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EE 230 saturation – 1
SaturationConsider the circuit below. We have seen this one already. As before, assume that the BJT is on and in forward active operation.
3 V
10 V
RBVBB
VCC
ib
ic
–
+vce
–+vbe
RC10 kΩ
47 kΩ
βF = 100
LE =9%% � YEH
5%=�9� �.�9��Nż = ��µ$
LF = �)LE = (���) (��µ$) = �.� P$
!!
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 reverse-biased. The collector current that we calculate is much too big.
Y&( = 9&& � L&5&
= �� 9� (�.� P$) (�� Nۙ) = ��� 9
EE 230 saturation – 2
n p
+–
vBE
holes
emitter base
n
collector
+ – vBC
holes
electrons
electrons
SaturationWhen 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 emitter.
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 – 3
n p
+–
vBE
holes
emitter base
n
collector
+ – vBC
holes
electrons
electronsiE = ip1 + in1 – in2 iC = –ip2 + in1 – in2
iB = ip1+ip2
1 2
21
in1 = ISN1�exp
�vBEkT/q
�� 1
�� ISN1 exp
�vBEkT/q
�electrons injected emitter to base:
ip1 = ISP1�exp
�vBEkT/q
�� 1
�� ISP1 exp
�vBEkT/q
�holes injected base to emitter:
ip2 = ISP1�exp
�vBCkT/q
�� 1
�� ISP2 exp
�vBCkT/q
�holes injected base to collector:
in2 = ISN1�exp
�vBCkT/q
�� 1
�� ISN2 exp
�vBCkT/q
�electrons injected collector to base:
EE 230 saturation – 4
n p
+–
vBE
holes
emitter base
n
collector
+ – vBC
holes
electrons
electronsiE = ip1 + in1 – in2 iC = –ip2 + in1 – in2
iB = ip1+ip2
1 2
21
iB = ISP1 exp
�vBEkT/q
�+ ISP2 exp
�vBCkT/q
�
iC = ISN1 exp
�vBEkT/q
�� ISN2 exp
�vBCkT/q
�� ISP2 exp
�vBCkT/q
�
iE = ISN1 exp
�vBEkT/q
�� ISN2 exp
�vBCkT/q
�+ ISP1 exp
�vBEkT/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 – 5
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
�vBEkT/q
��1� exp
�� vCE
kT/q
��� ISP2 exp
�vBCkT/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:
iB = ISP1 exp
�vBEkT/q
�Better, but still not easy.
iC = ISN1 exp
�vBEkT/q
��1� exp
�� vCE
kT/q
��= βFib
�1� exp
�� vCE
kT/q
��
EE 230 saturation – 6
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.
iB = 40 µA (vBE = 0.6477 V)iB
iC
–
+vce
–+vbe
forward active
saturation For the transistor:
ISN1 = 5x10–14 A,
ISP1 = 5x10–16 A,
ISP2 = 1x10–14 A,
EE 230 saturation – 7
iC - vCE family of curves for npn transistor.
EE 230 saturation – 8
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.)
0
20
40
60
80
100
120
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4
I C/I B
VCE
(V)
IC/I
B = β
F
forward active
saturation
Saturation only occurs when vCE < 0.3 V.
A simple approximation then is to set vCE ≈ 0.2 V in saturation.
The confirmation that the BJT is in saturation is that iC/iB < βF.
EE 230 saturation – 9
forward active saturation off
VBE ≈ 0.7 V VBE ≈ 0.7 V IC = IB = IE = 0
IC = βF IB VCE ≈ 0.2 V checkVBE < 0
IE = IB + IC IE = IB +ICcheck
VBC < 0
check:VCE > 0.2 V
check IC < βF IB.
Summary of npn BJT operation.
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 – 10
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. 3 V
10 V
RBVBB
VCC
ib
ic
–
+vce
–+vbe
RC10 kΩ
47 kΩ
βF = 100
Using the approximation for saturation: vCE = 0.2 V. Then,
Because of the bad initial guess, we know that the BJT must be in saturation. But we can still apply the confirmation check:
iC =VCC � vCE
RC=10 V� 0.2 V
10 kΩ = 0.98 mA
iCiB
=0.98 mA0.049 mA = 20 < βF
The calculation is quite simple. The approximation for saturation helps us avoid a pile of computational messiness.
iB =VBB � vBE
RB=3 V� 0.7 V47 kΩ = 49 μA
