Electronic (1 ) :
Lecture (8):
DC Biasing—BJTs
Transistor Biasing:
Transistor Biasing is the process of setting a transistor DC operating
voltage or current conditions to the correct levels so that AC input signal
can be amplified correctly by the transistor.
DC Biasing Types:
Remember:
Important basic relationships for a transistor
OPERATING POINT:
Fixed-Bias Configuration:
For the dc analysis the network can be isolated from the indicated ac
levels by replacing the capacitors with an open-circuit equivalent.
The dc supply Vcc can be separated into two supplies.
Base-Emitter Loop :
Collector-Emitter Loop
The voltages VB VC VE with respect to the ground
Determining Icsat for the fixed-bias configuration
Saturation conditions are normally avoided because the base – collector
junction is no longer reverse – biased and the output amplified signal
will be distorted.
Load-Line Analysis for the fixed – bias configuration
Example (1) :
Determine the following for the fixed-bias configuration.
Solution:
Example (2) : Given the load line of and the defined Q-point, determine
the required values of VCC, RC, and RB for a fixed-bias configuration.
Solution:
Determine values of VCC , RC , and RB
Example (3) : Given the information appearing in Fig, determine:
Solution:
a. IC.
Ic = βIB = 80 (40 µA) = 3.2ma
b. RC.
c. RB.
d. VCE.
VCE = Vc =6V
Example (4) : Determine the saturation level for the network of
Fig.
Solution:
Emitter – Bias Configuration :
The dc bias network of contains an emitter resistor to improve the
stability level over that of the fixed-bias configuration.
The analysis of Emitter – Bias Configuration:
Base–emitter loop
Collector–Emitter Loop
The voltages VB , VC , and VE with respect to the ground :
Example (5) : For the emitter-bias network , determine:
Solution:
a)
b)
c)
d)
e)
f)
g)
Determining Icsat for the Emitter – Bias configuration
Example (6) : Determine the saturation current for the network
of Example 5.
Solution:
Note: which is about three times the level of ICQ for Example 5.
Load – line Analysis for the Emitter – bias configuration
Example (7) :
a. Draw the load line for the network of Fig .(a) on the characteristics for
the transistor appearing in Fig. (b).
b. For a Q-point at the intersection of the load line with a base current of
15 mA, find the values of ICQ and VCEQ.
c. Determine the dc beta at the Q-point.
Solution:
Voltage-Divider Bias Configuration.
If the circuit parameters are properly chosen, the resulting levels of IcEQ
can be almost totally independent of beta.
The level of IBQ will change with the change in β,but the operating point
on the characteristics defined by ICQ and VCEQ can remain fixed.
Exact Analysis:
The Thévenin equivalent network for the network to the left of the base
terminal can then be found in the following manner:
RTh The voltage source is replaced by a short-circuit equivalent.
ETh The voltage source VCC is returned to the network and the open-
circuit Thévenin voltage
The Thevenin network
IQB can be determine by first applying kirchhoff’s voltage law in the
clockwise direction for the loop indicated:
Once IB is known, the remaining quantities of the network can be found
in the same manner as developed for the emitter-bias configuration. That
is,
The remaining equations for VE, VC, and VB are also the same as
obtained for the emitter-bias configuration.
Example (8) : Determine the dc bias voltage VCE and the current IC for
the voltage-divider configuration of Fig.
Solution:
Example (9) :
Solution:
Example (10) :
Determine the levels of ICQ and VCEQ for the voltage-divider
configuration of Fig. using the exact technique.
Solution:
Exercises from (1-10):