Recall Last LectureVoltage Transfer CharacteristicA plot of Vo versus ViUse BE loop to obtain a current equation, IB in terms of ViUse CE loop to get IC in terms of VoChange IC in terms of IBEquate the two equations to link Vi with VoBipolar Transistor BiasingFixed Bias Biasing Circuit

= 4.8 = 4.3x

Biasing using Collector to Base Feedback ResistorFind RB and RC such that IE = 1mA , VCE = 2.3 V, VCC = 10 V and b=100.

ICIEIBIC + IB = IENOTE: Proposed to use branch current equations and node voltages

Biasing using Collector to Base Feedback Resistor(VC VB ) / RB= IB but VC = VCEand VB = VBE = 0.7 V(2.3 0.7) / RB = (IE / (b+1) RB = 161.6 kW

(VCC VC ) / RC = IE RC = 7.7 kW

IE = 1mA , VCE = 2.3 V, VCC = 10 V and b=100.VCVB

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

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

VTh: open-circuit Thevenin voltage is determined.Inserting the Thevenin equivalent circuitAnalysisUse voltage divider

The Thevenin equivalent circuitAnalysis

BJT Biasing in Amplifier ExampleFind VCE ,IE, IC and IB given b=100, VCC=10V, R1 = 56 kW, R2 = 12.2 kW, RC = 2 kW and RE = 0.4 kW

VTH= R2 /(R1 + R2 )VCCVTH = 12.2k/(56k+12.2k).(10)VTH = 1.79V

RTH = R1 // R2 = 10 kW

BJT Biasing in Amplifier Circuits VTH = RTH IB + VBE + RE IE 1.79 = 10k IB + 0.7 + 0.4k (b+1)IB IB = 21.62mA

IC = bIB = 100(21.62m)=2.16mAIE = IC + IB = 2.18mAVCC = RC IC + VCE + RE IE 10 = 2k(2.16m)+VCE +0.4(2.18m)VCE = 4.8 V

Basic Transistor Application

Digital Logic NOT GATEIn the simple inverter circuit, if the input is approximately zero volts, the transistor is in cutoff and the output is high and equal to VCC.If the input is high and equal to VCC, the transistor is driven into saturation, and the output is low and equal to VCE (sat).

Digital Logic NOR GateIf the two inputs are zero, both transistors Q1 and Q2 are in cutoff, and VO = 5 V.

When V1 = 5 V and V2 = 0, transistor Q1 can be driven into saturation, and Q2 remains in cutoff. With Q1 in saturation, the output voltage VO = VCE (sat). If V1 = 0 and V2 = 5 V, then Q1 is in cutoff, and Q2 can be driven in saturation, and VO = VCE (sat).

If both inputs are high, meaning V1 = V2 = 5 V, then both transistors can be driven into saturation, and VO = VCE (sat).

In a positive logic system, meaning that the larger voltage is a logic 1 and the lower voltage is a logic 0, the circuit performs the NOR logic function.

The circuit is then a two-input bipolar NOR logic circuit.

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