Week 8 transistors

The Eber Molls model discussed below makes one critical point. If you are designing a transistor circuit then in some regions of operation the simple βIb=IC relationship doesn’t hold for constant β. β doesn’t remain constant. But you can predict the current IC knowing the Vbe voltage using Eber Molls. The transistor appears to operate as a diode determined by the base-­‐emitter voltage but with a current IC. The base current is not always proportional. When operating in regions near Vbe ~ 0.6 V usually β is fairly constant. In many well-­‐
designed circuits this is sufficient to understand operation. Well-­‐designed circuits do not rely on the value of β. [The voltage amplifier is a great example. The resistor Re acts as a type of feedback controlling Vbe. It adjusts the voltage (see derivation below) to maintain the 0.6 turn-­‐on thus following the voltage up and down which produces a current that ultimately changes voltage across RC. A small ΔVbe creates a large change in IC. This variation ξ may be neglected because a typical result is a 60 mV change, ΔVbe, can increase the current 10-­‐fold, 10IC. This is consistent with both models. ] C2 The emitter capacitor is included to hold the emitter voltage at the quiescent point. If an AC current is added through Re then the voltage at the emitter will change. However if this additional current simply flows to the capacitor then Ve stays about the same. C2 is a bypass capacitor. It provides the AC ground (common).
Bypassed emitter (Horowitz).
This capacitor is important when one wants to maximize the gain and therefore use a small Re. The
simplest case to consider is Re = 0. However if Re is too small then there could be large Vbe which results
in more temperature variation and potentially too much current depending on the input. So you limit this
by using a reasonable Re. The capacitor in this case acts as a small resistance over a range of frequencies.
The varying signal in to the base then can generate a large change in IC at those frequencies because the
capacitor is the dominant element absorbing and releasing charge to sustain the current. So it is like two
currents under the control of the transistor. There is the quiescent current determined by the zero input
analysis and the oscillating current that is able to be much larger for a given ΔVbe than you would predict
based on Re.
The bypass capacitor is chosen to be small wrt re for frequencies of interest. In fact, the real relevant
resistance is the effective resistance of the emitter diode. Fix the voltage on the base and ask how much
ΔVbe will impact the current Ie. Because of the amplification only a small change is required to produce a
large change in Ie . As a load resistor connected to the emitter you would see a votage source in series with
re)
(re is the effective impedance looking into the emitter for the base held at fixed Vb. Using eber
molls and the diode equation [see below] we can take the derivative of Vbe wrt Ie you get
approximately 25/Ie(mA)=re (Ohms).
rin Input impedance (emitter follower where ΔVbe= ΔVe ) èMeasure of ΔVbe to ΔIb . The
amplification β means that a much larger current is generated on output. So the base current can be
a factor β smaller to generate the Ie. This leads to an input impedance roughly rin =Reβ)
Choosing R and V at the inputs I can set a value of Vb 𝑉! = 𝑉!! − 𝐼! 𝑅! 𝑉!! − 𝑉!
𝐼! =
𝑅!
𝑉! = 𝑉!" − 0.6 (𝑉!" − [0.6 + 𝜉])
𝐼! =
𝑅!
𝑉!! − 𝑉! (𝑉!" − [0.6 + 𝜉])
=
𝑅!
𝑅!
𝑅! (𝑉!" − [0.6 + 𝜉])
𝑉! = 𝑉!! −
𝑅!
The capacitors will eliminate any constant voltages that appear in the above equation. Or if we examine only the time dependent part 𝑅! 𝑣!"
𝜉
𝑣! = −
+ 𝑅!
𝑅!
𝜉 ~0, one can ignore it for 1st order operation but it implies that if the Vbe varies a bit due to a less than infinite rise in the diode current it has a very small effect unless Re is small. ied property 4:
-3 (Section
d to
by
are
at room
q is the eleccoulombs), is
tant (1.38 x
e absolute temperature
and
n current of the particends on T). Then the
also depends on
d by
=
+
t"
is typically in
1000, but depends on
and temperature.
everse leakage current.
and
ion
rm can be neglected in
e exponential.
r
is known as the
n. It also approximaterent versus voltage for
multiplied by a correceen 1 and 2. For tranant to realize that the
accurately determined
r voltage, rather than
t (the base current is
mined by
and that
w is accurate over an
currents, typically from
iamps.
Figure 2.32
TRANSISTORS
Chapter 2
80
Eber Molls model. as a current amplifier whose input circuit makes the point graphically. If you meaand collector
Figure 2.32. Transistor basebehaved
sure ethe
base currentthe at various
collector of a diode. That's
roughly
cor-begin There ay be ofquite arect,
bit and
ovoltage
f like
confusion when you xamining properties currents asmfunctions
base-to-emitter
vercurrents, you will get a graph of
for some applications it's good
VBE.
transistors enough. But to understand differential amsus
like that in Figure 2.33.
log scale
10
10
10
10
plifiers, logarithmic converters, temperature compensation, and other important
applications, you must think of the transistor as a transconductance device - collector
current is determined by base-to-emitter
voltage.
Here's the modified property 4:
4. When rules 1-3 (Section
are
10
10
10is related to
obeyed,
by
=
exp
Figure 2.33. Typical transistor current gain
versus collector current.
=
where
=
at room
temperature
q is the elecAlthough the Ebers-Moll tron
equation
tells
charge (1.60 x
coulombs), is
Figure 2.32. Transistor base and collector
us that the base-emitter voltage
"proBoltzmann's constant (1.38 x
currents as functions of base-to-emitter voltage
grams" the collector current, this property
T is the absolute temperature
VBE.
may not be directly usable ininpractice
(bidegrees
Kelvin show t+
These two plots from aH
orowitz hat β is and
not a constant over the operating asing a transistor
by applying
base
voltis the saturation current of the particage) because
of the
large temperature
range of the transistor. For cothe region 0.5è0.75 the two currents are fairly be= ular transistor
(depends
on V
T).
Then
the
efficient of base-emitter voltage.
You willwhich also depends on
base
current,
parallel wthe
hich means (on log they are proportional and so the β=constant equation
pro-scale) see later how
Ebers-Moll
can
be approximated
by
vides insight
problem.
model is Oand
K solutions
[hfe is βto] this
but in the extremes the model breaks down. However the relationship between design
the voltage on the base-­‐emitter Vbe in the diode equation log scale
Rules of thumb for transistor
is typically in
where the "constant"
works well to predict t
he c
urrent I
. S
o i
n s
ome r
egions o
f o
peration w
here T 10may C but depends on
the range 20 to 1000,
10
10
10
10
10
10
From the Ebers-Moll equation
we cantype,
get
and
temperature.
transistor
be varying or Vbe is not represents
near the nominal 0.6 volts the base current is not a good the reverse leakage current.
predictor for the collector current Vbe a2.33.
la ETypical
ber M
olls is current
a good and
In the active
regionIC but the voltage Figure
transistor
gain
therefore the -1 term can be neglected in
versus collector current.
predictor. comparison with the exponential.
The equation for
is known as the
Although the Ebers-Moll equation tells
Ebers-Moll equation. It also approximateus that the base-emitter voltage "pro ly describes the current versus voltage for
grams" the collector current, this property
multiplied by a correca diode, if
may not be directly usable in practice (bi tion factor m between 1 and 2. For transistors it is important to realize that the
collector current is accurately determined
by the base-emitter voltage, rather than
by the base current (the base current is
then roughly determined by
and that
this exponential law is accurate over an
enormous range of currents, typically from
nanoamps to milliamps.
Figure 2.32
asing a transistor by applying a base voltage) because of the large temperature coefficient of base-emitter voltage. You will
see later how the Ebers-Moll equation provides insight and solutions to this problem.
Rules of thumb for transistor design
From the Ebers-Moll equation we can get
To get the currents above the transistor is designed to get the electrons from the emitter through the base…. You minimize the percentage of electrons that recombine before reaching the collector–base junction, the transistor's base region must be thin enough that electrons can diffuse across it in much less time than the semiconductor's minority carrier lifetime (for base + majority carriers so electrons are minority carriers). In particular, the thickness of the base must be much less than the diffusion length of the electrons. The collector–base junction is reverse-­‐biased, and so little electron injection occurs from the collector to the base, but electrons that diffuse through the base towards the collector are swept into the collector by the electric field in the depletion region of the collector–base junction. The thin shared base and asymmetric collector–emitter doping is what differentiates a bipolar transistor from two separate and oppositely biased diodes connected in series. The collector–emitter current can be viewed as being controlled by the base–
emitter current (current control), or by the base–emitter voltage (voltage control). These views are related by the current–voltage relation of the base–emitter junction, which is just the usual exponential current–voltage curve of a p-­‐n junction (diode).[1] Transistors
For current arrows must be reversed because we are following electrons and current convention is opposite (figure below diode and i-­‐source reversed) We will follow the electrons which is the grey line on the right picture. This flow is all of the electrons that leave the emitter (purple + blue in left diagram). This current is treated as a diode current. 𝑖! = 𝐼!" 𝑒
!!"
!!
!
!!"
!"
!!
− 1 diode equation. ln (! ! + 1) =
depends on • base emitter Voltage Vbe • temperature dependent constant that is about 25.85 mV at room temperature VT • the reverse saturation current Ies which is the current value obtain at reverse voltages. This is the leakage current when the diode is “off” i.e. in the reverse biased direction. Thus the diode equation needs to predict the total flow through the diode. This must include both the current to the base and to the collector. The Ies must be the reverse current including the enhanced flow due to the diffusion to the collector. [Originally this was a bit confusing because I preferred to view the diode part as only the current that reached the base and was therefore much smaller. For the case where β is constant you assume that there is simply a proportionality that then generates an additional current then this current is also following the diode equ. i.e. 𝑖! = 𝐼!" 𝑒
!!!
!!
𝑖! = 𝛽𝑖! − 1 this is a diode that sends all its current to the base 𝑖! = 𝑖! + 𝛽𝑖! = 1 + 𝛽 𝐼!" 𝑒
!!"
!!
− 1 = 𝐼!" 𝑒
!!"
!!
− 1 So if the current that reaches the collector is a constant factor β times the base current then treating all the current as following a diode equation is possible if one has the correct reverse saturation current. The model itself was no doubt based on the best modeling of the underlying processes i.e. the diffusion of majority and minority carriers. I did not attempt to follow these arguments. I just tried to understand the model and that led me to consider the base emitter as the diode rather than the emitter –(base+collector).] In the end the best model assumes that the collector current and the Vbe are related via the diode equation and the base current is not always proportional to IC so this is not a good variable to use when treating the transistor behavior in a circuit. This is the more complete picture which allows one to consider various biases. When the transistor for example is saturated the VC drops below Vb and so the base collector is forward biased. In normal operation (voltage amplifier in the linear region you can simplify to my original Approximated Ebers–Moll Model for an NPN transistor in the forward active mode. The collector diode is reverse-­‐biased so ICD is virtually zero. Most of the emitter diode current (αF is nearly 1) is drawn from the collector, providing the amplification of the base current. where
I is the diode current,
IS is the reverse bias saturation current (or scale current),
part of the reverse current in a diode caused by diffusion of minority carriers from the neutral regions to the depletion region. This current is almost independent of the reverse voltage." reverse saturation current, the current that flows when the diode is reverse biased (that is, VD is large and negative). More below [1e-­‐16 A è 1e-­‐8 A]
VD is the voltage across the diode,
VT is the thermal voltage, and
n is the ideality factor, also known as the quality factor or sometimes emission coefficient. The
ideality factor n typically varies from 1 to 2 (though can in some cases be higher), depending on
the fabrication process and semiconductor material and in many cases is assumed to be
approximately equal to 1 (thus the notation n is omitted). The ideality factor does not form part of
the Shockley ideal diode equation, and was added to account for imperfect junctions as observed
in real transistors. By setting n = 1 above, the equation reduces to the Shockley ideal diode
equation.
The thermal voltage VT is approximately 25.85 mV at 300 K, a temperature close to "room temperature"
diode eq VT= 25.85 0.03 current (Arb.) 0.025 0.02 0.015 0.01 0.005 0 0 100 200 300 400 500 600 700 800 V (MV) Exponentials look the same on V ranges. The y-­‐axis range changes. So turn on occurs based on Is and what current is considered meaningful. Above if mA currents are significant then the diode starts to provide mAmps at 600 mV. For the transistor there is then a Vbe where βIbe=IC is large enough to set Ve. If Vb changes (asume +δ) Vb +δ then more current will be delivered β(Ibe+ε). This will raise Ve which lowers Vbe and therfore δ. There will be a dance to steady state where the final Vbe, Ve, IC… are adjusted so that they are self consistent. Because the current available by slightly changing the diode is large current (Arb.) diode eq VT= 25.85 0.03 0.025 0.02 0.015 0.01 0.005 0 0 200 400 600 800 V (MV) same as above but smaller y range. The first plot is more characteristic of our expected behavior so the saturation current is perhaps smaller for the base-­‐emmiter diode. Diode Characteristics
Reverse saturation current
When negative voltages are applied to the diode the current becomes constant at IS as
the exponential term in Equation 1 quickly approaches zero. That is why it is referred to
as the reverse saturation current. The current is independent of applied voltage once a
small voltage magnitude is exceeded. This current is very small and is typically in the
low nanoampere region. The reverse saturation current is a strong function of
temperature as illustrated in Figure 3. The following equation is a simplified model for
the reverse saturation current and provides excellent results in normal operating regions.
Reverse saturation current
When negative voltages are applied to the diode the current
becomes constant at –IS as the exponential term in Equation 1
Iquickly
= I * exp approaches zero. That is why it is referred
Eq. 4to as the reverse
saturation current. The current is independent of applied voltage
Where:
I is the reverse saturation current
once a small voltage magnitude is exceeded. This current is very
I is a constant derived from n and I at a known temperature
Eg
is the bandgap
voltage
for silicon (ranges
from
aboutnanoampere
1.20 to 1.28 volts) region. The reverse
small
and is
typically
in the
low
n is the junction constant (typically around 2 for diodes, 1 for transistors)
saturation
is a strong
VT
is the thermal current
voltage as previously
discussed function of temperature as illustrated
in reverse
Figure
3. The
equation
is a simplified
The
saturation
currentfollowing
should not be confused
with an imperfection
in diodes model for the
known
as leakage
current from acurrent
high value shunt
across the
diode jucntion.
reverse
saturation
andresistance
provides
excellent
results in normal
Leakage current is often many times larger than I . Thus, I can not be directly measured
o
and
must be computed
using data(room
from the forward
operating
regions.
tempbias
25region
C) (see the section on
S
-Eg/nVT
K
S
K
S
S
S
measuring diode characteristics).
Reverse Saturation Current versus Temperature
10.0E-6
1.0E-6
100.0E-9
Current
10.0E-9
1.0E-9
100.0E-12
10.0E-12
1.0E-12
100.0E-15
10.0E-15
-60
-50
-40
-30
-20
-10
0
10
20
30
40
50
60
70
80
90
100
110
120
130
deg. C
Figure 3: Reverse saturation current, IS, versus Temperature
4
For first attempt at behavior let the diode have an almost direct rise. diode eq VT= 25.85 0.045 0.04 0.035 current (Arb.) 0.03 0.025 0.02 0.015 0.01 0.005 0 -­‐0.005 -­‐0.01 0 100 200 300 400 500 600 700 800 V (MV) This means the drop Vbe cannot be greater than 600 mV or the current will blow up. If you need more current to through Re or RC a negligible change in Vbe creates an increase to any amount. This will make Re critical for the voltage amplifier. This limits the current through the diode. The current amplification enhances the effect by amplifying the current by the factor β. The diode equation now has extra factor β contibuting to its current. Horowitz explains that an increase in a factor of 10 for the collector IC requires about 60 mV cange in the base voltage. eber molls The key is that the voltage on the collector, base and emitter could have the 1. both diodes forward biase Vb>Ve, Vb>VC 2. collector forward emitter reverse Vb<Ve, Vb>VC 3. emitter forward collector reverse Vb>Ve, Vb<VC 4. both reverse Vb<Ve, Vb<VC 3 is th standard, 1 is saturation, 4 is off (normal), 2 using the transistor in a funny mode making the collector the emitter and the emitter the collector. The equations really turn the diodes off when they are reverse biased. Impedance First review impedance as a tool to see how a load impacts a source and vice versa. We assume that we can develop a source as a Thevenin voltage Vth or Vo and Thevenin resistance Rth or Ro . Ro will reduce the available voltage and limit available current. Its impact will be small if the current is small which is the case when the load is greater than Ro. The emitter follower (common collector) can be used to primarily change impedance. The circuit that drives the base has a large load even if the emitter drives a small actual load. The notion of input and output impedance is important for several reasons. In addition to the impact on current and voltage we will also see that signal propagation depends on maintaining static impedance. In order for a signal to propagate correctly the ratio of the voltage/current should be maintained as the signal propagates through the medium. We will return to this issue in the future. In analyzing the emitter follower one can examine how the ΔVb and ΔIb are related given ΔVe and ΔIe. Δ is used to examine AC parts of the signal as opposed to the DC parts. For example, the base may be biased at some VQb and a capacitively coupled input varies around this bias. ΔVb and ΔIb are changes wrt VQb. Put a voltage on the collector, use a resistor attached to a lower voltage Vapp on the emitter side (Vapp=0 in many circuits, Re is the resistor shown in above circuit.) and add some current to base above or below the quiescent bias. Assuming β is large è ΔIb=ib ~ ΔIe/ β ; ΔVb~ ΔVe ; ΔVe=Re ΔIe ib ~ ΔVb/ β Re rin= β Re which means that to effect a change at the base you need voltages and currents that are related by the effective resistance rin. To change the base values you need to change the emitter values which change based on Re. But since there is a gain the inputs change β times smaller. Thus making the ibrin~ ΔVb . Output (impedance looking into the emitter ie the Rth of a diagram above showing a source for the base). Now Vs looks like Vth and one needs to find the effective resistance in series with this voltage because it will limit how much current can be drawn from this source etc. And the resistance is the resistance of the source applied at the base So one can neglect Re if it is much larger than RL but how much current can be drawn through RL for a given applies VS depends on the resistance Rs but again because the actual current that can be drawn is amplified there is a need for much smaller current from the source and this reduces the impact of RS. 1
Δ𝐼! =
Δ𝐼 𝛽 + 1 !"#
𝑍!
𝑍!"# =
𝛽+1
Thus Zout is reduced. i.e. you can pull current out of the emitter with less effect by Zs. other notes iF that’s the total current through the b-­‐e diode in the Forward direction (majority carriers) i.e. αF iF through the collector and joins with the forward current from the base. iR that’s the current through the b-­‐C diode in the Reverse direction (minority carriers) 5.5
The Ebers-Moll Model
287
αF fraction of the emmiter current that comes from the c5.5
ollector The Ebers-Moll
Model
287
βF is the ratio of collector to base currents iF/ βF is the diode current satisfying the diode equation in the standard way. There is a voltage Vbe a saturation current 5.5.1 Forward
Characteristics of the npn Transistor
5.5.1 Forward Characteristics of the npn Transistor
The total current
crossing the emitter-base junction in the forward direction in Fig. 5.3 is described
by Eq. (5.5) and The
can be
rewritten
as
total
current crossing
the emitter-base junction in the forward direction in Fig. 5.3 is described
!by Eq.
" (5.5)#and can
$ be rewritten
!
" as #
$
IS
vB E
vB E
IS
iE =
exp
− 1! = "I E S exp
−
1
(5.20)
#
$
!
" where
# I E S$=
αF
VT I S
VT
αF
vB E
vB E
IS
iE =
exp
− 1 = I E S exp
−1
where I E S =
(5.20)
αF
VT
VT
αF
in which the new parameter I E S represents the reverse saturation current of the base-emitter diode.
The collector current
in Eq.
be rewritten
in terms the
of Ireverse
E S as saturation current of the base-emitter diode.
in which
the(5.1)
new can
parameter
I represents
ES
The collector
! current
" in#Eq. (5.1)
$ can be !rewritten
" in#terms$of I E S as
vB E
vB E
i C = I S exp
− 1! = "
α F I E S# exp $
− 1!
"
# (5.21)
$
VT
VT
vB E
vB E
i C = I S exp
− 1 = α F I E S exp −1
(5.21)
VT
VT
The forward common-base
current
gain
α
represents
the
fraction
of
the
emitter
current
that
F
crosses the base
and appears in the collector terminal.
The forward common-base current gain α F represents the fraction of the emitter current that
crosses the base and appears in the collector terminal.
5.5.2 Reverse Characteristics of the npn Transistor
5.5.2 depicted
Reverse
Characteristics
npn Transistor
For the reverse direction
in Fig.
5.4, the current crossingof
thethe
collector-base
junction is
described by Eq.For
(5.11)
can be
writtendepicted
as
the and
reverse
direction
in Fig. 5.4, the current crossing the collector-base junction is
by
!described
"
# Eq. $(5.11) and!can be
" written
# as$
v BC
v BC
IS
IS
exp
− 1! = "
−IC S #exp $
− 1!
(5.22)
" where
# IC S$= α
αR
VT I S
V
T
R
v BC
v BC
IS
iC = −
exp
− 1 = −IC S exp
−1
where IC S =
(5.22)
αR
VT
VT
αR
The new parameter IC S represents the reverse saturation current of the base-emitter diode. The
emitter current from
can be Irewritten
in terms
of IC S assaturation current of the base-emitter diode. The
The Eq.
new(5.9)
parameter
represents
the reverse
iC = −
CS
emitter current
! from
" Eq.
# (5.9)$can be rewritten
! in
"terms#of IC$S as
v BC
v BC
i E = −I S exp
− 1! = "
−α R IC#S exp
− 1!
$
"
# (5.23)
$
VT
VT
v BC
v BC
i E = −I S exp
− 1 = −α R IC S exp
−1
(5.23)
VT
VT
The reverse common-base current gain α R represents the fraction of the collector current that
crosses the base from
the emitter
terminal. current gain α represents the fraction of the collector current that
The reverse
common-base
R
crosses the base from the emitter terminal.
5.5.3 The Ebers-Moll Model for the npn Transistor
5.5.3
Theare
Ebers-Moll
Model
npn Transistor
The full Ebers-Moll
equations
obtained by combining
Eqs.for
(5.20)the
to (5.23):
The full Ebers-Moll
Eqs.
!
" equations
#
$are obtained
! by"combining
#
$ (5.20) to (5.23):
vB E
v BC
i E = I E S exp
− 1! − "
α R IC S #exp $
− 1!
"
#
$
VT
VT
vB E
v BC
−(5.24)
1
! i E"= I E#S exp$ V ! − 1" − α R#IC S $exp V
T
T
vB E
v BC
(5.24)
i C = α F I E S exp
− 1! − "
IC S exp
− 1!
#
"
#
$
VT
V$T
vB E
v BC
i C = α F I E S exp
− 1 − IC S exp
−1
VT
VT
This model contains four parameters, I E S , IC S , α F , and α R . From the definitions of I E S and IC S ,
we can obtain theThis
important
relation
model auxillary
contains four
parameters, I E S , IC S , α F , and α R . From the definitions of I E S and IC S ,
we can obtain the important auxillary relation
α F I E S = α R IC S
(5.25)
α F I E S = α R IC S
(5.25)