1. The Currents of a BJT The collector current. The base Current.
The Emitter Current.

2. The Currents of a BJT Collector Current
It can be expressed as IC=ISe(VBE/VT)
Is=saturation current or current scale factor.

3. The Currents of a BJT The Base Current It is given by IB=IC/β
Where β is a constant for particular transistor and is known as common emitter current gain, since
IC/IB=β.

4. The Currents of a BJT The Emitter Current
Since the total current which enters a transistor must leave it, hence we may write
IE=IB+IC
But IB=IC/ β
Therefore,
IE=IC+IC/ β

5. The Currents of a BJT Or IE= IC(β+1)/ β IE= Is e(VBE/VT) (β+1)/ β
Alternatively, we may write
IC=αIE α= β/ (β+1)
Or IE=(Is/α) e(VBE/VT)

6. The Currents of a BJT Also we may write β=α/(α-1)
α is known as common base current gain.

7. Example 4.3: Show the transistor is in saturation.

8. For input loop, applying KVL
6-VBE-IE(3.3k)=0
Rearranging IE=(6-0.7)/3.3k=1.61mA

9. Therefore
IB=IE/101=1.61m/101=15.94A
Also VE=IERE=1.61m(3.3k)=5.3V
& IC=15.94 (100)=1.59mA.
Therefore KVL Equation for the out put loop is
m(4.7k)-VC=0
VC= =2.50V
VC<VB,
hence the transistor is in saturation.

10. DC BIASING-BJT
Fixed bias circuit.
Emitter stabilized biased circuit.
Voltage divider bias.

11. DC BIASING-BJT Fixed Bias Circuit

12. DC equivalent of the circuit.

13. Forward bias of Base Emitter
KVl Equation for the loop is
VCC-IBRB-VBE=0 or IB=(VCC-VBE)/RB

14. Collector Emitter Loop
Applying KVL
VCE+ICRC-VCC=0 or VCE=VCC-ICRC

15. Also we may write
VCE=VC-VE
VE=0 in this case
 VCE=VC
Also VBE=VB-VE
And VE=0, thus
VBE=VB

16. Example
Find IBQ and ICQ, VCEQ, VB, Vc, and VBC.
Consider β=50.

18. Also we know that

20. LOAD LINE ANALYSIS
The output equation of fixed bias circuit is given as
VCE = VCC – ICRC
Which is an equation of a straight line. The corresponding x and y axes intercepts can be calculated by inserting appropriate values equal to zero. Therefore by putting

21. LOAD LINE ANALYSIS IC = 0 We have VCE = VCC – (0)RC VCE = VCC
Y intercept can be found by putting
VCE = 0 volts so that
0 = VCC – ICRC
Ic = VCC/RC
Which defines the slop of the load line.

22. LOAD LINE ANALYSIS

23. LOAD LINE ANALYSIS

24. LOAD LINE ANALYSIS

25. LOAD LINE ANALYSIS

26. Example: Given a curve find
Vcc,Rc,and RB for fixed bias circuit.

27. Solution From the graph we have, for Ic=0 VCE=VCC=20V Also at VCE=0
IC=VCC/RC
RC=VCC/IC=20/10m=2k
Now IB=(VCC-VBE)/RB
Or RB=(Vcc-VBE)/IB
=(20-0.7)/25=772k .

28. DC BIASING-BJT Emitter-Stabilized Bias Circuit

29. Base Emitter Loop

30. Base Emitter Loop KVl Equation for the loop is. VCC-IBRB-VBE-IERE=0
Base Emitter Loop KVl Equation for the loop is VCC-IBRB-VBE-IERE=0 But IE=(β+1)IB Therefore, VCC-IBRB-VBE-(β+1)IBRE=0 -IB(RB+(β+1)RE)+VCC-VBE=0 IB=(VCC-VBE)/(RB+(β+1)RE)

31. Collector-Emitter Loop

32. Collector-Emitter Loop
IERE+VCE+ICRC-VCC=0
But IE=Ic
VCE-Vcc+Ic(Rc+RE)=0
Vce=Vcc-Ic(Rc+RE)
Now VE is the voltage between the Emitter terminal and the ground, given by
VE=IERE
Also Collector to ground voltage will be
VCE=VC-VE
VC=VCE+VE
Also Vc=VCC-IcRC

33. Collector-Emitter Loop
Voltage at the base w.r.t ground will be
VB=VCC-IBRB
VB=VBE+VE
Stabilization
Due to addition of the emitter resistance the values of DC bias currents and voltages remain constant with any changes in the Temperature and transistor beta.