---
title: "Output resistance of unbypassed-emitter CE amplifier"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:ce-unbypassed-output-resistance"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-unbypassed-output-resistance
site: BCT Engineering
---

# Output resistance of unbypassed-emitter CE amplifier

Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor

- Asked in: 2068 Baisakh, 5 marks
- Video: [CE Amplifier with Unbypassed RE - Small Signal Analysis using Hybrid Equivalent Model](https://www.youtube.com/watch?v=XcDXJ1S2Sfk) — Padmasri Naban
- 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/ce-unbypassed-output-resistance
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Derive an expression to find output resistance for emitter unbypassed common emitter amplifier circuit.

## Short answer

The collector-side resistance becomes ro(1 + gm R'E) + R'E with R'E = RE||(rπ+RB'). Since this is much larger than RC, the amplifier's output resistance is Rout ≈ RC.

## Answer

The output resistance is found by setting the input source to zero (vs = 0), removing the load and applying a test voltage vx at the output (collector); Rout = vx/ix.

#### Small-signal circuit (hybrid-π with ro)
```text
 test vx --> C ----+----------+
 ix              ro         gm vπ
                   |          |
 B--rπ--+----------E----------+
 |      |          |
 RB'    vπ         RE
 |                 |
GND               GND
```
RB' = Rs || R1 || R2 is the resistance from the base to ground (vs = 0). The emitter node sees RE in parallel with the base path (rπ + RB'):

```text
R'E = RE || (rπ + RB')
```

#### Derivation
The collector current ix returns to ground through the emitter, so
```text
ve = ix R'E
```
The base voltage is ve × RB'/(rπ + RB'), so vπ = vb - ve = -a ve, where a = rπ/(rπ + RB') ≤ 1. KCL at the collector:
```text
ix = gm vπ + (vx - ve)/ro = -gm a ve + (vx - ve)/ro
   = -gm a ix R'E + (vx - ix R'E)/ro

ix [1 + gm a R'E + R'E/ro] = vx/ro
```
Hence the resistance looking into the collector is
```text
Rout(collector) = vx/ix = ro (1 + gm a R'E) + R'E
                ≈ ro (1 + gm R'E)        (for RB' ≪ rπ, so a ≈ 1)
```
The unbypassed RE (series feedback) raises the collector resistance far above ro.

#### Output resistance of the amplifier
The collector resistance is in parallel with RC:
```text
Rout = RC || [ro (1 + gm a R'E) + R'E]  ≈ RC
```
because the bracket is very large (for example ro = 50 kΩ, gm R'E = 10 gives 550 kΩ). If ro is ignored, Rout = RC.

**Result: Rout ≈ RC**, the same as for the bypassed amplifier; RE does not lower the output resistance.

## Check yourself

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

1. How is the output resistance found in this derivation?
   A. Set vs = 0, keep the load, measure the open-circuit voltage
   B. Set vs = 0, remove the load, apply a test voltage vx at the collector and find vx/ix
   C. Short the collector to ground and measure ix
   D. Set RE = 0 and apply vx at the base
   Answer: B. Rout = vx/ix with the source zeroed and the load removed. (Key point: Method for Rout)

2. What is R'E, the resistance seen at the emitter node?
   A. RE only
   B. RE + rπ
   C. RE in parallel with (rπ + RB')
   D. rπ + RB'
   Answer: C. At the emitter, RE is in parallel with the path through rπ and the base resistance RB': R'E = RE || (rπ + RB'). (Key point: Effective emitter resistance)

3. What is the resistance looking into the collector?
   A. ro (1 + gm a R'E) + R'E, about ro(1 + gm R'E)
   B. ro only
   C. ro/(1 + gm R'E)
   D. RC (1 + gm RE)
   Answer: A. Series feedback from the unbypassed emitter resistance multiplies ro by (1 + gm R'E). (Key point: Collector resistance result)

4. What is the output resistance of the whole amplifier and why?
   A. Rout is much smaller than RC because RE adds feedback
   B. Rout = RE
   C. Rout is about RC, because the collector resistance is very large and in parallel with RC
   D. Rout = ro(1 + gm RE)
   Answer: C. RC is in parallel with a very large collector resistance, so Rout is approximately RC, same as the bypassed case. (Key point: Final Rout)

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