---
title: "Current mirror advantages and Widlar output resistance"
subject: "Advanced Electronics"
chapter: 1
key: "semester_4/advanced_electronics:current-mirror-advantages-widlar"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/current-mirror-advantages-widlar
site: BCT Engineering
---

# Current mirror advantages and Widlar output resistance

Advanced Electronics · Chapter 1: Operational Amplifier Circuits

- Asked in: 2074 Chaitra, 3+5 marks
- 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/current-mirror-advantages-widlar
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-1

## Question

What is a Current Mirror and its advantages? Find the output resistance of Widlar Current Source.

## Short answer

A current mirror copies a reference current to an output branch using matched VBE. Advantages: constant current, saves chip area, enables biasing many stages, good temperature tracking, and active‑load capability. The Widlar source adds emitter degeneration, giving Ro ≈ ro(1+gmRE) for RE<<rπ, far exceeding a simple mirror’s ro.

## Answer

#### Current mirror and its advantages
A **current mirror** is a circuit in which a reference current IREF set in one branch (through a diode-connected transistor Q1) is copied to an output branch (Q2) because both transistors have the same VBE.

```text
  +VCC                 IO (from load)
    |                    |
    R                    |
    | IREF               |
    +------+             |
    |      |             |
    C1     |             C2
  Q1  B----+-------------B  Q2
    E1   (Q1: C tied to B)   E2
    |                    |
   GND                  GND
```

```text
IREF = (VCC - VBE)/R
IREF = IC + 2IC/β  →  IO = IC = IREF/(1 + 2/β) ≈ IREF
```

**Advantages:**
- Gives a constant current that does not depend on the load (high output resistance ro).
- Uses transistors instead of large resistors, saving chip area in ICs.
- One reference can bias many stages (current steering).
- Good temperature tracking because the matched transistors are on the same chip.
- Used as an active load, giving very high gain in differential amplifiers.
- No coupling or bypass capacitors needed; works down to dc.

#### 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
```

The emitter resistor RE makes IO much smaller than IREF: 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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is the output current copied in a current mirror?
   A. Both transistors have the same VBE
   B. The load is the same
   C. Both have the same collector resistor
   D. Q2 is diode-connected
   Answer: A. Q1 is diode-connected and shares its VBE with Q2, so IC2 follows the reference current. (Key point: Definition)

2. What is the output current of a simple mirror with finite β?
   A. IO = IREF/(1 + 2/β)
   B. IO = IREF(1 + 2/β)
   C. IO = IREF·β
   D. IO = IREF/β
   Answer: A. IREF = IC + 2IC/β, so IO = IREF/(1 + 2/β). (Key point: Finite β)

3. Why is a mirror advantageous in ICs?
   A. It needs matched resistors only
   B. It needs large capacitors
   C. It uses transistors instead of large resistors, saving chip area, and one reference biases many stages
   D. It works only at high frequency
   Answer: C. Transistors are cheap and matched on-chip; mirrors replace big resistors and allow current steering. (Key point: Advantage)

4. 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)

## Related in Advanced Electronics

- [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 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×
- [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
- [Benefits of Widlar source and output resistance](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Awidlar-benefits-output-resistance) — Advanced Electronics, chapter 1, asked 2×
- [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
- [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