--- title: "Basic logarithmic amplifier circuit" subject: "Advanced Electronics" chapter: 5 key: "semester_4/advanced_electronics:basic-log-amplifier" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/basic-log-amplifier site: BCT Engineering --- # Basic logarithmic amplifier circuit Advanced Electronics · Chapter 5: Op-Amp Bipolar Transistor Logarithmic Amplifier - Asked in: 2072 Kartik, 4 marks - Video: [Log Amplifier using OpAmp (Basics, Circuit, Working & Application) Explained](https://www.youtube.com/watch?v=Ob1EjwT08jM) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/basic-log-amplifier - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-5 ## Question Draw the circuit diagram of a basic logarithmic amplifier and explain it. ## Short answer A basic log amplifier uses a BJT or diode in the feedback path of an op-amp. For Vin>0, the output is Vo = -VT ln(Vin/(R Is)), where VT ≈ 26 mV at 300 K. It compresses a wide input range, but Is and VT drift with temperature. ## Answer A log amplifier gives an output voltage proportional to the logarithm of the input voltage. The basic circuit places a diode or a transistor (transdiode connection) in the feedback path of an inverting op-amp. It uses the exponential current–voltage law of the PN junction. #### Circuit ```text Q1 (npn) C +----|<|----+ E | B=GND | Vin-[R]--+--(-)\ | A >----+--> Vo GND--(+)/ ``` (A diode with its anode at the inverting input and its cathode at the output can replace Q1.) #### Working and derivation 1. The inverting input is at virtual ground, so the input current is I = Vin/R. All of it flows into the collector of Q1. 2. The base and collector are both at 0 V, so VBE = −Vo. The transistor law gives: ```text IC = Is · exp(VBE/VT) ``` 3. Equating the two currents: ```text Vin/R = Is · exp(−Vo/VT) Vo = −VT ln( Vin / (R·Is) ) = −2.303 VT log10( Vin / (R·Is) ) ``` At 300 K, VT = kT/q ≈ 26 mV. So the output changes by about −60 mV for each tenfold (decade) increase in Vin. #### Features and limitations - It works only for one input polarity (positive Vin for an npn transistor). A protective diode is often added. - The output is small (hundreds of mV). A following amplifier scales it up. - Is and VT both depend on temperature, so the basic circuit drifts. Matched transistor pairs and temperature-compensating resistors are used in practical log amps. - The transistor version works over a wider current range (about 7 decades) than the diode version. **Uses:** compressing wide-range signals, dB meters, and analog multiplication and division. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the input current of the basic log amp equal to Vin/R? A. The op-amp output is at 0 V B. R equals Is C. The inverting input is at virtual ground, so the full Vin appears across R D. The transistor draws no current Answer: C. With the inverting input at 0 V, I = Vin/R, and it all flows into the collector of Q1. (Key point: Virtual ground current) 2. Which is the transfer function of the basic log amp? A. Vo = Vin R Is B. Vo = -VT ln(Vin/(R Is)) C. Vo = VT exp(Vin/R) D. Vo = -R Is exp(Vin/VT) Answer: B. Setting Vin/R = Is exp(-Vo/VT) gives Vo = -VT ln(Vin/(R Is)). (Key point: Output expression) 3. Why is VBE equal to -Vo? A. Base and collector are at 0 V and the emitter is at Vo B. The emitter is grounded C. The base is at Vo D. The collector is at Vo Answer: A. VBE = VB - VE = 0 - Vo. (Key point: Why VBE = -Vo) 4. Why is the transistor version preferred over the diode version? A. It follows the exponential law over a wider current range (about 7 decades) B. It works for both polarities C. It has no temperature drift D. It gives a larger output Answer: A. The transdiode connection uses only IC, which is very ideal, so the range is wider. 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