--- title: "Current mirror output current, finite β effects" subject: "Advanced Electronics" chapter: 1 key: "semester_4/advanced_electronics:current-mirror-finite-beta" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/current-mirror-finite-beta site: BCT Engineering --- # Current mirror output current, finite β effects Advanced Electronics · Chapter 1: Operational Amplifier Circuits - Asked in: 2069 Asar, 1+2+2+2 marks - Video: [#327: Bipolar Transistor Current Mirror output impedance, Early Effect, Wilson Current Mirror](https://www.youtube.com/watch?v=2IbWvq7Abv4) — w2aew - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/current-mirror-finite-beta - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-1 ## Question Draw a current mirror circuit and determine the expression for output current. Explain the effect of finite β and finite output resistance on output current of a current mirror circuit. ## Short answer The simple mirror output current is IO = IREF/(1+2/β) = IREF·β/(β+2). Finite β reduces IO (e.g., ~2% error for β=100). Finite output resistance ro due to Early effect causes IO to increase with VCE2: IO ∝ [1+(VCE2−VBE)/VA], with output resistance Ro = VA/IO. ## Answer #### Current mirror circuit A current mirror copies a reference current to an output branch using two matched transistors with equal VBE. Q1 is diode-connected (collector shorted to base). ```text +VCC IO (from load) | | R | | IREF | +------+ | | | | C1 | C2 Q1 B----+-------------B Q2 E1 (Q1: C tied to B) E2 | | GND GND ``` #### Expression for output current Q1 and Q2 are matched and have the same VBE, so IC1 = IC2 = IC and IB1 = IB2 = IC/β. ```text IREF = (VCC - VBE)/R KCL at the collector of Q1: IREF = IC1 + IB1 + IB2 = IC + 2IC/β IO = IC2 = IREF / (1 + 2/β) = IREF·β/(β + 2) ``` For β → ∞, **IO = IREF = (VCC − VBE)/R**. #### Effect of finite β Part of IREF is used to supply the base currents of both transistors, so IO is smaller than IREF: ```text Error = (IREF - IO)/IREF = 2/(β + 2) β = 100 → IO = 0.980 IREF (2% error) β = 50 → IO = 0.962 IREF (3.8% error) ``` The error also changes with temperature because β does. It is reduced by a buffered mirror (emitter-follower Q3 supplies the base currents, IO = IREF/(1 + 2/β²)) or a Wilson mirror. #### Effect of finite output resistance Because of the Early effect, IC2 depends on VCE2. Q1 has VCE1 = VBE, but VCE2 is set by the load: ```text IO = IREF/(1 + 2/β) · [1 + (VCE2 - VBE)/VA] Output resistance Ro = ΔVCE2/ΔIO = ro = VA/IO ``` So IO is not constant; it rises as the output voltage rises, and the source is not ideal (ideal Ro = ∞). For VA = 100 V and IO = 1 mA, ro = 100 kΩ, so a 10 V change in VCE2 changes IO by about 10%. Widlar, cascode and Wilson sources raise Ro to reduce this effect. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is IO of the simple mirror with finite β? A. IREF·β B. IREF β/(β + 2) C. IREF/β D. IREF (β + 2)/β Answer: B. IO = IREF/(1 + 2/β) = IREF β/(β + 2). (Key point: Expression) 2. What is the fractional error (IREF − IO)/IREF? A. β/2 B. 1/β C. 2/β² D. 2/(β + 2) Answer: D. 1 − β/(β + 2) = 2/(β + 2); for β = 100 it is about 2%. (Key point: Error) 3. What is the output resistance of the simple mirror and what does it mean? A. Ro = infinite; IO is constant B. Ro = β ro C. Ro = 1/gm; IO falls with voltage D. Ro = ro = VA/IO; IO rises with output voltage Answer: D. Finite ro makes IO = IREF/(1+2/β)[1 + (VCE2 − VBE)/VA], varying with VCE2. (Key point: Early effect) 4. Which circuit reduces the β error? A. Adding a larger load B. Removing Q1 C. Using the same VCE D. Buffered (emitter-follower) mirror or Wilson mirror Answer: D. The emitter follower supplies base currents, giving IO = IREF/(1 + 2/β²). 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