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7/27/2019 Transistors I
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Introduction to Transistors• A transistor is a device with three separate layers of
semiconductor material stacked together – The layers are made of n –type or p –type material in the
order pnp or npn
– The layers change abruptly to form the pn or np junctions
– A terminal is attached to each layer (The Art of Electronics, Horowitzand Hill, 2nd Ed.)
(Introductory Electronics, Simpson, 2nd Ed.)
7/27/2019 Transistors I
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Introduction to Transistors• Thus when a transistor is
off it behaves like atwo –diode circuit
• A transistor operates (or
turns on) when the base –emitter junction is forward
biased and the base –collector junction is reversedbiased (“biasing”)
(Electronic Devices and
Circuits, Bogart, 1986)
(The Art of Electronics,
Horowitz and Hill, 2nd Ed.)
(Lab 4 –1)
7/27/2019 Transistors I
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Transistor Biasing (npn Transistor)• Electrons are constantly
supplied to the emitter bythe battery with voltage V EE
• These electrons can:1. Recombine with holes in
the base, giving rise to I B 2. Diffuse across base and be swept (by electric field at
base –emitter junction) into collector, then diffuse around
and eventually recombine with holes injected into
collector, giving rise to I C • Since the base region is designed so thin, process 2
dominates (no time for #1 to occur as often) – In an actual npn transistor, 98 or 99% of the electrons that
diffuse into the base will be swept into the collector
7/27/2019 Transistors I
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Current Flow Inside a Transistor • Current flow for an npn transistor (reverse for pnp):
– From conservation of current( I E = I B + I C ) we can obtain the following
expressions relating the currents:
where b ≈ 20 – 200
(depends on emitter current)
• b increases as I E increases (for very small I E ) since there is less
chance that recombination will occur in the base
• b decreases slightly (10 –20%) as I E increases beyond several mA
due to increased base conductivity resulting from larger number of
charge carriers in the base
• Thus b is not a constant for a given transistor!
• An average value of 100 is typically used
(Electronic Devices and Circuits,
Bogart, 1986)
BC I I b B E I I 1 b
(and thus I C ≈ I E )
(Lab 4 –5)
7/27/2019 Transistors I
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Transistor Current Amplification• If the “input” current is I B and the “output” current is
I C , then we have a current amplification or gain – Happens because base –emitter junction is forward-biased
– Forward bias ensures that the base –emitter junction
conducts (transistor is turned on)
– Reverse bias ensures that most of the large increase inthe base –emitter current shows up as collector current
(Student Manual for The Art
of Electronics, Hayes and
Horowitz, 2nd Ed.)
Thus small gains in I B
result in large gains in I E
and hence I C
7/27/2019 Transistors I
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Basic Transistor Switch Circuit• Transistor “switch” circuit:
– With switch open, transistor is off and lamp is off
– With switch closed, I B = (10 – 0.6) V / 1k = 9.4 mA
– However, I C = b I B 940 mA (assuming b = 100)• When collector current I C = 100 mA, lamp has 10V across it
• To get a higher current, collector would need to be below ground
• Transistor can’t do this, so it goes into saturation
• Collector voltage gets as close to emitter voltage as it can (about0.2 V higher) and I C remains constant ( I C is “maxed out”)
(The Art of Electronics, Horowitz and Hill, 2nd Ed.) 0 V
0.6 V0.2 V
(BC junction forward biased)
V B V C
V E
(Lab 4 –9)
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Emitter Follower • Output “follows” the input: only
difference is a 0.6 V diode drop – True for V in > 0.6 V
– If V in < 0.6 V, transistor turns off (no
current – “valve” is closed) and V out = 0
– Data with R E = 3.3k : (The Art of Electronics, Horowitz
and Hill, 2nd Ed.)
BE
C
E
(Lab 4 –2)
V in
V out
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Emitter Follower • By returning the emitter
resistor to a negativesupply voltage, you can
obtain negative voltage
swings as well
– Data with R E = 3.3k :(The Art of Electronics,
Horowitz and Hill, 2nd Ed.)
7/27/2019 Transistors I
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Emitter Follower Biasing
• You must always provide a DC path for base bias
current, even if it is just through a resistor to ground – HW Problem 2.5
(The Art of Electronics, Horowitz and Hill, 2nd Ed.)
7/27/2019 Transistors I
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Emitter Follower Biasing• With R B included in the previous circuit:
f = 1 kHz
7/27/2019 Transistors I
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Emitter Follower Biasing• Without R B included in the previous circuit:
(Here there is no DC base bias current, so transistor is off.)
7/27/2019 Transistors I
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Emitter Follower Biasing• To obtain symmetric output waveforms without
“clipping,” provide constant DC bias using a voltagedivider
– Capacitors block “outside” DC current, which may affect
quiescent (no input) values (“AC-coupled follower”)
(The Art of Electronics, Horowitz and Hill, 2nd Ed.)
(Lab 4 –4)
7/27/2019 Transistors I
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Emitter Follower Impedance• The usefulness of the emitter follower can be seen
by determining its input and output impedance: – Input impedance (i.e. the impedance looking into the base
of the transistor):
• Details of proof given in class – Output impedance (i.e. the impedance looking into the
emitter of the transistor):
• Details provided by you in the homework!
• Thus the input impedance is much larger than the
output impedance
loadloadin 1 Z Z Z b b
b b
sourcesourceout
1
Z Z Z
7/27/2019 Transistors I
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Emitter Follower Impedance• Thus the input and output “sees” what it wants to
see on the other side of the transistor:
• Using an emitter follower, a given signal sourcerequires less power to drive a load than if the sourcewere to drive the load directly – Very good, since in general we want
Z out (stage n) << Z in (stage n + 1) (by at least a factor of 10)
– An emitter follower has current gain, even though it has novoltage gain
– The emitter follower has power gain
(Student Manual for The Art
of Electronics, Hayes and
Horowitz, 2nd Ed.)
7/27/2019 Transistors I
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Emitter Follower Impedance• When measuring the input and output impedance of
the emitter follower, it is useful to think about theThévenin equivalent circuit as “seen” at the input
and the output:
– Input impedance seen by the source:
– Output impedance seen by the load:
V in
Z source
~
Z out Z load
Z in
V B
V out, load V out, no load
in
insource
inB V
Z Z
Z V
loadnoout,
loadout
loadloadout, V
Z Z
Z V
(Student Manual for The Art
of Electronics, Hayes andHorowitz, 2nd Ed.)
(Lab 4 –3)
7/27/2019 Transistors I
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Emitter Follower With Load• Consider the following circuit:
– V out and V in waveforms:
V in (V) V out (V) I E (mA)
+9.4 8.8 27.6
5 4.4 18.8
0 – 0.6 8.8 – 3 – 3.6 2.8
– 4.4 – 5.0 0.0
– 5 – 5.0 0.0
– 10 – 5.0 0.0
(The Art of Electronics,
Horowitz and Hill, 2nd Ed.)
V in
V out
V in V out I E
(HW 2.2)
7/27/2019 Transistors I
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Emitter Follower With Load• Thus the npn emitter follower can only “source”
current (supply current to something like a load)
• It cannot “sink” current (draw current from somethinglike a load)
• In this example, the transistor turns off when
V in = –
4.4 V (V out = –
5.0 V) – Then I E = 0 and the base –emitter junction becomes reversebiased
– As V in increases further, a rather large reverse biasdevelops across this junction which could result in
breakdown
• The output could swing more negative than –5 V byreducing the R E = 1k resistor, but this increasespower consumption in both the resistor and transistor
7/27/2019 Transistors I
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Zener Diodes as Voltage Regulators• Zener diodes “like” to break down at
a particular reverse bias: – When reverse biased, they provide a
constant voltage drop over a wide range
of currents
• Zeners thus provide a means of voltage regulation
– We choose the specifications for the zener based on:
(max)outoutminin, I
R
V V
zener minout,
min
outmaxin,zener V I R
V V P
(The Art of Electronics, Horowitz
and Hill, 2nd Ed.)
(Student Manual for The Art
of Electronics, Hayes and
Horowitz, 2nd Ed.)
7/27/2019 Transistors I
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Example Problem 2.3
Solution details given in class.
Design a +10 V regulated supply for loadcurrents from 0 to 100 mA; the input
voltage is +20 to +25 V. Allow at least
10 mA zener current under all (worst-case)
conditions. What power rating must thezener have?
7/27/2019 Transistors I
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Emitter Followers as Voltage Regulators
• However, the zener current can change significantly
depending on the load, affecting regulation
performance
• A better voltage regulator would incorporate an
emitter follower:
– Here the zener current is more constant, relatively
independent of load current since changes in I E (or I load)
produce only small changes in I B
• Load current determined from (V B – 0.6 V) / Rload
(The Art of
Electronics, Horowitz
and Hill, 2nd Ed.)
(HW 2.4)
7/27/2019 Transistors I
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Transistors as Current Sources• A transistor can be used as a
current source with the setup at
right:
– Note that I C is independent of V C as long as V C > V E + 0.2 V(i.e., the transistor is not saturated)
• The output voltage (V load or V C ) range over which I load
(= I C ) is (nearly) constant is called the output
compliance
E
B
E
E
C E
R
V
R
V I I
6.0
(The Art of Electronics, Horowitz and Hill, 2nd Ed.)
V C
V E
(Lab 4 –6)
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Deficiencies of Current Sources• The load current will still vary somewhat, even when
the transistor is “on” and not in saturation
• There are two kinds of effects that cause this: – V BE varies somewhat with collector-to-emitter voltage for a
given collector current (Early effect), as does b • DV BE ≈ – 0.0001 DV CE
• We assume V BE = constant = 0.6 V in the basic transistor model
– V BE and b depend on temperature• DV BE ≈ – 2.1 mV/0C
• We neglect changes in b by assuming I C = I E
• To minimize DV BE from both effects, choose V E largeenough ( 1V) so that DV BE 10 mV will not result inlarge fractional changes in the voltage across R E – V E too large will result in decreased output compliance,
however (V C range for transistor “on” state decreases)
7/27/2019 Transistors I
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Common –Emitter Amplifier • Consider a transistor current
source with a resistor RC
as
load, and block unwanted
DC at the base input (V in is
an AC signal):
so where
– Now imagine we apply a base wiggle v B via the input signal
– The emitter follows the wiggle so v E = v B
– Then the wiggle in the emitter current is:
C R f
eq
dB32
1
eq32
1
R f C
dB
E R R R R b 21eq
C
E
B
E
E
E i R
v
R
v
i
(The Art of Electronics, Horowitz and Hill, 2nd Ed.)
(lower-case letters
represent smallchanges, or “wiggles”)
(Note DC quiescent
output voltage of 10 V)
(Lab 4 –7)
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Common –Emitter Amplifier – Now V C = V CC – I C RC so vC = – iC RC = – v B( RC / R E )
– Since vin
= v B
and vout
= vC
, we have a voltage amplifier,
with a voltage gain of:
– Minus sign means that a positive wiggle at the input getsturned into a negative wiggle at the output
• Input and output impedance:
– Z in = R1 R2 b R E ≈ 8k (see figure on previous slide)
– Z out = RC (impedance looking into collector) = RC (high Z current source) ≈ RC = 10k (see figure on previous slide)
• Be careful to choose R1 and R2 correctly so that
design is not b dependent ( R1 R2 << b R E )
E
C
R
R
v
vG
in
out