Download Recall Lecture 12

Recall Last Lecture
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Voltage Transfer Characteristic
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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
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Fixed Bias Biasing Circuit
Vo (V)
Cutoff
5
= 4.8
Active
Saturation
0.2
0.7
x
5 V (V)
i
x
= 4.3
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 b=100.
NOTE: Proposed to use branch current equations and node voltages
BiasingI =using
Collector
to
Base
Feedback
1mA
,
V
=
2.3
V,
V
=
10
V
and
b=100.
E
CE
CC
Resistor
VC
VB
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•
•
•
•
(VC – VB ) / RB= IB
but VC = VCE
and 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
Voltage Divider Biasing Circuit
This is a very stable bias circuit.
The currents and voltages are almost independent of variations in b.
Analysis
Redrawing the input side of the network by
changing it into Thevenin Equivalent
RTh: the voltage source is replaced
by a short-circuit equivalent
Analysis
VTh: open-circuit Thevenin voltage
is determined.
VTH
VTH
Use voltage divider
Inserting the Thevenin
equivalent circuit
Analysis
The Thevenin equivalent circuit
BJT Biasing in Amplifier
Example
Find 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 )VCC
VTH = 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)=2.16mA
IE = IC + IB = 2.18mA
VCC = RC IC + VCE + RE IE
10 = 2k(2.16m)+VCE +0.4(2.18m)
VCE = 4.8 V
Basic Transistor
Application
Digital Logic – NOT GATE
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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.
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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 Gate
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If the two inputs are zero,
both transistors Q1 and Q2
are in cutoff, and VO = 5 V.
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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).
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If V1 = 0 and V2 = 5 V, then Q1 is in cutoff,
and Q2 can be driven in saturation, and
VO = VCE (sat).
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If both inputs are high,
meaning V1 = V2 = 5 V,
then both transistors can
be driven into saturation,
and VO = VCE (sat).
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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.
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The circuit is then a two-input bipolar
NOR logic circuit.