BJT
Transistors
Transistor Analogy 1: Rules
Rules: (Which Always Apply!)
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First Transistor: Bell Labs 1947
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IB is the sense or control current
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IB can only flow in one direction
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IE = IC + IB
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Results in 3 operating modes:
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What Is It & Why (and
Where) Would I Use One?
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Transistor Analogy 1: Three
Operating Modes
Active Components (as opposed to
passive components)
1. (Cut) Off:
Current Source / Amplifier
2. Active:
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2
(Application: Amplifier)
IC = β IB
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Change Input / Output impedances
(“Transform “resistances” =>
TRANS-(res)ISTOR)
(Application: Switch)
IB = 0, so IC = 0 and IE = 0
β is the (current) gain
β is sometimes also called hfe
β is typically between 100 to 200
IE = IC +IB = IC +IC /β ≅ IC
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Transistor Analogy 1:
Operating Modes (continued)
Water Transistor Analogy
3. Saturated:
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IC ≠ β IB
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(Application: Switch)
The transistor has reached its limits in
regulating IC ; therefore, the relations ship
IC = β IB is NO LONGER VALID!
In this case, IB << IC is NO LONGER VALID!
IE = IC +IB (still holds!)
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1
Transistor Circuit Solving: DC
Analysis
BJT* Transistor Symbols
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*BJT = Bipolar
Junction Transistor
(as opposed to a
“Field Effect”
Transistor, i.e.,
“FET”)
“npn” mnemonic:
“never points in”
Two pn-diodes do
not make a
transistor!
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Transistor Circuit Solving
Approach: 5 Steps
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Steps 2, 3 & 4:
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Step 5:
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Replace all capacitors with open circuits
Short all (independent) AC voltage sources
Replace the transistor with its appropriate
DC equivalent circuit model
Solve (hm…)
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For Active and Cutoff Mode Only
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VBE = VB – VE ≅ 0.6 V in active (forward
biased) mode
DC & AC Analysis and then Superposition
Check your results; if wrong, choose a different
model and start at step 1 again.
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Pick a Transistor Operating Mode (Cutoff, Active,
Saturated)
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DC Analysis
DC Equivalent Transistor
Circuit Model
Note: It’s an iterative approach!
Step 1:
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Transistor Circuit Solving: DC,
AC Analysis and Superposition
DC Equivalent Transistor
Circuit Model
Superposition:
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Solve the transistor circuit with only DC
bias applied, i.e., find the quiescent
voltages and currents
Solve the transistor circuit with only AC
signals applied.
Finally combine the results:
VTotal = VDC + vac
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For Saturated Mode Only
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VSat = 0.1 to 0.3 V typically
VBE ≅ 0.6 V in forward biased mode
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2
Transistor Circuit Solving: AC
Analysis
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AC Analysis
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Again replace the transistor with its appropriate
AC equivalent circuit model.
Also replace all capacitors with short circuits
Short all (independent) DC voltage sources
Trick: for now short circuit rπ.
Only active AC model is provided since for cutoff
or saturation AC signal no longer propagates.
Solve (hm…)
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Use appropriate DC equivalent model; short all AC
sources; open all capacitors.
VB =
VE =
VC =
VOUT =
IE =
IC =
IB =
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Use AC equivalent model & short all DC sources and all
capacitors. For simplicity, assume rπ = 0.
rπ = β VTh/IC
= β / (40 IC)
Let vocos(ωt) = δvin = vin
Example:
β = 100; IC = 1mA;
then rπ ≅ 1 kΩ
For the rest ignore
(short) rπ, i.e., rπ ≅
0!
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Example 1: Common Collector
Emitter Follower Analysis:
Find all voltages
and currents
vo < 1.0 V
β = 100
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vb =
ve =
vc =
ie =
ic =
ib =
vgain = vout / vin =
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Example 1 Analysis: Step 4:
Superposition
Combine results from step 2 and 3. Again, vocos(ωt) = δvin
Step 1: Assume
circuit is in
active mode!
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Example 1 Analysis: Step 3:
AC Analysis
AC Equivalent Transistor Circuit
Model For ACTIVE Mode ONLY!
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Example 1 Analysis: Step 2:
DC Analysis
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VB =
VE =
VC =
VOUT =
IE =
IC =
IB =
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3
Example 1 Analysis: Step 5:
Check Model
Example 2: C. C. E. Follower
AC Output Impedance
Find Limits of DC Model: Adjust DC Bias Voltage
Source VDC ; Try VDC =1V and 20V and δvin = 1 VAC.
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Example 1 Analysis: Step 5:
Check Model: Limits
Saturated:
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Conclusion(s):
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Use appropriate AC equivalent model; assume rπ = 0
and short all DC sources and short all capacitors.
zout = vout/iout
iout = i1 – ie
i1 = vout/RE
ie = ib (β+1)
ib = -vout / RB
zout=RB/[(β+1)+ RB/RE)]
≅ RB/(β
β +1)
VB >
Transistor Model Used:
Purpose of Circuit:
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Example 1: C. C. E. Follower
AC Input Impedance
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Common Collector Emitter
Follower Conclusions
zin = vin/iin
=
=
=
Conclusion:
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VB <
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Solve use AC
small signal
model
Example 2: C. C. E. Follower
AC Output Impedance
Limits:
! CutOff:
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zout = vout/iout
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1.
AC Voltage Gain:
2.
zin:
3.
zout:
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Biasing the Common Collector
Emitter Follower 1
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Requires 2 DC Power Supplies
δvin must be “floating”
Biasing the Common Collector
Emitter Follower 3
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Requires 1 DC Power Supply
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δvin not required to be floating
Determine the Optimal Vout Quiescent
Point for Max. Symmetrical Output
Calculate all other Voltages
Find IB
Calculate R1 and R2
1.
2.
3.
4.
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Biasing the Common Collector
Emitter Follower 2
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Biasing the Common Collector
Emitter Follower 4
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Optimized Follower (Larger Dynamic Range!)
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Circuit 1
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Biasing Considerations
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Best Input / Output Range
Transistor Operating Mode Range
Quiescent Vout Power Consumption
Select the Voltage Divider Resistance Values
Based on (Quiescent) Power Consumption
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Follower Definition
Select the Voltages Based on an Optimal Vout
Quiescent Point Based On:
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Determine CIN and
COUT if:
20 Hz < fin < 20 kHz
Circuit 2
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Find zin and zout
Consider Zout, i.e., RLoad
Factor of 10 Rule for Voltage Dividers
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1.
No Voltage Gain
2.
Large Zin
3.
Low Zout
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Where and why would I use one?
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Common Emitter Amplifier
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Common Emitter Amplifier 4
What’s the difference between the follower circuit?
“How” does it amplify the input voltage? (Hint: Use small
signal model.)
1.
2.
3.
4.
5.
6.
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Example: Design a Common Emitter Amplifier
Circuit with the following characteristics:
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AC Voltage Gain: -20
zout = 20 kΩ (Hint: zout Common Emitter Amplifier
is RC.)
Optimal (Largest) Output Voltage Range
7.
8.
9.
10.
11.
12.
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Transistor Conclusions
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+15V
input 1
RC
2N3904
7.5k
out
2N3904
100
100
input 2
RE
7.5k
RE
R1
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-15V
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Find all voltages, currents, resistances and vin
range.
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VB :
IB:
IR1:
R1:
R2:
vin Range:
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Common Emitter Amplifier 2
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VC:
RC:
IC:
IE:
RE:
VE:
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Circuits so far only
amplify an AC voltage
signal.
Need a “differential”
amplifier to amplify DC
Common Mode Gain vs.
Differential Gain;
Common Mode
Rejection Ratio (CMRR)
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Common Emitter Amplifier 3
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