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Transistors. Review : What is a Transistor ?. Collector. NPN Transistor Rules: V C > V E V B = V E + 0.6 Volts I C = I B  . I C : Collector Current. Base. I B : Base Current. I E : Emitter Current. I E = I C + I B  I E = (1 +  ) I B.

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Transistors

Transistors

Tim Sumner, Imperial College, Rm: 1009, x47552


Review what is a transistor
Review : What is a Transistor ?

Collector

NPN Transistor Rules:

VC > VE

VB = VE + 0.6 Volts

I C = I B 

IC : Collector

Current

Base

IB : Base Current

IE : Emitter

Current

I E = I C + I B 

I E = (1 + )I B

Emitter

FOR MORE INFO...

Relevant Books: Art of Electronics + Lab: Harowitz and Hill

Electronic Circuits: E.C. Lowenberg, SCHAUM SERIES

Electric Circuits: J.A. EDMINISTER SCHAUM SERIES

Tim Sumner, Imperial College, Rm: 1009, x47552


Some transistor applications i
Some Transistor Applications I

+ V

  • Follower:

  • Current Source

+ V

Load

IC

Vin

IC

Vin= 5.6 V

Vout= Vin+ 0.6 Volts

Vout= 5.0 V

IE

R = 5 K

IE = 1 mA

- V

Tim Sumner, Imperial College, Rm: 1009, x47552


Some transistor applications ii
Some Transistor Applications II

  • Push-Pull :

  • Common Emitter Amplifier :

+ V

+ V

RC >> Vout >>

Qup

Vout

Vin = VB

Vin = VE

RE <<  IE = IC >>

Qdown

- V

Tim Sumner, Imperial College, Rm: 1009, x47552


The emitter follower i
The Emitter Follower I

Given : Vcc = 15 Volts

Ic = 0.5 mA;  = 100

f(100 Hz) = 3 db

What about C1 ?

Vcc

Source

R1

Ic

Load

C1

1F

10 K

~

4.7 K

R2

R3

Rbase= VB/IB = 1.64 M

(2) RTH(divider) = 1/10 RB

RB = 164 K 

R1 = R2= 328 K

(1) R3=7.5 V/0.5 mA = 15K

Tim Sumner, Imperial College, Rm: 1009, x47552


The emitter follower ii
The Emitter Follower II

(3) Think of an RC filter (high-pass) .

15K is parallel to 4.7 K and the whole think times 

358 K ; parallel with 328 K 

112 K; So f=(1/2RC)  0.014 F

Vcc

Source

328 K

Ic

Load

C1

1F

10 K

~

4.7 K

328K

15 K

 = 100

f(100 Hz) = 3 db

Tim Sumner, Imperial College, Rm: 1009, x47552


Emitter follower a bit diff
Emitter Follower (a bit diff.)

Rin=Rload(1+)

Rout=Rsource/(1+)

Tim Sumner, Imperial College, Rm: 1009, x47552


Common emitter amplifier
Common Emitter Amplifier

Gain = R5/R2

Rout=R5

Tim Sumner, Imperial College, Rm: 1009, x47552


Common emitter amplifier ii
Common Emitter Amplifier II

Tim Sumner, Imperial College, Rm: 1009, x47552


The ebers moll model for bjt
The Ebers-Moll Model for BJT

Tim Sumner, Imperial College, Rm: 1009, x47552


Temperature effects
Temperature Effects

Tim Sumner, Imperial College, Rm: 1009, x47552


Temperature effects and current mirrors
Temperature Effects and Current Mirrors

Tim Sumner, Imperial College, Rm: 1009, x47552


Exercises
Exercises

  • Make a Current Source using a transistor

  • Make common emitter follower

  • Make a Common Emitter Amplifier

Tim Sumner, Imperial College, Rm: 1009, x47552


Making gates using transistors i
Making Gates using Transistors I

+5 Volts

InPut A

A .OR. B

InPut B

Tim Sumner, Imperial College, Rm: 1009, x47552


Making gates using transistors ii
Making Gates using Transistors II

InPut A

A .OR. B

InPut B

Here is what we have made

Tim Sumner, Imperial College, Rm: 1009, x47552


Exercise i
Exercise I

Design and construct an AND gate

using transistors and resistors :

InPut A

A .AND. B

InPut B

The AND gate TRUTH table

Tim Sumner, Imperial College, Rm: 1009, x47552


Exercise i continued
Exercise I continued

Verify the truth table ! You can set an input to ‘1’ by connecting it to +5 V by a resistor of 1K and you set it to ‘0’ by grounding it.

5 V

‘0’

InPut B

‘0’

A .AND. B

InPut A

‘1’

5 V

Tim Sumner, Imperial College, Rm: 1009, x47552


Answer almost
Answer (almost):

TTL

CMOS

Not quite because this is a NAND gate but we do know

by now how to invert using a transistor

Tim Sumner, Imperial College, Rm: 1009, x47552


Important conclusion
Important conclusion

You may understand by now that the inputs have no ability to set signals high or low. In contrast to this the outputs can drive signals to

High or Low

Tim Sumner, Imperial College, Rm: 1009, x47552


Remember
Remember:

Inputs draw no current. 1K resistors to +5 V can set inputs to high or they can be set low by connecting to ground.

Outputs can draw current and can force lines to low or high.

Tim Sumner, Imperial College, Rm: 1009, x47552


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