Recall Last Lecture Voltage Transfer Characteristic A plot of Vo versus Vi Use BE loop to obtain a current equation, IB in terms of Vi Use CE loop to get IC in terms of Vo Change IC in terms of IB Equate the two equations to link Vi with Vo

Vo (V) Vi (V) 5 0.7 Cutoff Active 0.2 x Saturation = 4.8 x = 4.3

Bipolar Transistor Biasing

Bipolar Transistor Biasing Biasing refers to the DC voltages applied to the transistor for it to turn on and operate in the forward active region, so that it can amplify the input AC signal

Proper Biasing Effect Ref: Neamen

Effect of Improper Biasing on Amplified Signal Waveform Ref: Neamen

Three types of biasing Fixed Bias Biasing Circuit Biasing using Collector to Base Feedback Resistor Voltage Divider Biasing Circuit

Biasing Circuits – Fixed Bias Biasing Circuit The circuit is one of the simplest transistor circuits is known as fixed-bias biasing circuit. There is a single dc power supply, and the quiescent base current is established through the resistor RB. The coupling capacitor C1 acts as an open circuit to dc, isolating the signal source from the base current. Typical values of C1 are in the rage of 1 to 10 μF, although the actual value depends on the frequency range of interest.

Example – Fixed Bias Biasing Circuit Determine the following: (a) IB and IC (b) VCE Assume VBE(on) = 0.7 V, and  = 50 (a) VB – VE = 0.7 VB = 0.7V IB = (12 – 0.7) / 240k = 0.0471 mA IC = 50 (0.0471) = 2.355 mA VB = ?? VE = 0V (b) IC = 2.355 mA = (12 – VC) / 2.2k VC = 6.819 V VCE = 6.819 – 0 = 6.819 V NOTE: Proposed to use branch current equations and node voltages

Biasing using Collector to Base Feedback Resistor IC + IB = IE IB IC IE Find RB and RC such that IE = 1mA , VCE = 2.3 V, VCC = 10 V and  =100. NOTE: Proposed to use branch current equations and node voltages

Biasing using Collector to Base Feedback Resistor IE = 1mA , VCE = 2.3 V, VCC = 10 V and  =100. IB = (IE / (+1) = 0.0099 mA (VC – VB ) / RB= IB but VC = VCE and VB = VBE = 0.7 V (2.3 – 0.7) / RB = 0.0099 mA RB = 161.6 k (VCC – VC ) / RC = IE RC = 7.7 k VC VB VE = 0V

Voltage Divider Biasing Circuit This is a very stable bias circuit. The currents and voltages are almost independent of variations in .

Analysis Redrawing the input side of the network by changing it into Thevenin Equivalent RTH RTh: the voltage source is replaced by a short-circuit equivalent

Analysis VTH VTh: open-circuit Thevenin voltage is determined. VTH Inserting the Thevenin equivalent circuit Use voltage divider

Analysis The Thevenin equivalent circuit

Example Find VCE ,IE, IC and IB given β=100, VCC=10V, R1 = 56 k, R2 = 12.2 k, RC = 2 k and RE = 0.4 k VTH= R2 /(R1 + R2 )VCC VTH = 12.2k/(56k+12.2k).(10) VTH = 1.79V RTH = R1 // R2 = 10 k

BJT Biasing in Amplifier Circuits VTH = RTH IB + VBE + RE IE 1.79 = 10k IB + 0.7 + 0.4k ( +1)IB IB = 21.62mA IC = bIB = 100(21.62m)=2.16mA IE = IC + IB = 2.18mA VCC = RC IC + VCE + RE IE 10 = 2k(2.16m)+VCE +0.4(2.18m) VCE = 4.8 V

Example – A Design Question VCC = 10 V IE = 1 mA =100 RE = 0.470 k RTH = 33 k Determine the values of R1 and R2

R2 = 38.8 k BE Loop: IBRTH + 0.7 + IERE – VTH = 0 VTH = 1.5 V We know:       1.5 (220 + R2) = 10R2 330 = 8.5 R2 R2 = 38.8 k

QUIZ 2 91 k 110 k 3.3 k 18 V Find the currents, IE , IB and calculate the value of  given that VCE = 4.8 V. Assume VBEon = 0.7 V