1. Recall Lecture 10 Introduction to BJT 3 modes of operation
Cut-off
Active
Saturation
Active mode operation of NPN

2. IC =  IB IC =  IE IE = IB(  + 1)  = [ / 1 -  ]  = [  /  + 1 ]
NPN
PNP
IE = IS [ e VBE / VT ]
IE = IS [ e VEB / VT]
IC =  IB
IC =  IE
IE = IB(  + 1)
 = [ / 1 -  ]
 = [  /  + 1 ]
Based on KCL: IE = IC + IB

3. DC Analysis of BJT Circuit and Load Lines

4. DC analysis of BJT BE Loop (EB Loop) – VBE for npn and VEB for pnp
CE Loop (EC Loop) - VCE for npn and VEC for pnp
When node voltages are known, branch current equations can be used.

5. Common-Emitter Circuit
The figures below is showing a common-emitter circuit with an npn transistor and the DC equivalent circuit.
Assume that the B-E junction is forward biased, so the voltage drop across that junction is the cut-in or turn-on voltage VBE (on).
Unless given , always assume VBE = 0.7V

6. KVL at B-E loop You can calculate IC using IC = βIB Calculate IE using
IE = IB + IC

7. Input Load Line IB versus VBE

8. Input Load Line – IB versus VBE
Derived using B-E loop
The input load line is obtained from Kirchhoff’s voltage law equation around the B-E loop, written as follows:
VBE – VBB + IBRB = 0
y = mx + c
Both the load line and the quiescent base current change as either or both VBB and RB change.

9. For example;
 = 200
The input load line is essentially the same as the load line characteristics for diode circuits.
VBE – 4 + IB(220k) = 0
IB = -VBE
220k
y = mx + c
IBQ = 15 μA

10. KVL at C-E loop of the circuit
You have calculated the value of IC from the value of IB

11. Output Load Line IC versus VCE

12. Output Load Line – IC versus VCE
Derived using C-E loop
For the C-E portion of the circuit, the load line is found by writing Kirchhoff’s voltage law around the C-E loop. We obtain:
y = mx + c

13. For example; IC = -VCE + 10 2k
y = mx + c
 = 200
To find the intersection points setting IC = 0,
VCE = VCC = 10 V
setting VCE = 0
IC = VCC / RC = 5 mA
Q-point is the intersection of the load line with the iC vs vCE curve, corresponding to the appropriate base current

14. Examples
BJT DC Analysis

15. Common-Emitter Circuit
Example 1
Calculate the base, collector and emitter currents and the C-E voltage for a common-emitter circuit by considering VBB = 4 V, RB = 220kΩ, RC = 2 kΩ, VCC = 10 V, VBE (on) = 0.7 V and β = 200.

16. Example 2
 = 0.99
KVL at BE loop: IERE – 4 = 0
IE = 3.3 / 3.3 = 1 mA
Hence, IC =  IE = 0.99 mA
IB = IE – IC = 0.01 mA
KVL at CE loop: ICRC + VCE + IERE – 10 = 0
VCE = 10 – 3.3 – = V

17. Common-Emitter Circuit - PNP
Example 3
Find IB, IC, IE and RC such that VEC = 2.5 V for a common-emitter circuit .
Consider:
VBB = 1.5 V, RB = 580 kΩ, VCC = 5 V, VEB (on) = 0.6 V,  = 100.

18. Example 4: DC Analysis and Load Line
Calculate the characteristics of a circuit containing an emitter resistor and plot the output load line.
For the circuit, let VBE (on) = 0.7 V and β = 75.

19. Output Load Line
Use KVL at B-E loop to find the value of IB

20. Use KVL at C-E loop – to obtain the linear equation
IC RC + VCE + IE RE – 12 = 0
IC RC + VCE + (IC / )RE – 12 = 0

21. VCE = 12 – IC (1.01) IC = - VCE + 12 1.01 IB = 75.1 A
VCE = 6.31 V

22. Example: DC Analysis – Common Base (npn)

23.  = 100
15V
5k
10 k 5V
BE Loop
0.7 + IERE – 5 = 0
IE = 0.43 mA
CE Loop
VCE + ICRC + IERE – 5 – 15 = 0
VCE = 20 – 4.3 – 2.13
VCE = V