RECALL LECTURE 9 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 CircuitDC Analysis of BJT Circuit

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

Implicitly assuming that VBB > VBE (on), which means that IB > 0. When VBB < VBE (on), the transistor is cut off and IB = 0.

Common-Emitter CircuitCommon-Emitter Circuit The base current: KVL at B-E loop

Implicitly assuming that the transistor is biased in the forward-active mode.

Common-Emitter CircuitCommon-Emitter Circuit

In the C-E loop of the circuit, we can use: and

ExamplesBJT DC Analysis

Example Calculate 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

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

Example Find IB, IC, IE and RC such that VEC = ½ VCC for a common-emitter circuit .Consider: VBB = 1.5 V, RB = 580 Ω, VCC = 5 V,

VEB (on) = 0.6 V, β = 100.

Common-Emitter Circuit - PNPCommon-Emitter Circuit - PNP

IE

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