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

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

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

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 – 4.653 = 2.047 V

BJT Circuits at DC  = 0.99 VC = 5.347 V Hence, VCE = VC – VE = 5.347 – 3.3 = 2.047 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

LOAD LINE AND VOLTAGE TRANSFER CHARACTERISTIC

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

LOAD LINE

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.

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 + 4 220k y = mx + c IBQ = 15 μA

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:

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

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.

Load Line Use KVL at B-E loop

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

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