Saturation
Consider the circuit below. We have seen this one already. As before,
assume that the BJT is on and in forward active operation.
VCC 10 V
VBB
ib
RC
10 k!
RB
3V
47 k!
+
vbe
–
ic
+
vce βF = 100
–
=
=
.
=
=(
)(
=
µ
µ )= .
=
=
( .
)(
!!
)=
Red alert! Red Alert! Something is seriously wrong here.
VCE is extremely negative, which is not at all compatible with the B-C
junction being reverse-biased. Furthermore, there is no way that we
could get 39 V across any two points in the circuit when there is only
one 10-V source and one 3-V source.
The problem lies in the assumption that the base-collector is reversebiased. The collector current that we calculate is much too big.
EE 230
saturation – 1
Saturation
When both junctions are forward-biased, the transistor is said to be in
saturation. The B-E junction is still forward-biased, and electrons and
holes are still injected across it. The electrons will still cross the base
and travel into the collector, as we saw in the forward-active case.
But now, the base-collector is also forward-biased. Holes are being
injected from base to collector. Also electrons are injected from
collector to base. These electrons will cross the base and travel into the
vBE
emitter.
vBC
–
+
+
–
n
holes
p
n
holes
electrons
electrons
emitter
base
collector
In saturation, there are lots of carriers flowing across the two junctions.
There is more hole current (i.e. more base current). The two electron
currents flow in opposite directions and tend to cancel each other.
EE 230
saturation – 2
vBE
vBC
–
+
+
–
iB = ip1+ip2
n
iE = ip1 + in1 – in2
1
holes
p
holes
2
electrons
1
base
collector
electrons injected emitter to base: in1 = ISN1 exp
vBE
kT /
q
holes injected base to emitter: ip1 = ISP1 exp
vBE
kT /
q
electrons injected collector to base: in2 = ISN1 exp
vBC
kT /
q
holes injected base to collector: ip2 = ISP1 exp
vBC
kT /
q
EE 230
iC = –ip2 + in1 – in2
2
electrons
emitter
n
1
1
1
1
ISN1 exp
vBE
kT /
q
ISP1 exp
vBE
kT /
q
ISN2 exp
vBC
kT /
q
ISP2 exp
vBC
kT /
q
saturation – 3
vBE
vBC
–
+
+
–
iB = ip1+ip2
n
iE = ip1 + in1 – in2
1
holes
p
holes
2
electrons
iB = ISP1 exp
vBE
kT /
q
iC = ISN1 exp
vBE
kT /
q
iE = ISN1 exp
vBE
kT /
q
2
electrons
1
emitter
n
base
iC = –ip2 + in1 – in2
collector
+ ISP2 exp
vBC
kT /
q
ISN2 exp
vBC
kT /
q
ISN2 exp
vBC
kT /
q
ISP2 exp
vBC
kT /
q
+ ISP1 exp
vBE
kT /
q
Wow - big mess. Note that iB has increased (extra holes injected out of
base) and iC has decreased (electron currents are partially canceling), so
iC < βF iB. We don’t even have that simplification in this case!
EE 230
saturation – 4
With some knowledge from a future class (EE 332) and an approximation
or two, we can simplify the equations slightly.
First, ICBS (it can be shown – and will be in EE 332) that for all BJTs, ISN1
= ISN2. Secondly, we can always express the base-collector voltage in
terms of the base-emitter and collector-emitter voltages: vBC = vBE – vCE.
Using these two bit of info, the collector current equation can be written:
iC = ISN1 exp
vBE
kT /
q
1
exp
vCE
kT /
q
ISP2 exp
vBC
kT /
q
Thirdly, we might make the assumption that the hole current injected from
base to collector (represented by the terms with the ISP2 coefficient) is
small in the base and collector current equations. (This approximation
may be “iffy” since ISP2 >> ISP1 for most BJTs. However, let’s go with it for
now.) With this, the base and collector currents can be written:
EE 230
iB = ISP1 exp
vBE
kT /
q
iC = ISN1 exp
vBE
kT /
q
Better, but still not easy.
1
exp
vCE
kT /
q
= β F ib 1
exp
vCE
kT /
q
saturation – 5
A standard way of presenting a BJT i-v relationship is to fix the base current
at some value (same as fixing VBE) and then measuring (and plotting) the
collector current while varying the collector-emitter voltage, VCE.
iC
forward active
saturation
iB
iB = 40 µA (vBE = 0.6477 V)
+
vbe –
+
vce
–
For the transistor:
ISN1 = 5x10–14 A,
ISP1 = 5x10–16 A,
ISP2 = 1x10–14 A,
EE 230
saturation – 6
iC - vCE family of curves for npn transistor.
EE 230
saturation – 7
The equations for saturation look very complicated. How do we cope?
As always, we make a simplifying approximation. Look more closely at
the saturation region, plotting iC/iB vs. vCE. (In this way, we get the same
curve for all values of B-E bias.)
Saturation only occurs when
vCE < 0.3 V.
120
I /I = β
C B
F
100
forward active
C
I /I
B
80
60
A simple approximation then
is to set vCE ≈ 0.2 V in
saturation.
saturation
40
The confirmation that the BJT is
in saturation is that iC/iB < βF.
20
0
0
0.05
0.1
0.15
0.2
V
EE 230
CE
0.25
0.3
0.35
0.4
(V)
saturation – 8
Summary of npn BJT operation.
forward active
saturation
off
VBE ≈ 0.7 V
VBE ≈ 0.7 V
IC = IB = IE = 0
VCE ≈ 0.2 V
check
VBE < 0
IE = IB + IC
IE = IB +IC
check
VBC < 0
check:
VCE > 0.2 V
check
IC < βF IB.
IC = βF IB
Note that we have not mentioned the case where the base-emitter is
reverse biased and the base-collector is forward biased. This is called
reverse active. Generally, BJTs are not operated in reverse-active mode.
If your transistor ends up in RA mode, it is probably because you made a
mistake.
EE 230
saturation – 9
Re-consider the circuit from the first
slide. We know that it is not in forward
active, so it must be in saturation. Start
by calculating ib – the calculation is
unchanged, because our assumptions
about the base-emitter junction are the
same for saturation as for forward active.
VCC 10 V
VBB
ib
RC
10 k!
RB
3V
47 k!
+
vbe
–
ic
+
vce βF = 100
–
VBB vBE
3 V 0.7 V
iB =
=
= 49 μA
RB
47 kΩ
Using the approximation for saturation: vCE = 0.2 V. Then,
VCC vCE
10 V 0.2 V
iC =
=
= 0.98 mA
RC
10 kΩ
Because of the bad initial guess, we know that the BJT must be in
saturation. But we can still apply the confirmation check:
iC
0.98 mA
=
= 20 < βF
iB
0.049 mA
The calculation is quite simple. The approximation for saturation helps
us avoid a pile of computational messiness.
EE 230
saturation – 10