Recall Last Lecture

Introduction to BJT 3 modes of operation

Cut-off Active Saturation

Active mode operation of NPN

IE = IS [ e VBE / VT ]

Based on KCL: IE = IC + IB

IC = IB

IC = IE

= [ / + 1 ]

IE = IB( + 1)

IE = IS [ e VEB / VT]

NPN PNP

= [ / 1 - ]

DC Analysis of BJT Circuit DC Analysis of BJT Circuit and Load Linesand Load Lines

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.

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).

Common-Emitter CircuitCommon-Emitter Circuit

The figures below is showing a common-emitter circuit with an npn transistor and the DC equivalent circuit.

Unless given , always assume VBE = 0.7V

KVL at B-E loop

You can calculate IC using IC = βIB

Calculate IE using IE = IB + IC

Input Load LineInput Load LineIIBB versus V versus VBEBE

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

y = mx + c

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

= 200

KVL at C-E loop of the circuit

You have calculated the value of IC from the value of IB

Output Load LineOutput Load LineIICC versus V versus VCECE

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:

y = mx + c

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

= 200

Examples

BJT DC Analysis

Example 1Calculate 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.

Common-Emitter CircuitCommon-Emitter Circuit

= 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

Example 2

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, β = 100.

Common-Emitter Circuit - PNPCommon-Emitter Circuit - PNP

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.

Output Load LineOutput Load LineUse KVL at B-E loop to find the value of IB

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

IB = 75.1 A

VCE = 12 – 5.63 (1.01)VCE = 6.31 V

VCE = 12 – IC (1.01)IC = - VCE + 12

1.01