1. Recall Lecture 10 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

2. EXAMPLE - PNP IE
Given  = 75 and VEC = 6V. Find the values of the labelled parameters, RC and IE,

3. DC analysis of BJT (Va – Vb) / R = I
When node voltages is known, branch current equations can be used. R a b I
(Va – Vb) / R = I

4. BJT Circuits at DC  = 0.99 KVL at BE loop: 0.7 + 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

5. BJT Circuits at DC  = 0.99 VC = 5.347 V Hence, VCE = VC – VE
= – 3.3
= V
IC = IE
10 - VC = IC
4.7 k
IB = IE - IC
VBE = 0.7V
VB – VE = 0.7V
VE = 4 – 0.7 = 3.3 V since VB = 4V
VE – 0 = IE
3.3 k

6. LOAD LINE AND VOLTAGE TRANSFER CHARACTERISTIC

7. Load Line and Voltage Transfer Characteristic can be used to visualize the characteristic of the transistor circuits.

8. LOAD LINE

9. 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
Both the load line and the quiescent base current change as either or both VBB and RB change.

10. For example;
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

11. 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:

12. For example; IC = -VCE + 10 2k
y = mx + c
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

13. Example: Calculation 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.

14. Load Line
Use KVL at B-E loop

15. Use KVL at C-E loop – to plot the load line

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