---
title: "Design CE voltage divider bias, beta 200"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:ce-divider-bias-design-beta200"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/ce-divider-bias-design-beta200
site: BCT Engineering
---

# Design CE voltage divider bias, beta 200

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

- Asked in: 2081 Bhadra, 8 marks
- Video: [AEC#13 Problems with voltage divider biasing using BJT || EC Academy](https://www.youtube.com/watch?v=-1t1f4rtfB8) — EC Academy
- 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-divider-bias-design-beta200
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Design common emitter BJT voltage divider bias circuit with high input impedance. Given: β= 200, IC = 2 mA, and VCC = +12 V.

## Short answer

Design: VE=1.2V, VCE=6V, RE=597Ω, RC=2.4kΩ, R1=168kΩ, R2=38kΩ. RE unbypassed yields Rin≈24.7kΩ, Av≈-3.9, ensuring high input impedance with moderate gain.

## Answer

#### Design guidelines used
1. **Emitter voltage** VE ≈ VCC/10 (large enough for thermal and bias stability, small enough to leave room for the output swing).
2. **Collector-emitter voltage** VCE ≈ VCC/2, so the output can swing equally up and down (Q-point at the centre of the load line).
3. **Divider current** I2 = 5 IB, the lower end of the 5 to 10 IB range, so that R1 || R2 is as large as possible (high input impedance) while the base voltage stays fixed and the bias remains independent of β.
4. VB = VE + 0.7 V.
5. For a high input impedance, RE is left unbypassed, so that (β+1)RE appears in the input resistance.

#### Design
```text
IC  = 2 mA        IB = IC/β = 2 mA/200 = 10 µA
IE  = IC + IB = 2.01 mA
VE  = VCC/10 = 1.2 V     RE = VE/IE = 1.2/2.01 mA = 597 Ω
VCE = VCC/2  = 6 V
RC  = (VCC - VE - VCE)/IC = (12 - 1.2 - 6)/2 mA = 2.4 kΩ
VB  = VE + 0.7 = 1.9 V
I2  = 5 IB = 50 µA
R2  = VB/I2 = 1.9/50 µA = 38 kΩ
R1  = (VCC - VB)/(I2 + IB) = 10.1/60 µA = 168 kΩ
```
Calculated values: R1 ≈ 168 kΩ, R2 ≈ 38 kΩ, RE ≈ 597 Ω, RC ≈ 2.4 kΩ (in practice, use the nearest standard resistor values). C1 and C2 are the coupling capacitors.

#### Check and small-signal results (RE unbypassed)
```text
re  = VT/IE = 25 mV/2.01 mA = 12.4 Ω
R1 || R2 = 31 kΩ
Rib = (β+1)(re + RE) = 201 × (12.4 + 597) = 122.5 kΩ
Rin = R1 || R2 || Rib = 31 k || 122.5 k = 24.7 kΩ
Av  = -RC/(re + RE) = -2400/609.4 = -3.9
```
Ro ≈ RC = 2.4 kΩ.

Because the base current is small (β = 200), a divider current of only 5 IB (50 µA) still keeps VB stiff and gives R1 || R2 ≈ 31 kΩ, so the amplifier's input resistance is high (≈ 24.7 kΩ). If RE were bypassed by a capacitor, the gain would rise to -gm RC = -80 mA/V × 2.4 kΩ = -192, but Rin would drop to R1 || R2 || rπ = 31 k || 2.49 k ≈ 2.3 kΩ (rπ = β re = 2488 Ω).

## Check yourself

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

1. For VCC = 12 V, IC = 2 mA, beta = 200, VE and RE are:
   A. VE = 1.2 V, RE ≈ 5.97 kΩ
   B. VE = 1.2 V, RE ≈ 597 Ω
   C. VE = 6 V, RE ≈ 3 kΩ
   D. VE = 1.9 V, RE ≈ 945 Ω
   Answer: B. VE = VCC/10 = 1.2 V; RE = 1.2/2.01 mA ≈ 597 Ω. (Key point: VE and RE)

2. What is RC?
   A. 6 kΩ from 12 V/2 mA
   B. 2.4 kΩ from (12 - 1.2 - 6)/2 mA
   C. 5.4 kΩ from 10.8/2 mA
   D. 3 kΩ from 6 V/2 mA
   Answer: B. RC = (VCC - VE - VCE)/IC = 4.8 V / 2 mA = 2.4 kΩ. (Key point: RC)

3. With VB = 1.9 V and I2 = 5 IB = 50 µA, R2 and R1 are about:
   A. R2 = 3.8 kΩ, R1 = 16.8 kΩ
   B. R2 = 19 kΩ, R1 = 101 kΩ
   C. R2 = 38 kΩ, R1 = 168 kΩ
   D. R2 = 168 kΩ, R1 = 38 kΩ
   Answer: C. R2 = 1.9/50 µA = 38 kΩ; R1 = 10.1 V/(50+10) µA = 168 kΩ. (Key point: Divider values)

4. Why is a divider current of only 5 IB acceptable here?
   A. Because VCC is low
   B. Beta is high so IB is small (10 µA), and a 50 µA divider current still keeps VB stiff with a large R1||R2
   C. Because VBE is 0.7 V
   D. Because RE is bypassed
   Answer: B. A small IB lets a modest divider current hold VB fixed, giving high input resistance. (Key point: Why 5 IB is acceptable)

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