--- title: "Output of log-ratio amplifier with RT compensation" subject: "Advanced Electronics" chapter: 5 key: "semester_4/advanced_electronics:log-ratio-temperature-compensated" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/log-ratio-temperature-compensated site: BCT Engineering --- # Output of log-ratio amplifier with RT compensation Advanced Electronics · Chapter 5: Op-Amp Bipolar Transistor Logarithmic Amplifier - Asked in: 2068 Chaitra, 4 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/log-ratio-temperature-compensated - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-5 ## Question Find expression of output voltage for the circuit shown in figure below (Q1 and Q2 are matched transistors). [Figure: V1 through R1 to the inverting input of op-amp A1 with transistor Q1 (base grounded) as feedback; VR through R1 to the inverting input of op-amp A2 with transistor Q2 (base grounded) as feedback; non-inverting inputs grounded. The A1 output goes through R to the inverting input of op-amp A3 (feedback R); the A2 output goes through R to the non-inverting input of A3, which is connected to ground through R. The A3 output feeds the non-inverting input of op-amp A4, which has R2 from its output V0 to its inverting input and RT from its inverting input to ground.] ## Short answer Vo = (1 + R2/RT)·VT·ln(V1/VR). The matched Q1, Q2 cancel Is, and the PTC resistor RT compensates VT (kT/q), making the output stable over temperature. ## Answer The circuit has two log amplifiers (A1, A2), a unity-gain difference amplifier (A3) and a non-inverting amplifier (A4) with a temperature-compensating resistor RT. Its output is proportional to the log of the ratio V1/VR. #### Step 1: log amplifiers Q1 and Q2 have their bases grounded and their collectors at virtual ground. Using VBE = VT ln(IC/Is): ```text Vo1 = −VBE1 = −VT ln( V1 / (R1·Is) ) Vo2 = −VBE2 = −VT ln( VR / (R1·Is) ) ``` #### Step 2: difference amplifier A3 (all resistors R) Vo1 goes to the inverting input through R (feedback R). Vo2 goes to the non-inverting input through R, which has R to ground, so V+ = Vo2/2. ```text Vo3 = (1 + R/R)(Vo2/2) − (R/R)·Vo1 = Vo2 − Vo1 = −VT ln( VR/(R1·Is) ) + VT ln( V1/(R1·Is) ) = VT ln( V1 / VR ) ``` Is cancels because Q1 and Q2 are matched. #### Step 3: non-inverting amplifier A4 ```text V0 = (1 + R2/RT) · Vo3 ``` #### Result ```text V0 = (1 + R2/RT) · VT · ln( V1 / VR ) = 2.303 (1 + R2/RT) · VT · log10( V1 / VR ) ``` where VT = kT/q (about 25.9 mV at 300 K). #### Notes - V0 = 0 when V1 = VR. It rises by 2.303(1 + R2/RT)VT for every decade of V1/VR. - Is has been removed by the matched pair. The remaining temperature dependence is in VT. RT is a positive-temperature-coefficient resistor, so (1 + R2/RT) falls as T rises, keeping (1 + R2/RT)·VT nearly constant. - **Example:** to get 1 V per decade at 300 K, 2.303 × 0.0259 × (1 + R2/RT) = 1, so 1 + R2/RT ≈ 16.8. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the unity-gain difference amplifier A3 give? A. Vo3 = -VT ln(V1/(R1 Is)) B. Vo3 = VT ln(V1/VR) C. Vo3 = VT ln(V1 VR) D. Vo3 = V1 - VR Answer: B. Vo3 = Vo2 - Vo1, and the Is terms cancel. (Key point: A3 output) 2. In A3 the non-inverting input receives Vo2/2. Why? A. Vo2 is divided by two equal resistors R and R to ground B. The op-amp halves its input C. Q2 gives half the voltage D. R1 divides it Answer: A. The R-R divider halves Vo2, and the gain (1+R/R) = 2 restores it. (Key point: Why V+ = Vo2/2) 3. What is the overall output V0? A. V0 = (R2/RT) VT ln(V1 VR) B. V0 = VT ln(V1/VR) C. V0 = (1 + RT/R2) Is ln(V1) D. V0 = (1 + R2/RT) VT ln(V1/VR) Answer: D. A4 is a non-inverting amplifier of gain (1 + R2/RT). (Key point: Final output) 4. How does RT compensate for temperature? A. RT cancels Is B. RT is matched to Q1 C. RT keeps V1 constant D. Positive-coefficient RT rises with T, so (1 + R2/RT) falls and offsets the rise in VT Answer: D. The product (1 + R2/RT) VT stays nearly constant. 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