--- title: "Current mirror requirements, Widlar resistance, best source" subject: "Advanced Electronics" chapter: 1 key: "semester_4/advanced_electronics:current-mirror-best-source" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/current-mirror-best-source site: BCT Engineering --- # Current mirror requirements, Widlar resistance, best source Advanced Electronics · Chapter 1: Operational Amplifier Circuits - Asked in: 2073 Chaitra, 2+4+1 marks - Video: [Widlar Current Source | Introduction to Operational Amplifiers | Linear Integrated Circuits](https://www.youtube.com/watch?v=3ayAPiE2weA) — Ekeeda - 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-best-source - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-1 ## Question List out the basic requirements of current mirror circuits. Derive and express the output resistance of Widlar current source. Which one is best current source and why? ## Short answer Basic requirements: matched, same-temperature, equal-VBE, active-region, high-beta transistors. Widlar output resistance derived by small-signal analysis gives Ro ≈ ro(1+gmRE) due to emitter degeneration. The best source is the Wilson mirror for its very high output resistance (~βro/2) and minimal β error. ## Answer #### Basic requirements of current mirror circuits - **Matched transistors:** Q1 and Q2 must have the same IS (same VBE–IC curve), same β and same Early voltage. This is easy in an IC because both are made in the same process steps, side by side. - **Same temperature:** both transistors must be at the same temperature (close together on one chip), so that VBE changes track each other. - **Equal VBE:** the bases are tied together and the emitters are tied together (or returned to the same rail), so VBE1 = VBE2 and therefore IC2 = IC1. - **Output transistor in the active region:** Q2 must have VCE2 > VCE(sat), i.e. the load must leave enough voltage across Q2. - **High β:** so that the base currents taken from the reference branch are negligible. #### Output resistance of the Widlar current source ```text +VCC IO (from load) | | R | | IREF | +------+ | | | | C1 | C2 Q1 B----+-------------B Q2 E1 E2 | | | RE | | GND GND ``` DC design relation: IO·RE = VT ln(IREF/IO). Small-signal model: Q1 is diode-connected, so its resistance is only about 1/gm1 and the base of Q2 is close to ac ground. Apply a test voltage *vx* at the collector of Q2 and let *ix* flow into it. ```text Emitter node: ix flows into RE ‖ rπ (rπ goes to the grounded base), so ve = ix (RE ‖ rπ) = ix R'E Base is grounded: vbe = 0 - ve = -ix R'E Current into collector: ix = gm vbe + (vx - ve)/ro = -gm ix R'E + (vx - ix R'E)/ro Multiply by ro and collect terms: vx = ix ro + ix R'E (1 + gm ro) Ro = vx/ix = ro + R'E (1 + gm ro) Since gm ro >> 1: Ro ≈ ro (1 + gm R'E), R'E = RE ‖ rπ ``` If RE << rπ, then **Ro ≈ ro(1 + gm RE)**, where gm = IO/VT and ro = VA/IO. A simple mirror has Ro = ro only, so the Widlar source has an output resistance larger by the factor (1 + gm RE). Since gm·RE = IO·RE/VT = ln(IREF/IO), this factor is typically about 4 to 6 (for example, 5.6 when IREF/IO = 100). This is the result of the current-series negative feedback provided by the unbypassed RE: any rise in IO raises the drop across RE, lowers VBE2 and pulls IO back down. #### Best current source Among the common sources, the **Wilson current source** (and the cascode mirror) is usually the best: - Its output resistance is about β·ro/2, much higher than ro (simple mirror) and ro(1 + gm RE) (Widlar). - Its β error is very small: IO = IREF[1 − 2/(β² + 2β + 2)], compared with IREF/(1 + 2/β) in the simple mirror. Between the two derived here, the Widlar source is better than the simple mirror because it gives a higher output resistance and small currents with small resistors. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which is NOT a requirement of a current mirror? A. Matched transistors B. Same temperature C. High β D. Output transistor in saturation Answer: D. Q2 must remain in the active region (VCE2 > VCE(sat)), not saturation. (Key point: Requirements) 2. What is the output resistance of the Widlar current source (RE << rπ)? A. gm RE B. ro/(1 + gm RE) C. ro D. ro(1 + gm RE) Answer: D. The unbypassed RE gives current-series feedback, raising Ro from ro to about ro(1 + gm RE). (Key point: Widlar output resistance) 3. Which source is usually the best of the common ones? A. Simple mirror B. Resistor bias C. Wilson (or cascode) source D. Widlar source Answer: C. The Wilson source has Ro about β·ro/2 and a very small β error. (Key point: Best source) 4. What is the output resistance of the Wilson source? A. About 1/gm B. ro C. ro(1 + gm RE) D. About β ro / 2 Answer: D. Wilson Ro ≈ β·ro/2, much higher than ro or the Widlar value. 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