transfer function of op-amp phase-shift circuit](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-opamp-allpass-r1-c-r2) — Control Systems, chapter 2, asked 2× - [Transfer function of op-amp integrator-type circuit](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-opamp-r1-r2-c) — Control Systems, chapter 2 - [Voltage ratio transfer function of two-section LC network](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Avoltage-ratio-tf-two-1h-shunt) — Electrical Circuit Theory, chapter 5