Chi tiết về transistor BJT

Chi tiết về transistor BJT. dành cho sinh viên kỹ thuật điện-điện tử.

Intro to Systems

Transfer function

Transfer functions

Poles, Zeros factorization

Response function of poles

Response function

Stable operation ==> No poles on right half-plane

Block Diagrams

Output is integral of inputIntegrator

Laplace Transforms

Inverse Laplace Transforms

Application to Linear Systems

Frequency Domain Analysis

• Step 1: Obtain Transfer function of circuit H(s)

• Step 2: Apply Laplace Transform to vin(t) to obtain Vin(s)

• Step 3: Multiply Vin(s) by H(s) to obtain Vout(s)

• Step 4: Apply Inverse Laplace Transfom to Vout(s) to obtain vout(t)

BJT – Bipolar Junction Transistors

• p+ > p, n+ > n  emitter, collector NOT interchangeable!!!

• base width small compared to minority

carrier diffusion length

 NOT just back to back diodes:

Transistor Band Structure:

Applying VBC:

Forward Biasing Base-Emitter Diode:

Currents in Forward Active Mode:

IE = IB + IC

VEB + VBC + VCE = 0

Currents in Forward Active Mode:

pnptransistor

Configurations:

Transistor Uses:

Switch: cut-off, saturation Amplifier: forward-active

VCE (V) 0

5 10 15

Transistor Ratings:

• Power Dissipation

– Power generated from transistor currents

• AC Current Gain

– Gain calculated under AC signal conditions

• Maximum Emitter-Base Voltage

– Maximum reverse voltage before breakdown

• Maximum Collector Current

– Maximum allowable continuous current

• Maximum Collector-Emitter Voltage

– Maximum current before breakdown

• Maximum Collector-Base Voltage

– Maximum reverse bias before breakdown

Common Packagings:

Looking from below:TO-92 examples when seen as to the left:

forward voltage drop

0.7 Vreverse current

~ nAreverse breakdown voltage

I = I0 e

e kT

Diode - Dynamic Impedance:

V >> nTk/e Voltage input signal change  impedance:

Room temperature  kT/e
25mV

V n

V n = I enkT

dV

dI = Zd = kT

eI = 25mV

I

What region?

Given: npn with
= 100

1.) IB = 50 μA, IC = 3 mA?

 saturation: IC <
IB2.) IB = 50 μA, VCE = 5V?

 active: IB > 0 and VCE > 0.2 V3.) VBE = -2V, VCE = -1V?

cut off: VBE < 0 and VBC = VBE –VCE = -1V < 0

i.e both reverse biased…

BJT circuit analysis:

guess operating mode: active, saturated, cut off?

i.e.: is BE forward or reverse biased?

reverse biased  cut off

forward biased  assume active: vBE = 0.7V

use: iE = iC + iB, iC =
iB; compute: vCE

if vCE < 0.2V  saturation; redo calculation with vCE = 0.2V

vBE = 0.8V

BJT amplifier biasing:

1.) choose a collector current IC:

typical: 1 – 10mA (small signal: 0.1 mA)2.) determine the supply voltage vs (5V, 9V, 12V, …)

amplifier  active region!!!

3.) choose an operating point:

typical: 10% of vs  RE = 0.1vs/IC4.) base voltage: VB = ICRE + 0.7V (Si)

 RC = (vs – VC)/IC

Review of BJT Equations:

iE = iC + iB VCE = VCB + VBEKirchhoff:

iC = ISe

VBE

VT

Cutoff and Saturation

Cutoff

Transistor as amplifier

Vary Input voltage > Vary IbaseCauses VCE to move along load line > Amplification

Transistor as a switch