RECALL%20LECTURE%209 - PowerPoint PPT Presentation

About This Presentation
Title:

RECALL%20LECTURE%209

Description:

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 ... – PowerPoint PPT presentation

Number of Views:88
Avg rating:3.0/5.0
Slides: 13
Provided by: edum77
Category:
Tags: 20lecture | recall

less

Transcript and Presenter's Notes

Title: RECALL%20LECTURE%209


1
RECALL LECTURE 9
  • Introduction to BJT
  • 3 modes of operation
  • Cut-off
  • Active
  • Saturation
  • Active mode operation of NPN

2
NPN
PNP
IE IS e VBE / VT
IE IS e VEB / VT
IC ? IB
IC ? IE
IE IB( ? 1)
? ? / 1 - ?
? ? / ? 1
Based on KCL IE IC IB
3
DC Analysis of BJT Circuit
4
  • 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.

5
Common-Emitter Circuit
  • The figures below is showing a common-emitter
    circuit with an npn transistor and the dc
    equivalent circuit.
  • 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).

Unless given , always assume VBE 0.7V
6
Common-Emitter Circuit
  • The base current KVL at B-E loop
  • Implicitly assuming that VBB gt VBE (on), which
    means that IB gt 0. When VBB lt VBE (on), the
    transistor is cut off and IB 0.

7
Common-Emitter Circuit
  • In the C-E loop of the circuit, we can use
    and
  • Implicitly assuming that the transistor is biased
    in the forward-active mode.

8
Examples
  • BJT DC Analysis

9
Common-Emitter Circuit
Example Calculate the base, collector and
emitter currents and the C-E voltage for a
common-emitter circuit by considering VBB 4 V,
RB 220kO, RC 2 kO, VCC 10 V, VBE (on) 0.7
V and ß 200.
10
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
11
Common-Emitter Circuit - PNP
Example Find IB, IC, IE and RC such that VEC
½ VCC for a common-emitter circuit . Consider
VBB 1.5 V, RB 580 O, VCC 5 V, VEB (on)
0.6 V, ß 100.
12
EXAMPLE
Given ? 75 and VEC 6V. Find the values of the
labelled parameters, RC and IE,
Write a Comment
User Comments (0)
About PowerShow.com