11/03/2014
DC Biasing of BJT’s
Biasing
Biasing refers to the DC voltages
applied to the transistor to turn it on
so that it can amplify the AC signal.
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Operating Point
The DC input establishes an operating or quiescent point called the Q point.
Biasing and the Three States of Operation
• Active or Linear Region Operation
Base – Emitter junction is forward biased
Base – Collector junction is reverse biased
• Cutoff Region Operation
Base – Emitter junction is reverse biased
Base – Collector junction is reverse biased
• Saturation Region Operation
Base – Emitter junction is forward biased
Base – Collector junction is forward biased
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BJT DC Analysis
1.
Draw the DC equivalent circuit
a.
b.
c.
2.
Write KVL for the loop which contains B-E junction
a.
b.
3.
Capacitors are open circuit (f = 0, XC = )
Inductors are short circuit (f = 0, XL = 0) or replaced by
their DC resistance (winding resistance) if given.
Kill the AC power sources (short AC voltage sources and
remove AC current sources)
Take VBE = VBE(ON) to ensure the transistor is on (or not in
the cut-off (OFF) state)
Determine the base current IBQ (or emitter current IEQ)
Write KVL for the loop which contains C-E terminals
a.
b.
c.
Assume the transistor is in the forward active (FA) state and
take ICQ= IBQ (or ICQ= IEQ)
Calculate VCEQ
If VCEQ  VCE(SAT) then the transistor is in the saturation
(SAT) state (i.e., ICQ  IBQ ), so take
VCEQ = VCE(SAT) and calculate ICQ (this is the saturation
current IC(SAT)).
DC Biasing Circuits
a. Fixed-Bias Circuit
b. Emitter-Stabilized Bias Circuit
c. Voltage Divider Bias Circuit
d. DC Bias with Voltage Feedback
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a. Fixed-Bias Circuit
Base-Emitter Loop
Using Kirchoff’s voltage law:
VCC – IBRB – VBE = 0
Solving for IB:
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Collector-Emitter Loop
Knowing:
Using Kirchoff’s voltage law:
Because:
Since VE = 0V, then:
And knowing:
And VE = 0V, then:
Example
IBQ , ICQ=?
IBQ = 47.08 A
ICQ= 2.35 mA
VCEQ = ?
VB , VC = ?
VCEQ = 6.83 V
VB = 0.7 V
VBC = ?
VC = 6.83 V
VBC = -6.13 V
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Transistor Saturation Level
When the transistor is operating in the Saturation Region
it is conducting at maximum current flow through the transistor.
Example
IBQ = 47.08 A
ICQ= 2.35 mA
VCEQ = 6.83 V
VB = 0.7 V
VC = 6.83 V
VBC = -6.13 V
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Load Line Analysis
VCE = VCC - RC IC
Load Line Analysis
The end points of the line are: IC(sat) and VCE(cutoff)
IC(sat):
VCE(cutoff):
Q-point is the particular operating point:
Value of RB
Sets the value of IBQ
Where IBQ and Load Line intersect
Sets the values of VCEQ and ICQ.
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Circuit values effect Q-point
Circuit values effect Q-point (continued)
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Circuit values effect Q-point (continued)
Example
VCC , RC , RB = ?
VCC = 20 V
RC = 2 k
RB = 772 k
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b. Emitter-Stabilized Bias Circuit
Adding a resistor to the emitter circuit stabilizes the bias circuit.
Base-Emitter Loop
Applying Kirchoffs voltage law:
Knowing:
Combining these two formulas:
Grouping terms and solving for IB:
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Collector-Emitter Loop
Applying Kirchoff’s voltage law:
Knowing that IE  IC and solving for VCE:
Finding VE:
Finding VC:
or
Finding VB:
or
Load Line Analysis
The load line end points can be calculated:
VCE(cutoff) :
IC(sat) :
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Example:
IBQ , ICQ=?
IBQ = 40.01 A
VCEQ = ?
VCEQ = 13.97 V
VC , VE = ?
VC = 15.98 V
VB , VBC = ?
VB = 2.71
ICsat =?
ICQ= 2.01 mA
VE = 2.01 V
VBC = -13.27 V
ICsat = 6.67 mA
c. Voltage Divider Bias
This is a very stable bias circuit.
The currents and voltages are almost independent of variations in .
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Exact Analysis
Example:
RB = 3.55 k
VBB = 2 V
IBQ , ICQ=?
IBQ = 6.05 A
VCEQ = ?
VCEQ = 12.22 V
ICQ= 0.85 mA
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Approximate Analysis
For Voltage Divider bias circuits in which:
IB << I1 and I2 and I1  I2 then the following approximations will do.
if
then
and
Applying Kirchoff’s voltage law:
IE  IC then
Example:
ICQ=?
ICQ= 0.867 mA
VCEQ = ?
VCEQ = 12.03 V
ICsat = ?
ICsat = 1.91 mA
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Load Line Analysis
The load line end points can be calculated:
VCEcutoff :
ICsat :
d. DC Bias with Voltage Feedback
Another way to improve the stability of a bias circuit is to add a feedback path from
collector to base. In this bias circuit the Q-point is only slightly dependent on the transistor
Beta .
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Base – Emitter Loop
VCC – ICRC – IBRB – VBE – IERE = 0
Applying Kirchoff’s voltage law:
Note:
IC = IC + IB
-- but usually IB << IC
Knowing IC = IB and IE  IC then:
-- so IC  IC
VCC – IB RC – IBRB – VBE – IBRE = 0
Simplifying and solving for IB:
Collector Emitter Loop
Applying Kirchoff’s voltage law:
IE + VCE + ICRC – VCC = 0
Since IC  IC and IC = IB:
IC(RC + RE) + VCE – VCC =0
Solving for VCE:
VCE = VCC – IC(RC + RE)
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Troubleshooting Techniques
A few rules of thumb:
VBE  0.7V for silicon transistors
VCE  25% to 75% of VCC
PNP Transistors
The analysis for PNP bias transistor circuits is the same as that for NPN transistor circuits.
The only difference is that the currents are flowing in the opposite direction.
Various Different Bias Circuits
Example: Determine the IEQ and VCEQ for the figure above?
IEQ = 4.16 mA
VCEQ = 11.68 V
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Bias Stabilization
Variation of BJT (Si) Parametes with Temperature
T (C)
ICO (nA)

VBE (V)
-65
0.2 x 10-3
20
0.85
25
0.1
50
0.65
100
20
80
0.48
175
3.3 x 103
120
0.3
T
ICO 

VBE 
ICO (reverse saturation current)
– doubles in value for every 10 C

– increases with temperature
VBE (base-emitter junction forward bias potential)
– decreases about 7.5mV per 1 C increase in temperature
Shift in dc bias point (Q-point) due to change in temperature
for the fixed-bias circuit
a) at 25C
b) at 100C
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Stability factors
Stability of the collector current IC depends on the stability of several parameters like
• ICO, , VBE, VCC, RB, RC etc.
Stability factors (Voltage-divider bias circuit)

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Stability factors (Voltage-divider bias circuit) (continued)
Stability factors (Simplification)
Variation of stability factor S (ICO) with the resistor ration RB/RE for the emitter-bias configuration

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Stability factors (Voltage-feedback bias circuit)

Example
T (C)
ICO (nA)

VBE (V)
-65
0.2 x 10-3
20
0.85
0.65
25
0.1
50
100
20
80
0.48
175
3.3 x 103
120
0.3
Find and compare the collector current change IC for a fixed-bias circuit and voltagedivider bias circuit when the temperature rises from 25C to 100C where RB = 240k,
IC1= 2mA and RTh / RE = 2, RE = 4.7k. Use the table above.
ICO = 19.9 nA
 = 30
VBE = -0.17 V
For fixed-bias circuit
= 51
= 0.04 mA
= -0.21 m-1
IC = 1.236 mA
For voltage-divider bias circuit
=3
= 1.5 A
= -0.2 m-1
IC = 80 A
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Stability of Transistor Circuits with Active Components
• VBE compensation
– by using a reverse-biased diode at the emitter
• ICO compensation
– by replacing R2 with a reverse-biased diode
• IC compensation (without RE)
– by using a current mirror
VBE compensation
– by using a reverse-biased diode at the emitter
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ICO compensation (esp. for Ge transistors)
– by replacing R2 with a reverse-biased diode
IC compensation (without RE)
– by using a current mirror
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Practical Applications
• Relay Driver
• Transistor Switch
• Logic Gates
• Current Mirror
• Voltage Level Indicator
Relay Driver
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Transistor Switch
Transistor Switching Networks
Transistors with only the DC source applied can be used as electronic switches.
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Switching Circuit Calculations
The transistor will switch between saturation and cutoff regions.
You must ensure that:
Also consider the resistance across the transistor at saturation:
The resistance across the transistor in cutoff is theoretically infinite, but it is calculated:
Switching Circuit Calculations
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Switching Time
Another consideration is the time it takes the transistor to switch.
Logic Gates
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Current Mirror
Voltage Level Indicator
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