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

IE

EXAMPLE - PNP

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

DC analysis of BJT 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 = 0IE = 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

VBE = 0.7VVB – VE = 0.7VVE = 4 – 0.7 = 3.3 V since VB = 4V

= 0.99

IC = IE

IB = IE - IC

VE – 0 = IE

3.3 k

10 - VC = IC

4.7 k

VC = 5.347 VHence, VCE = VC – VE

= 5.347 – 3.3= 2.047 V

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

The input load line is obtained from Kirchhoff’s voltage law equation around the B-E loop, written as follows:

Input Load Line Input Load Line – I– IBB versus V versus VBEBE

Derived using B-E loopDerived using B-E loop

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

VBE – VBB + IBRB = 0

The input load line is essentially the same as the load line characteristics for diode circuits.

For example;For example;

IBQ = 15 μA

VBE – 4 + IB(220k) = 0

IB = -VBE + 4

220k 220k

y = mx + c

Output Load Line Output Load Line – I– ICC versus V versus VCECE

Derived using C-E loopDerived 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;For example;

To find the intersection points setting IC = 0,VCE = VCC = 10 V

setting VCE = 0IC = VCC / RC = 5 mA

IC = -VCE + 10

2k 2k

y = mx + c

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 LineLoad LineUse KVL at B-E loop

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

IB = 75.1 A

VCE = 12 – 5.63 (1.01)VCE = 6.31 V

VCE = 12 – IC (1.01)IC = - VCE + 12

1.01