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

2. I E = I S [ e VBE / VT ] Based on KCL: I E = I C + I B I C =  I B I C =  I E  = [  /  + 1 ] I E = I B (  + 1) I E = I S [ e VEB / VT ] NPNPNP  = [  / 1 -  ]

3. DC Analysis of BJT Circuit and Load Lines

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

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

6.  KVL at B-E loop You can calculate I C using I C = βI B Calculate I E using I E = I B + I C

7. Input Load Line I B versus V BE

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

9.  The input load line is essentially the same as the load line characteristics for diode circuits. For example;  I BQ = 15 μA V BE – 4 + I B (220k) = 0 I B = -V BE + 4 220k y = mx + c  = 200

10.  KVL at C-E loop of the circuit You have calculated the value of I C from the value of I B

11. Output Load Line I C versus V CE

12. Output Load Line – I C versus V CE 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;  To find the intersection points setting I C = 0, V CE = V CC = 10 V  setting V CE = 0 I C = V CC / R C = 5 mA I C = -V CE + 10 2k y = mx + c Q-point is the intersection of the load line with the i C vs v CE curve, corresponding to the appropriate base current  = 200

14. Examples BJT DC Analysis

15. Example 1 Calculate the base, collector and emitter currents and the C-E voltage for a common-emitter circuit by considering V BB = 4 V, R B = 220kΩ, R C = 2 kΩ, V CC = 10 V, V BE ( on ) = 0.7 V and β = 200. Common-Emitter Circuit

16.  = 0.99 KVL at BE loop:0.7 + I E R E – 4 = 0 I E = 3.3 / 3.3 = 1 mA Hence, I C =  I E = 0.99 mA I B = I E – I C = 0.01 mA KVL at CE loop: I C R C + V CE + I E R E – 10 = 0 V CE = 10 – 3.3 – 4.653 = 2.047 V Example 2

17. Example 3 Find I B, I C, I E and R C such that V EC = 2.5 V for a common-emitter circuit. Consider: V BB = 1.5 V, R B = 580 kΩ, V CC = 5 V, V EB (on) = 0.6 V, β = 100. Common-Emitter Circuit - PNP

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 V BE (on) = 0.7 V and β = 75.

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

20. Use KVL at C-E loop – to obtain the linear equation I C R C + V CE + I E R E – 12 = 0 I C R C + V CE + (I C /  )R E – 12 = 0

21. I B = 75.1  A V CE = 12 – 5.63 (1.01) V CE = 6.31 V V CE = 12 – I C (1.01) I C = - V CE + 12 1.01