--- title: "Bandgap reference voltage source" subject: "Electronics Devices and Circuit" chapter: 6 key: "semester_3/electronics_devices_and_circuit:bandgap-reference" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/bandgap-reference site: BCT Engineering --- # Bandgap reference voltage source Electronics Devices and Circuit · Chapter 6: Power Supplies, Breakdown Diodes, and Voltage Regulators - Asked in: 2082 Shrawan, 6 marks; 2075 Ashwin, 6 marks; 2068 Baisakh, 3 marks (asked 3 times) - Video: [bandgap reference - voltage reference - voltage source - start-up circuit](https://www.youtube.com/watch?v=UUoxYuPRXhA) — Computer&Electronics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/bandgap-reference - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-6 ## Question Explain the operation of bandgap reference voltage source. ## Short answer It combines a negative‑temperature‑coefficient VBE with a positive‑TC voltage K·VT to cancel drift. Vref = VBE + 2 (R2/R1) VT ln n; with proper K, Vref ≈ 1.25 V, the bandgap voltage of silicon, nearly independent of temperature and supply. ## Answer A bandgap reference produces a stable dc voltage, about 1.25 V, which is almost independent of temperature, supply voltage and load. It is used in voltage regulators, ADCs, DACs and power supplies (like in the LM317). #### Principle It adds two voltages with opposite temperature coefficients so that they cancel: - **VBE of a transistor** has a negative temperature coefficient (CTAT): about −2 mV/°C. - **The thermal voltage VT = kT/q** has a positive temperature coefficient (PTAT): +0.086 mV/°C (VT = 25.85 mV at 300 K). Vref = VBE + K · VT, where K is a constant set by resistors and the area ratio. #### Circuit ```text +VCC [R3] [R3] | | +--+ +---+ op-amp inputs taken from the | | | | two collectors; its output Vref C1 | C2 | drives both bases Q1 (n x) Q2 (1 x) | B1-----+----B2 <---+--- Vref (op-amp output) E1 E2 [R1] | +---------+ | [R2] | GND ``` The op-amp (polarity chosen for negative feedback) makes the two collector voltages, and so the two collector currents, equal: I1 = I2 = I. Q1 has n times the emitter area of Q2. Both bases are at Vref. Applying KVL: VBE1 + I R1 + 2I R2 = VBE2 + 2I R2, so I R1 = VBE2 − VBE1 = VT ln n (the PTAT current I = VT ln n / R1) The reference voltage is the VBE of Q2 plus the drop across R2 caused by the total current 2I: **Vref = VBE2 + 2 I R2 = VBE2 + 2 (R2/R1) VT ln n** #### Temperature compensation d(Vref)/dT = dVBE/dT + K · dVT/dT = −2 mV/°C + K (0.0862 mV/°C) = 0 K = 2 (R2/R1) ln n = 2/0.0862 ≈ 23.2 Then Vref = VBE + 23.2 × 25.85 mV = 0.65 + 0.60 ≈ **1.25 V** This is about the bandgap voltage of silicon (1.12 eV, extrapolated 1.205 V at 0 K, with the final value near 1.25 V). For example, with n = 8 (ln 8 = 2.08), 2 (R2/R1)(2.08) = 23.2 gives R2/R1 ≈ 5.6. The reference voltage does not depend on temperature to the first order, and the circuit also has good supply rejection because the currents are set by the stable Vref. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does a bandgap reference get a voltage that is almost independent of temperature? A. It subtracts two voltages that both rise with temperature B. It adds a VBE (negative temperature coefficient) and a PTAT voltage K*VT (positive coefficient) so the drifts cancel C. It uses a Zener diode biased at 6 V where drift is zero D. It keeps the transistor at a fixed temperature using a heater Answer: B. VBE falls about 2 mV/degC while VT rises, so a suitably scaled sum has zero first-order drift. (Key point: Principle of bandgap) 2. In the bandgap circuit the op-amp forces the two collector currents equal, with Q1 having n times the emitter area of Q2. What sets the current I? A. I R1 = VBE2 - VBE1 = VT ln n, so I = VT ln n / R1 B. I = VBE / R2 C. I = Vref / R3 only, independent of n D. I = VT / (n R1) Answer: A. Equal currents in transistors of different area give a VBE difference of VT ln n across R1, so the current is proportional to VT (PTAT). (Key point: PTAT current) 3. Which expression gives the reference voltage of the circuit? A. Vref = VBE2 + (R2/R1) VT ln n B. Vref = VBE2 - 2 (R2/R1) VT ln n C. Vref = VBE1 + I R1 D. Vref = VBE2 + 2 (R2/R1) VT ln n Answer: D. R2 carries the total current 2I, so its drop 2 I R2 = 2(R2/R1) VT ln n adds to VBE2. (Key point: Vref expression) 4. Using dVBE/dT = -2 mV/°C and dVT/dT = +0.0862 mV/°C, what value of K makes Vref independent of temperature, and what is Vref? A. K ≈ 2, Vref ≈ 0.7 V B. K ≈ 0.043, Vref ≈ 0.65 V C. K ≈ 23.2, Vref ≈ 0.65 + 0.60 ≈ 1.25 V D. K ≈ 58, Vref ≈ 2.2 V Answer: C. Setting -2 + K(0.0862) = 0 gives K = 23.2, so Vref = 0.65 V + 23.2 x 25.85 mV = 1.25 V. 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