recall last lecture voltage transfer characteristic a plot of v o versus v i use be loop to obtain a...

Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop to get I C in terms of V o Change I C in terms of I B Equate the two equations to link V i with V o Bipolar Transistor Biasing Fixed Bias Biasing Circuit

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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

Bipolar Transistor Biasing Fixed Bias Biasing Circuit

Page 2: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

= 4.8

= 4.3

Vo (V)

Vi (V)

0.7

Cutoff

Active

0.2

x 5

Saturation

Page 3: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

Biasing using Collector to Base Feedback Biasing using Collector to Base Feedback ResistorResistor

Find RB and RC such that IE = 1mA , VCE = 2.3 V, VCC = 10 V and =100.

IC + IB = IE

NOTE: Proposed to use branch current equations and node voltages

Page 4: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

Biasing using Collector to Base Feedback Resistor

• (VC – VB ) / RB= IB

• but VC = VCE

• and VB = VBE = 0.7 V

• (2.3 – 0.7) / RB = (IE / (+1)

• RB = 161.6 k

• (VCC – VC ) / RC = IE

• RC = 7.7 k

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

Page 5: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

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

Voltage Divider Biasing CircuitVoltage Divider Biasing Circuit

Page 6: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

Redrawing the input side of the network by changing it into Thevenin Equivalent

RTh: the voltage source is replaced by a short-circuit equivalent

AnalysisAnalysis

Page 7: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

VTh: open-circuit Thevenin voltage is determined.

Inserting the Thevenin equivalent circuit

AnalysisAnalysis

VTH

Use voltage divider

VTH

Page 8: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

The Thevenin equivalent circuit

AnalysisAnalysis

Page 9: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

BJT Biasing in Amplifier ExampleFind VCE ,IE, IC and IB given

=100, VCC=10V, R1 = 56 k, R2 = 12.2 k

RC = 2 kandRE = 0.4 k

VTH= R2 /(R1 + R2 )VCC

VTH = 12.2k/(56k+12.2k).(10)

VTH = 1.79V

RTH = R1 // R2

= 10 k

Page 10: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

BJT Biasing in Amplifier Circuits

VTH = RTH IB + VBE + RE IE

1.79 = 10k IB + 0.7 + 0.4k (+1)IB

IB = 21.62A

IC = IB = 100(21.62)=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

Page 11: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

Basic Transistor Basic Transistor ApplicationApplication

Page 12: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

Digital Logic – NOT GATEDigital Logic – NOT GATE

In 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).

Page 13: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

Digital Logic – NOR GateDigital Logic – NOR Gate

If 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).

Page 14: Recall Last Lecture Voltage Transfer Characteristic A plot of V o versus V i Use BE loop to obtain a current equation, I B in terms of V i Use CE loop

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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