BJT Biasing Circuit FixedBase Bias Litar Pincangan Tetap

BJT Biasing Circuit – Fixed/Base Bias (Litar Pincangan Tetap) – Fixed Bias with Emitter Resistor (Litar Pincangan Pemancar Terstabil) – Voltage-Divider Bias Circuit (Litar Pincangan Pembahagi Voltan) – Feedback Bias Circuit (Litar Pincangan Suap-Balik Voltan)

Base Bias

Base Bias This is a common emitter configuration Solve the circuit using Kirchoff Voltage Law Locate the 2 main circuit loop which is the BE and CE circuit

Base Emitter Loop - KVL Using KVL: ΣV = 0 -VCC +IBRB + VBE = 0 Thus: IB = (VCC - VBE) / RB

Common Emitter Loop KVL We know IC = βIB KVL : -VCE - ICRC + VCC = 0 Recall that VCE = VC - VE Note that VE = 0 V (Grounded) Therefore VCE = VC As VBE = VB - VE But VE = 0 Therefore VBE = VB

Exercise 1 4. 1 165 Determine a. IBQ and ICQ b. VCEQ c. VB and VC d. VBC

Answer 1 VB = 0. 7 V VC = 6. 83 V VBC = -6. 13 V

Disadvantage of Base Bias • Base or Fixed Bias is essentially not stable. Why? • Too dependent on β and produce a wide change on Q point. • For a improved bias stability, emitter resistor is added to the dc bias.

BJT bias circuit with emitter resistor.

Improved Bias Stability Fixed Bias IB ( A) 50 47. 08 100 47. 08 IB 0% IC (m. A) 2. 35 4. 70 IC 100% VCE 6. 38 1. 64 VCE 74% Emitter Bias IB ( A) 50 40. 1 100 36. 3 IB 9% IC (m. A) 2. 01 3. 63 IC 81% VCE 13. 97 9. 11 VCE 35%

Emitter Bias • An emitter resistor, RE to improve stability level order. • BJT bias circuit with emitter resistor on the left. • To solve the analysis, start with BE-KVL follow by CE-KVL

Base Emitter - KVL Writing Kirchhoff’s Voltage Law (KVL) : Recall Substituting for IE Solving for IB gives

Prove the circuit below

FIGURE 4 -20 Reflected impedance level of RE.

Base Collector - KVL In the clockwise direction, KVL gives Substituting Thus VCE:

Collector-emitter loop.

Exercise 2 • A BJT has a value of IB= 35 A and β = 100. Determine the value of IC and IE for the device. • Complete the following table. Beta IB IC - 50 A 12 m. A 440 - 35 m. A 175 45 A - - 120 A 84 m. A

Answer 2 • A BJT has a value of IB= 35 A and β = 100. Determine the value of IC and IE for the device. • Complete the following table. Beta IB IC 240 50 A 12 m. A 440 79. 5 A 35 m. A 175 45 A 7. 8 m. A 700 120 A 84 m. A

Exercise 3 Determine • • 4. 4 172 IB IC VCE VC VE VB VBC

Exercise 3 Determine • • 4. 4 172 IB IC VCE VC VE VB VBC

Exercise 4 ? Find ICsat and VCEcutoff for Emitter Bias Circuit ?

Emitter Bias Load Line Analysis

Voltage-divider bias configuration • Most stable. • Changes in β is small.

Redrawing the input side of the network

Determining RTH.

Determining Eth/ Vth

Inserting the Thévenin equivalent circuit.

Input Loop What is Theorem Thevenin is used for?

DC Analysis HVK: where Therefore

Output Loop • In the CE loop, VCE can be written as

Q-point for the voltage-divider bias configuration

Load line for voltage-divider bias Find ICSAT and VCEcutoff Load line for the emitter-bias configuration

Load line for voltage-divider bias and Emitter bias is the same

Exercise 4 Determine • RTH • ETH • IB and IC • VCE • VC and VE • VB and VBC 4. 7 177

Exercise 5 Determine • R 1 • RC Given that IC = 2 m. A and VCE=10 V 4. 20 192

Exercise 6 Determine • VC • VB 4. 18 189