--- title: "Benefits of Widlar source and output resistance" subject: "Advanced Electronics" chapter: 1 key: "semester_4/advanced_electronics:widlar-benefits-output-resistance" source: https://www.bctengineer.com/q/semester_4/advanced_electronics/widlar-benefits-output-resistance site: BCT Engineering --- # Benefits of Widlar source and output resistance Advanced Electronics · Chapter 1: Operational Amplifier Circuits - Asked in: 2074 Asoj, 3+5 marks; 2071 Shrawan, 2+5 marks (asked 2 times) - Video: [Lec-51 Wildar Current Source- Basics, Derivation, Output R, Advantages | A E | R K Classes | Hindi |](https://www.youtube.com/watch?v=DY7bkW4VsfQ) — R K Classes - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/widlar-benefits-output-resistance - Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-1 ## Question What are the benefits of Widlar current source as compared to a simple current mirror (current source)? Derive an expression for an output resistance of Widlar Current Source. ## Short answer The Widlar source provides small currents (µA range) with practical resistor values, saving chip area. It offers higher output resistance Ro ≈ ro(1+gmRE) ≈ ro(1+ln(IREF/IO)), lower sensitivity to supply voltage, and reduced dependence on β and temperature. ## Answer #### Benefits of the Widlar source over a simple mirror In a simple mirror IO ≈ IREF = (VCC − VBE)/R, so a small current needs a huge resistor. The Widlar source adds RE in the emitter of Q2, giving IO·RE = VT ln(IREF/IO). | Point | Simple mirror | Widlar source | |---|---|---| | Output current | IO ≈ IREF | IO << IREF (µA range) | | Resistor for 10 µA (VCC = 15 V) | R ≈ 1.4 MΩ | R ≈ 14 kΩ and RE ≈ 12 kΩ | | Chip area | very large (big R) | small | | Output resistance | ro | ≈ ro(1 + gm RE), several times higher | | Sensitivity of IO to VCC | IO ∝ VCC | much less (logarithmic) | | Sensitivity to β, temperature | higher | lower (emitter feedback) | So the Widlar source gives small, stable bias currents with practical resistor values, a higher output resistance (better current source) and lower dependence on supply voltage. It is used in the input-stage biasing of the 741. #### 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 ``` 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the Widlar source preferred over a simple mirror for a 10 µA current? A. It gives IO larger than IREF B. It needs a larger chip area C. It needs no resistor at all D. It needs only moderate resistors (about 14 kΩ and 12 kΩ) instead of about 1.4 MΩ Answer: D. Because IO << IREF with IO·RE = VT ln(IREF/IO), small resistors suffice, saving chip area. (Key point: Benefit: small resistors) 2. How does IO of the Widlar source depend on VCC compared with a simple mirror? A. Equally (IO ∝ VCC) B. Much less, because the relation is logarithmic C. Not at all, it depends on β only D. Much more strongly Answer: B. IO·RE = VT ln(IREF/IO) is logarithmic, so IO changes little when VCC (and IREF) changes. (Key point: Supply sensitivity) 3. 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) 4. In the small-signal analysis, why does RE raise the output resistance? A. RE increases β of Q2 B. RE makes Q1 and Q2 have different VBE C. A rise in IO raises the drop across RE, lowers VBE2 and pulls IO back down D. RE increases the Early voltage of Q2 Answer: C. Negative (current-series) feedback from RE opposes any change in IO caused by a change in output voltage. (Key point: Emitter feedback) ## Related in Advanced Electronics - [Mirror output current error and Widlar resistance](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Amirror-error-widlar-output-resistance) — Advanced Electronics, chapter 1 - [Current mirror conditions and Widlar output resistance](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Awidlar-essential-conditions-output-resistance) — Advanced Electronics, chapter 1, asked 3× - [Current mirror requirements, Widlar resistance, best source](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Acurrent-mirror-best-source) — Advanced Electronics, chapter 1 - [Current mirror advantages and Widlar output resistance](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Acurrent-mirror-advantages-widlar) — Advanced Electronics, chapter 1 - [Widlar emitter resistance for 10 μA](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Awidlar-emitter-resistance-design) — Advanced Electronics, chapter 1 - [Current mirror output current, finite β effects](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Acurrent-mirror-finite-beta) — Advanced Electronics, chapter 1 ## Related in other subjects - [Source current expression and time to reach 25 mA](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_circuit_theory%3Asource-current-25ma-time-classical) — Electrical Circuit Theory, chapter 3