---
title: "Design CE amplifier (BC547B, no bypass)"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:ce-design-bc547b-no-bypass"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-design-bc547b-no-bypass
site: BCT Engineering
---

# Design CE amplifier (BC547B, no bypass)

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

- Asked in: 2079 Baisakh, 5+2+3+2 marks
- 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-design-bc547b-no-bypass
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Design voltage divider CE amplifier (without emitter by pass capacitor). Given: Transistor BC 547B having β = 295, Ic = 1.5 mA and Vcc = +9V.
a) Is this the best Q point? Why?
b) Calculate its input impedance and voltage gain.
c) What is the maximum peak voltage of the signal that can be applied to the input of this amplifier to ensure the transistor is always in active region?

## Short answer

a) Yes, Q-point at VCE=4.5V, IC=1.5mA is mid-load-line for maximum symmetrical swing. b) Zin≈22.3kΩ, Av≈-3.9. c) vce/vi = -4.9, so max input peak ≈ 0.88V (limited by saturation).

## Answer

#### Design (VCC = 9 V, IC = 1.5 mA, β = 295, RE not bypassed)
Guidelines: VE ≈ VCC/10, VCE ≈ VCC/2, divider current I2 = 10 IB, VB = VE + 0.7 V.

```text
IB = IC/β = 1.5 mA/295 = 5.08 µA          IE = 1.505 mA
VE = 0.9 V      RE = 0.9/1.505 mA = 598 Ω  (use 600 Ω)
VCE = 4.5 V     RC = (9 - 0.9 - 4.5)/1.5 mA = 2.4 kΩ
VB = 1.6 V      I2 = 10 IB = 50.8 µA
R2 = 1.6/50.8 µA = 31.5 kΩ                 R1 = (9 - 1.6)/55.9 µA = 132 kΩ
```
Choose R1 = 130 kΩ, R2 = 31.5 kΩ (30 kΩ), RE = 600 Ω, RC = 2.4 kΩ.

#### a) Is this the best Q-point?
Yes, for the unloaded amplifier. The dc load line equation is VCC = IC(RC + RE) + VCE. The Q-point is at VCE = 4.5 V and IC(RC + RE) = 1.5 mA × 3 kΩ = 4.5 V, so it lies at the **middle of the load line** (VCEQ = VCC/2 and ICQ = half of the saturation current of 3 mA). Since RE is not bypassed and there is no load, the ac load line is the same as the dc load line, so the Q-point is also at the centre of the ac load line. The output can swing about equally towards cut-off and towards saturation, giving the maximum symmetrical undistorted swing. (The actual swing is slightly less than 4.5 V because VCE(sat) ≈ 0.2 V.)

#### b) Input impedance and voltage gain
```text
re = VT/IE = 25 mV/1.505 mA = 16.6 Ω
Rib = (β+1)(re + RE) = 296 × (16.6 + 598) = 181.9 kΩ
Zin = R1 || R2 || Rib = 132 k || 31.5 k || 181.9 k = 22.3 kΩ

Av = -RC/(re + RE) = -2400/(16.6 + 598) = -3.9
```
**Zin ≈ 22.3 kΩ, Av ≈ -3.9.**

#### c) Maximum input peak for active operation
Because RE is unbypassed, the signal current ic flows through both RC and RE, so the collector-emitter voltage changes by ic(RC + RE), not just ic RC:
```text
vce/vi = -(RC + RE)/(re + RE) = -3000/614.6 ≈ -4.9
```
- Saturation side: VCE can fall from 4.5 V to VCE(sat) ≈ 0.2 V, a swing of 4.3 V, so vi(max) = 4.3/4.9 ≈ 0.88 V.
- Cut-off side: ic can fall by at most ICQ = 1.5 mA, so vi(max) = 1.5 mA × (re + RE) ≈ 1.5 mA × 614.6 Ω ≈ 0.92 V.

The smaller value (saturation side) sets the limit:
```text
vi(max) ≈ 4.3/4.9 ≈ 0.88 V (peak)
```
**The input peak must be below about 0.88 V** (assuming a signal source with negligible resistance). The large unbypassed RE keeps the stage linear, which is why such a large input is possible.

## Check yourself

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

1. Why is this Q-point considered the best for the unloaded amplifier?
   A. VCE is near zero
   B. It maximises beta
   C. It is at the middle of the dc load line (VCEQ = VCC/2, ICQ = half of saturation current)
   D. IC is near zero
   Answer: C. The middle Q-point gives equal swing toward cut-off and saturation. (Key point: Q-point optimal)

2. How is Zin found with unbypassed RE?
   A. Zin = R1 || R2 || (beta+1)(re + RE) ≈ 22.3 kΩ
   B. Zin = (beta+1) re only
   C. Zin = R1 || R2 || rpi ≈ 1 kΩ
   D. Zin = RC ≈ 2.4 kΩ
   Answer: A. RE is in the ac path, so the base sees (beta+1)(re+RE) ≈ 182 kΩ in parallel with the divider. (Key point: Input impedance)

3. What is Av?
   A. +RE/(re+RE) ≈ +1
   B. -RC/(re+RE) ≈ -3.9
   C. -gm RC ≈ -216
   D. -beta ≈ -295
   Answer: B. With RE unbypassed Av = -2400/614.6 ≈ -3.9. (Key point: Voltage gain)

4. Why is the maximum input peak about 1.1 V?
   A. 4.5 V divided by 295
   B. VCC divided by beta
   C. 0.7 V forward drop
   D. The smaller output swing (4.3 V toward saturation) divided by |Av| = 3.9
   Answer: D. The saturation side limits the output swing to 4.5 - 0.2 = 4.3 V; 4.3/3.9 ≈ 1.1 V. (Key point: Maximum input peak)

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