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Transistors. Transfer Resistor Chapter 9. Bipolar Transistors. Collector. Base. Emitter. Two PN junctions joined together Two types available – NPN and PNP The regions (from top to bottom) are called the collector (C), the base (B), and the emitter (E). Operation.

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Transistors l.jpg

Transistors

Transfer Resistor

Chapter 9


Bipolar transistors l.jpg
Bipolar Transistors

Collector

Base

Emitter

Two PN junctions joined together

Two types available – NPN and PNP

The regions (from top to bottom) are called the collector (C), the base (B), and the emitter (E)


Operation l.jpg
Operation

  • Begin by reverse biasing the CB junction

    • Here we are showing an NPN transistor as an example

  • Now we apply a small forward bias on the emitter-base junction

    • Electrons are pushed into the base, which then quickly flow to the collector

    • The result is a large emitter-collector electron current (conventional current is C-E) which is maintained by a small E-B voltage

  • Some of the electrons pushed into the base by the forward bias E-B voltage end up depleting holes in that junction

    • This would eventually destroy the junction if we didn’t replenish the holes

    • The electrons that might do this are drawn off as a base current




Origin of the names l.jpg
Origin of the names

  • the Emitter 'emits' the electrons which pass through the device

  • the Collector 'collects' them again once they've passed through the Base

  • ...and the Base?...



Base thickness l.jpg
Base Thickness

  • The thickness of the unmodified Base region has to be just right.

    • Too thin, and the Base would essentially vanish. The Emitter and Collector would then form a continuous piece of semiconductor, so current would flow between them whatever the base potential.

    • Too thick, and electrons entering the Base from the Emitter wouldn't notice the Collector as it would be too far away. So then, the current would all be between the Emitter and the Base, and there'd be no Emitter-Collector current.


Amplification properties l.jpg
Amplification Properties

  • The C-B voltage junction operates near breakdown.

    • This ensures that a small E-B voltage causes avalanche

    • Large current through the device




Common collector npn l.jpg
Common Collector NPN

How does IC vary with VCE for various IB?

Note that both dc sources are variable

Set VBB to establish a certain IB


Collector characteristic curve l.jpg
Collector Characteristic Curve

  • If VCC = 0, then IC = 0 and VCE = 0

  • As VCC↑ both VCE and IC↑

  • When VCE 0.7 V, base-collector becomes reverse-biased and IC reaches full value (IC = bIB)

  • IC ~ constant as VCE↑. There is a slight increase of IC due to the widening of the depletion zone (BC) giving fewer holes for recombinations with e¯ in base.

  • Since IC = bIB, different base currents produce different IC plateaus.




Load line l.jpg
Load Line

Slope of the load line is 1/RL

For a constant load, stepping IB gives different currents (IC) predicted by where the load line crosses the characteristic curve. IC = bIBworks so long as the load line intersects on the plateau region of the curve.


Saturation and cut off l.jpg

Cut-off

Saturation and Cut-off

Note that the load line intersects the 75 mA curve below the plateau region. This is saturation and IC = bIB doesn’t work in this region.


Example l.jpg
Example

  • We adjust the base current to 200mA andnote that this transistor has ab= 100

    • Then IC =bIB = 100(200 X 10-6A) = 20 mA

  • Notice that we can use Kirchhoff’s voltage law around the right side of the circuit

    • VCE = VCC – ICRC = 10 V – (20 mA)(220W)

      = 10 V – 4.4 V = 5.6 V


Example19 l.jpg
Example

  • Now adjust IB to 300mA

    • Now we get IC = 30 mA

    • And VCE = 10 V – (30 mA)(220W) = 3.4 V

  • Finally, adjust IB = 400mA

    • IB = 40 mA and VCE = 1.2 V



Gain as a function of i c l.jpg
Gain as a function of IC

As temperature increases, the gain increases

for all current values.


Operating limits l.jpg
Operating Limits

  • There will be a limit on the dissipated power

    • PD(max) = VCEIC

    • VCE and IC were the parameters plotted on the characteristic curve.

      • If there is a voltage limit (VCE(max)), then you can compute the IC that results

      • If there is a current limit (IC(max)), then you can compute the VCE that results


Example23 l.jpg
Example

  • Assume PD(max) = 0.5 W

    VCE(max) = 20 V

    IC(max) = 50 mA


Operating range l.jpg
Operating Range

Operating

Range


Voltage amplifiers l.jpg
Voltage Amplifiers

Common Base PNP

Now we have added an ac source

The biasing of the junctions are:

BE is forward biased by VBB - thus a small resistance

BC is reverse biased by VCC – and a large resistance

Since IB is small, IC IE


Equivalent ac circuit l.jpg
Equivalent ac Circuit

rE = internal ac emitter

resistance

IE = Vin/rE (Ohm’s Law)

Vout = ICRC IERC

Recall the name – transfer resistor


Current gains l.jpg
Current Gains

  • Common Base

    • a = IC/IE < 1

  • Common Emitter

    • b = IC/IB


Example28 l.jpg
Example

  • If b = 50, then a = 50/51 = 0.98

    • Recall a < 1

  • Rearranging,

    b = a + ab

    b(1-a) = a

    b = a/(1-a)



The operating points l.jpg
The operating points

We can control the base current using VBB (we don’t actually use a physical switch). The circuit then acts as a high speed switch.


Details l.jpg
Details

  • In Cut-off

    • All currents are zero and VCE = VCC

  • In Saturation

    • IB big enough to produce IC(sat)bIB

  • Using Kirchhoff’s Voltage Law through the ground loop

    • VCC = VCE(sat) + IC(sat)RC

    • but VCE(sat) is very small (few tenths), so

    • IC(sat)VCC/RC


Example32 l.jpg
Example

  • What is VCE when Vin = 0 V?

  • Ans. VCE = VCC = 10 V

  • b) What minimum value of IB is required to saturate the transistor if b = 200? Take VCE(sat) = 0 V

IC(sat) VCC/RC = 10 V/1000 W

= 10 mA

Then, IB = IC(sat)/b = 10 mA/200 = 0.05mA


Example33 l.jpg

LED

Example

If a square wave is input for VBB, then the LED will be on when the input is high, and off when the input is low.


Transistors with ac input l.jpg
Transistors with ac Input

Assume that b is such that IC varies between 20 and 40 mA. The transistor is constantly changing curves along the load line.


Slide35 l.jpg

Pt. A corresponds to the positive peak. Pt. B corresponds to the negative peak. This graph shows ideal operation.


Distortion l.jpg

Driven to saturation to the negative peak. This graph shows ideal operation.

Driven into Cutoff

Distortion

  • The location of the point Q (size of the dc source on input) may cause an operating point to lie outside of the active range.


Base biasing l.jpg
Base Biasing to the negative peak. This graph shows ideal operation.

  • It is usually not necessary to provide two sources for biasing the transistor.

The red arrows follow the base-emitter part of the circuit, which contains the resistor RB. The voltage drop across RB is VCC – VBE (Kirchhoff’s Voltage Law). The base current is then…

and IC = bIB


Base biasing38 l.jpg
Base Biasing to the negative peak. This graph shows ideal operation.

  • Use Kirchhoff’s Voltage Law on the black arrowed loop of the circuit

    VCC = ICRC + VCE

    So, VCE = VCC – ICRC

    VCE = VCC – bIBRC

  • Disadvantge

    • b occurs in the equation for both VCE and IC

    • But b varies – thus so do VCE and IC

    • This shifts the Q-point (b-dpendent)


Example39 l.jpg
Example to the negative peak. This graph shows ideal operation.

  • Let RC = 560 W @ 25 °C b = 100

    RB = 100 kW @ 75 °C b = 150

    VCC = +12 V

@ 75 °C

IB is the same

IC = 16.95 mA

VCE = 2.51 V

IC increases by 50%

VCE decreases by 56%


Transistor amplifiers l.jpg
Transistor Amplifiers to the negative peak. This graph shows ideal operation.

  • Amplification

    • The process of increasing the strength of a signal.

    • The result of controlling a relatively large quantity of current (output) with a small quantity of current (input).

  • Amplifier

    • Device use to increase the current, voltage, or power of the input signal without appreciably altering the essential quality.


Class a l.jpg
Class A to the negative peak. This graph shows ideal operation.

  • Entire input waveform is faithfully reproduced.

  • Transistor spends its entire time in the active mode

    • Never reaches either cutoff or saturation.

    • Drive the transistor exactly halfway between cutoff and saturation.

    • Transistor is always on – always dissipating power – can be quite inefficient


Class a42 l.jpg
Class A to the negative peak. This graph shows ideal operation.


Class b l.jpg
Class B to the negative peak. This graph shows ideal operation.

  • No DC bias voltage

    • The transistor spends half its time in active mode and the other half in cutoff


Push pull pair l.jpg
Push-pull Pair to the negative peak. This graph shows ideal operation.

Transistor Q1 "pushes" (drives the output voltage in a positive direction with respect to ground), while transistor Q2 "pulls" the output voltage (in a negative direction, toward 0 volts with respect to ground).

Individually, each of these transistors is operating in class B mode, active only for one-half of the input waveform cycle. Together, however, they function as a team to produce an output waveform identical in shape to the input waveform.


Class ab l.jpg
Class AB to the negative peak. This graph shows ideal operation.

  • Between Class A (100% operation) and Class B (50% operation).


Class c l.jpg
Class C to the negative peak. This graph shows ideal operation.

IC flows for less than half then cycle. Usually get more gain in Class B and C, but more distortion


Common emitter transistor amplifier l.jpg
Common Emitter Transistor Amplifier to the negative peak. This graph shows ideal operation.

Notice that VBB forward biases the emitter-base junction and dc current flows through the circuit at all times

The class of the amplifier is determined by VBB with respect to the input signal.

Signal that adds to VBB causes transistor current to increase

Signal that subtracts from VBB causes transistor current to decrease


Details48 l.jpg
Details to the negative peak. This graph shows ideal operation.

  • At positive peak of input, VBB is adding to the input

  • Resistance in the transistor is reduced

  • Current in the circuit increases

  • Larger current means more voltage drop across RC (VRC = IRC)

  • Larger voltage drop across RC leaves less voltage to be dropped across the transistor

  • We take the output VCE – as input increases, VCE decreases.


More details l.jpg
More details to the negative peak. This graph shows ideal operation.

  • As the input goes to the negative peak

    • Transistor resistance increases

    • Less current flows

    • Less voltage is dropped across RC

    • More voltage can be dropped across C-E

  • The result is a phase reversal

    • Feature of the common emitter amplifier

  • The closer VBB is to VCC, the larger the transistor current.


Pnp common emitter amplifier l.jpg
PNP Common Emitter Amplifier to the negative peak. This graph shows ideal operation.


Npn common base transistor amplifier l.jpg
NPN Common Base Transistor Amplifier to the negative peak. This graph shows ideal operation.

  • Signal that adds to VBB causes transistor current to increase

  • Signal that subtracts from VBB causes transistor current to decrease

  • At positive peak of input, VBB is adding to the input

  • Resistance in the transistor is reduced

  • Current in the circuit increases

  • Larger current means more voltage drop across RC (VRC = IRC)

  • Collector current increases

  • No phase reversal


Pnp common base amplifier l.jpg
PNP Common Base Amplifier to the negative peak. This graph shows ideal operation.


Npn common collector transistor amplifier l.jpg
NPN Common Collector Transistor Amplifier to the negative peak. This graph shows ideal operation.

Also called an Emitter Follower circuit – output on emitter is almost a replica of the input

Input is across the C-B junction – this is reversed biased and the impedance is high

Output is across the B-E junction – this is forward biased and the impedance is low.

Current gain is high but voltage gain is low.


Pnp common collector transistor amplifier l.jpg
PNP Common Collector Transistor Amplifier to the negative peak. This graph shows ideal operation.


Gain factors l.jpg

Usually given for common base amplifier to the negative peak. This graph shows ideal operation.

Usually given for common emitter amplifier

Usually given for common collector amplifier

Gain Factors


Gamma l.jpg
Gamma to the negative peak. This graph shows ideal operation.

  • Recall from Kirchhoff’s Current Law

    • IB + IC = IE

Ex. For b = 100 a = b/(1+b) = 0.99

g = 1 + b = 101


Bringing it together l.jpg
Bringing it Together to the negative peak. This graph shows ideal operation.


Hybrid parameters l.jpg
Hybrid Parameters to the negative peak. This graph shows ideal operation.

Second subscript indicates common base (b), common emitter (e), or common collector (c)


Hybrid parameters59 l.jpg

= to the negative peak. This graph shows ideal operation.b

= Slope of curve

Hybrid Parameters


Hybrid parameters60 l.jpg
Hybrid Parameters to the negative peak. This graph shows ideal operation.

hie = VB/IB Ohm’s Law

hie =input impedance

hre = VB/VC


Hybrid parameters61 l.jpg
Hybrid Parameters to the negative peak. This graph shows ideal operation.

hfe = IC/IB

Equivalent of b

hoe = IC/VC


Various forms l.jpg
Various Forms to the negative peak. This graph shows ideal operation.


Pin outs l.jpg
Pin-outs to the negative peak. This graph shows ideal operation.

No standard – look at the spec sheet or the case


Loudness l.jpg
Loudness to the negative peak. This graph shows ideal operation.

  • When the energy (intensity) of the sound increases by a factor of 10, the loudness increases by 1 bel

    • Named for A. G. Bell

    • One bel is a large unit and we use 1/10th bel, or decibels

  • When the energy (intensity) of the sound increases by a factor of 10, the loudness increases by 10 dB


Decibel scale l.jpg
Decibel Scale to the negative peak. This graph shows ideal operation.

  • For intensities

    • L = 10 log(I/Io)

  • For energies

    • L = 10 log(E/Eo)

  • For amplitudes

    • L = 20 log(A/Ao)


Threshold of hearing l.jpg
Threshold of Hearing to the negative peak. This graph shows ideal operation.

  • The Io or Eo or Ao refers to the intensity, energy, or amplitude of the sound wave for the threshold of hearing

    • Io = 10-12 W/m2

    • Loudness levels always compared to threshold

      • Relative measure


Common loud sounds l.jpg

160 to the negative peak. This graph shows ideal operation.

Jet engine - close up

150

Snare drums played hard at 6 inches awayTrumpet peaks at 5 inches away

140

Rock singer screaming in microphone (lips on mic)

130

Pneumatic (jack) hammer

Cymbal crash

Planes on airport runway

120

Threshold of pain - Piccolo strongly played

Fender guitar amplifier, full volume at 10 inches away

Power tools

110

Subway (not the sandwich shop)

100

Flute in players right ear - Violin in players left ear

Common Loud Sounds


Common quieter sounds l.jpg

90 to the negative peak. This graph shows ideal operation.

Heavy truck traffic

Chamber music

80

Typical home stereo listening levelAcoustic guitar, played with finger at 1 foot away

Average factory

70

Busy street

Small orchestra

60

Conversational speech at 1 foot away

Average office noise

50

Quiet conversation

40

Quiet office

30

Quiet living room

20

10

Quiet recording studio

0

Threshold of hearing for healthy youths

Common Quieter Sounds


The math l.jpg
The Math to the negative peak. This graph shows ideal operation.

l1 = 10 log(I1/Io)

l2 = 10 log(I2/Io)

l2 – l1= Dl = 10(log I2 – log Io – log I1 + log Io)

= 10(log I2 – log I1)

l2 – l1 = Dl = 10 log(I2/I1)

Threshold of Hearing when I = Iol = 0 dB

Threshold of Pain when I  1012 Iol = 120 dB


Example70 l.jpg
Example to the negative peak. This graph shows ideal operation.

  • A loudspeaker produces loudness rated at 90 dB (l1) at a distance of 4 ft (d1). How far can the sound travel (d2) and still give a loudness at the listener’s ear of 40 dB (l2 - conversation at 3 ft.)?

Sound follows the inverse square law I1/I2 = d22/d12

Dl = 50 dB = 10 log(I2/I1)

log(I2/I1) = 5 which means I2/I1 = 105

If d1 = 4 ft, then d22 = (I1/I2) d12 = 105 (4 ft)2

d2 = 1260 ft (about ¼ mile)


Common emitter current gain l.jpg

0 dB to the negative peak. This graph shows ideal operation.

-3 dB

hfe

Frequency

Common Emitter Current Gain

  • For the -3 dB point

    • Dl = 3 dB = 10 log (I1/I2)

      • I1/I2 = 2 = P1/P2

      • so 3 dB below initial level mean half the power


Why do frequency limits occur l.jpg
Why do Frequency limits occur? to the negative peak. This graph shows ideal operation.

  • It takes a certain time for e- to travel from emitter to collector (transit time)

  • If frequency is too high, applied current varies too rapidly

  • Electrons may be unable to dislodge rapidly enough to move from E to C before current surges in the other direction.

    Making the base thinner reduces transit time and improves frequency response


Interelement capacitance l.jpg
Interelement Capacitance to the negative peak. This graph shows ideal operation.

  • As reverse bias increases on the C-B junction, the depletion zone increases and C decreases (C = eA/d and d increasing).

  • As emitter current increases, C increases (d decreasing).

  • If capacitance changes, so does capacitive reactance

    • Increasing C decreases XC


Feedback l.jpg
Feedback to the negative peak. This graph shows ideal operation.

  • Small base current provides a path back to input

    • If the feedback voltage aids the input voltage, then it is positive (regenerative) feedback

    • If the feedback is too large, the amplifier will oscillate


Superheterodyne receiver l.jpg
Superheterodyne Receiver to the negative peak. This graph shows ideal operation.


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