transistors transfer resistor chapter 9. bipolar transistors two pn junctions joined together two...
TRANSCRIPT
Transistors
Transfer Resistor
Chapter 9
Bipolar Transistors
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)
BaseCollector
Emitter
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
Currents
Conventional View
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?...
Original Manufacture
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
The C-B voltage junction operates near breakdown.This ensures that a small E-B voltage causes
avalancheLarge current through the device
Common Base NPN
Common Emitter NPN
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 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 = IB)
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 = IB, different base currents produce different IC plateaus.
NPN Characteristic Curves
PNP Characteristic Curves
Load Line
For a constant load, stepping IB gives different currents (IC) predicted by where the load line crosses the characteristic curve. IC = IBworks so long as the load line intersects on the plateau region of the curve.
Slope of the load line is 1/RL
Saturation and Cut-off
Note that the load line intersects the 75 mA curve below the plateau region. This is saturation and IC = IB doesn’t work in this region.
Cut-off
Example
We adjust the base current to 200 A and note that this transistor has a = 100Then IC = IB = 100(200 X 10-6A) = 20 mA
Notice that we can use Kirchhoff’s voltage law around the right side of the circuitVCE = VCC – ICRC = 10 V – (20 mA)(220 ) = 10 V – 4.4 V = 5.6 V
Example
Now adjust IB to 300 ANow we get IC = 30 mA
And VCE = 10 V – (30 mA)(220 ) = 3.4 V
Finally, adjust IB = 400 A IB = 40 mA and VCE = 1.2 V
Plot the load line
VCE IC
5.6 V 20 mA
3.4 V 30 mA
1.2 V 40 mA
Gain as a function of IC
As temperature increases, the gain increasesfor all current values.
Operating Limits
There will be a limit on the dissipated powerPD(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
Example
Assume PD(max) = 0.5 W
VCE(max) = 20 V
IC(max) = 50 mA
PD(max) VCE IC
0.5 W 5 V 100 mA
10 50
15 33
20 25
Operating Range
Operating
Range
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
gain voltage Vin
out AV
V
rE = internal ac emitter resistanceIE = Vin/rE (Ohm’s Law)
Vout = ICRC IERC
E
C
EE
CEV r
R
rI
RIA Recall the name – transfer resistor
Current Gains
Common Base = IC/IE < 1
Common Emitter = IC/IB
1
11
I
I1
I
I
III
Law Current sKirchhoff' From
C
B
C
E
BCE
1
)1(
11
Example
If = 50, then = 50/51 = 0.98Recall < 1
Rearranging, = + (1-) = = /(1-)
Transistors as Switches
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
In Cut-offAll currents are zero and VCE = VCC
In Saturation IB big enough to produce IC(sat) IB
Using Kirchhoff’s Voltage Law through the ground loopVCC = VCE(sat) + IC(sat)RC
but VCE(sat) is very small (few tenths), so IC(sat) VCC/RC
Examplea) 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 = 200? Take VCE(sat) = 0 V
IC(sat) VCC/RC = 10 V/1000
= 10 mA
Then, IB = IC(sat)/ = 10 mA/200 = 0.05mA
Example
LED
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
Assume that is such that IC varies between 20 and 40 mA. The transistor is constantly changing curves along the load line.
Pt. A corresponds to the positive peak. Pt. B corresponds to the negative peak. This graph shows
ideal operation.
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.
Driven to saturation
Driven into Cutoff
Base Biasing
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…
C
BECC
R
VV BI and IC = IB
Base Biasing
Use Kirchhoff’s Voltage Law on the black arrowed loop of the circuitVCC = ICRC + VCE
So, VCE = VCC – ICRC
VCE = VCC – IBRC
Disadvantge occurs in the equation for both VCE and IC
But varies – thus so do VCE and IC
This shifts the Q-point (-dpendent)
Example
Let RC = 560 @ 25 °C = 100
RB = 100 k @ 75 °C = 150
VCC = +12 V
mA 11.3 A) (100)(113 I I BC
V 5.67
) A)(560 (100)(113 - V 12
RI V V CBCCCE
@ 75 °C
IB is the same
IC = 16.95 mA
VCE = 2.51 V
IC increases by 50%
VCE decreases by 56%
A113 100,000
V 0.7 - V 12I
C 25 @
B
B
BECC
R
VV
Transistor Amplifiers
AmplificationThe 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).
AmplifierDevice use to increase the current, voltage, or
power of the input signal without appreciably altering the essential quality.
Class A
Entire input waveform is faithfully reproduced.
Transistor spends its entire time in the active modeNever 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 A
Class B
No DC bias voltageThe transistor spends half its time in active
mode and the other half in cutoff
Push-pull Pair
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
Between Class A (100% operation) and Class B (50% operation).
Class C
IC flows for less than half then cycle. Usually get more gain in Class B and C, but more distortion
Common Emitter Transistor Amplifier
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 increaseSignal that subtracts from VBB causes transistor current to decrease
Details
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
As the input goes to the negative peakTransistor resistance increasesLess current flowsLess voltage is dropped across RC
More voltage can be dropped across C-EThe result is a phase reversal
Feature of the common emitter amplifierThe closer VBB is to VCC, the larger the
transistor current.
PNP Common Emitter Amplifier
NPN Common Base Transistor Amplifier
Signal that adds to VBB causes transistor current to increaseSignal 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
NPN Common Collector Transistor Amplifier
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
Gain Factors
E
C
I
I Usually given for common base amplifier
B
C
I
I Usually given for common emitter amplifier
B
E
I
I Usually given for common collector amplifier
Gamma
Recall from Kirchhoff’s Current Law IB + IC = IE
1
I
I
I
I1 I
B
E
B
CB
-1
1
-1
-1 LCD
-1
1
-1 since And
Ex. For = 100 = /(1+) = 0.99
= 1 + = 101
Bringing it Together
Type Common Base
Common Emitter
Common Collector
Relation between input/output phase
0° 180° 0°
Voltage Gain High Medium Low
Current Gain Low () Medium () High ()
Power Gain Low High Medium
Input Z Low Medium High
Output Z High Medium Low
Hybrid Parameters
Condition
hi Input resistance Output shorted
hr Voltage feedback ratio Input open
hf Forward current gain Output shorted
ho Output conductance Input open
Second subscript indicates common base (b), common emitter (e), or common collector (c)
Hybrid Parameters
=
= Slope of curve
Hybrid Parameters
hie = VB/IB Ohm’s Law
hie =input impedance
hre = VB/VC
Hybrid Parameters
hfe = IC/IB
Equivalent of
hoe = IC/VC
Various Forms
Common Emitter (e)
Common Base (b)
Common Collector (c)
hi (ohms) VB/IB VE/IB VB/IB
hr (unitless) VB/VC VE/VC VB/VE
hf (unitless) IC/IB IC/IE IE/IB
ho (watts) ICVC ICVC IEVE
Pin-outs
No standard – look at the spec sheet or the case
Loudness
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
For intensitiesL = 10 log(I/Io)
For energiesL = 10 log(E/Eo)
For amplitudesL = 20 log(A/Ao)
Threshold of Hearing
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
160
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) 100Flute in players right ear - Violin in players left ear
Common Quieter Sounds
90
Heavy truck traffic
Chamber music 80Typical 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
The Math
l1 = 10 log(I1/Io)l2 = 10 log(I2/Io)
l2 – l1 = l = 10(log I2 – log Io – log I1 + log Io) = 10(log I2 – log I1)
l2 – l1 = l = 10 log(I2/I1)
Threshold of Hearing when I = Io l = 0 dB
Threshold of Pain when I 1012 Io l = 120 dB
Example
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/d1
2
l = 50 dB = 10 log(I2/I1)
log(I2/I1) = 5 which means I2/I1 = 105
If d1 = 4 ft, then d22 = (I1/I2) d1
2 = 105 (4 ft)2
d2 = 1260 ft (about ¼ mile)
Common Emitter Current Gain
For the -3 dB pointl = 3 dB = 10 log (I1/I2)
I1/I2 = 2 = P1/P2
so 3 dB below initial level mean half the power
Frequency
hfe
0 dB-3 dB
Why do Frequency limits occur?
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
As reverse bias increases on the C-B junction, the depletion zone increases and C decreases (C = A/d and d increasing).
As emitter current increases, C increases (d decreasing).
If capacitance changes, so does capacitive reactance
Increasing C decreases XC
C f 2
1X C
Feedback
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