COMSATS Institute of Information Technology
Virtual campus
Islamabad
Dr. Nasim Zafar
Electronics 1 EEE 231
Fall Semester – 2012
DC Analysis of Transistor CircuitsII
Lecture No:
17
References:
Ø Microelectronic Circuits:
Adel S. Sedra and Kenneth C. Smith.
Ø
Electronic Devices :
Thomas L. Floyd ( Prentice Hall ).
Ø
Integrated Electronics
Jacob Millman and Christos Halkias (McGrawHill).
Ø
Electronic Devices and Circuit Theory:
Robert Boylestad & Louis Nashelsky ( Prentice Hall ).
Ø
Introductory Electronic Devices and Circuits:
Robert T. Paynter.
Lecture No. 17
DC Analysis of Transistor CircuitsII
Reference:
Chapter 5.4
Microelectronic Circuits
Adel S. Sedra and Kenneth C. Smith.
DC Analysis of Transistor Circuits
Basic Transistor Operation
v
Consider this circuit as two separate
circuits:
Ø
The BaseEmitter Circuit
Ø
The CollectorEmitter Circuit
Ø
Ø
The amount of current flow in the base
emitter circuit controls the amount of
current that flows in the collector circuit.
Small changes in baseemitter current yields
a large change in collectorcurrent.
DC Analysis of Transistor Circuits
Analysis of this transistor circuit to predict:
v
DC Voltages and
v
Currents requires use of :
Ø
Ohm’s law,
Ø
Kirchhoff’s voltage law
Ø
and the ß for the transistor.
DC Analysis of Transistor Circuits
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Kirchhoff’s voltage law:
Ø
In the Base Circuit:
VBB is distributed across
the baseemitter junction
and RB
Ø
In the collector circuit:
We determine that VCC is
distributed proportionally
across RC and the transistor,
VCE.
Transistor Characteristics and Parameters
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There are three dc voltages and three dc currents to be considered.
IB: dc base current
IE: dc emitter current
IC: dc collector current
VBE: dc voltage across
baseemitter junction
VCB: dc voltage across
collectorbase junction
VCE: dc voltage from
collector to emitter
BJTCurrent and Voltage Analysis
Ø
Ø
For all circuits: Assume the NPN transistor operates in the linear region:
Ø
write BE voltage loop
Ø
write CE voltage loop
When the baseemitter junction, in an NPN transistor is forward biased,
it is like a forward biased diode and has a forwardvoltage drop of:
VBE = 0.7 V
NPN
BJTCurrent and Voltage Analysis:
Ø
Input Circuit:
Forward Biased E
NPN
B Junction.
BJTCurrent and Voltage Analysis:
Ø
Output Circuit: Reverse Biased BC Junction.
Ø
Using Kirchhoff’s voltage law, the voltages in the Output Circuit are:
VCC = VCE + VRL
VCE = VCC VRL
The voltage drop across RL is:
VRL = IC RL
The collector voltage is:
RL
VCE = VCC IC
DC Voltages for the Biased Transistor
v
v
Collector voltage:
VCE = VCC ICRC
Base voltage:
VBE = VCE – VCB
– IB = (VBB – VBE)/RB
– for silicon transistors, VBE = 0.7 V
– for germanium transistors, VBE = 0.3 V
DC Analysis of Transistor Circuits
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Transistor Currents: IE = IC + IB
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alpha ( DC): IC = DCIE
v
beta ( DC): IC = DCIB
–
DC typically has a value between 20 and 200
Examples and Exercises:
DC Analysis of Transistor Circuits
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v
Q: What is IB, IC, IE and
also VCE, VCB, VBE ??
A: I don’t know ! But, we
can find out—IF we
complete each of the four
steps required for BJT
DC analysis.
DC Analysis of Transistor Circuits
Ø
Step 1 – Assume an operating mode.
Let’s assume the BJT is in the linear region ! Remember,
this is just a guess; we have no way of knowing for sure what
mode the BJT is in at this point.
Ø
Step 2 Enforce the conditions of the assumed mode.
Ø
Step 3 – ANALYZE the circuit.
Ø
Step 4 “Write KVL equations for the baseemitter “leg”.
Modes of BJT Operation:
IC(mA)
Saturation Region
IB = 200 A
30
Active Region
IB = 150 A
22.5
IB = 100 A
15
IB = 50 A
7.5
Cutoff Region
IB = 0
0
0
5
10
15
Ø
Active: BJT acts like an amplifier (most common use).
Ø
Saturation: BJT acts like a short circuit.
Ø
Nasim Zafar
Cutoff: BJT acts like an open circuit.
20
18
VCE (V)
Transistor Characteristics and Parameters
The Cutoff Region
With no IB the transistor is in the cutoff region and just as the name
implies there is practically no current flow in the collector part of the
circuit. With the transistor in a cutoff state the full VCC can be
measured across the collector and emitter(VCE)
Example 5.4 Figure 5.34
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Consider the circuit shown in Fig. 5.34(a), which is redrawn in
Fig. 5.34(b)
Example 5.4
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Ø
Ø
Ø
Ø
Example: 5.4
We wish to analyze this circuit to determine:
all node voltages
and branch currents.
We will assume that β is specified to be 100.
Example 5.4 Figure 5.34
Solution Example 5.4
Solution Example 5.4
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v
Ø
Ø
Ø
Input Circuit: Forward Biased EB Junction:
Step 1:
The circuit in Fig. 5.34(b) shows that the base is
connected to +4 V and the emitter is connected to ground
through a resistance RE.
The base–emitter junction will be forward biased.
Since the emitter is grounded, by Kirchhoff’s voltage
law, the voltages in the input circuit are:
Solution Example 5.4
Ø
Ø
Ø
Assuming that VBE is approximately 0.7 V, it follows that the emitter
voltage will be:
Step 2:
We know the voltages at the two ends of RE and thus can determine the
current IE through it,