Bipolar Junction Transistors (BJT)
Symbols
Simplified structure
Input, transfer, and output characteristics
Operating principle
Operating regions
Currents through BJT
BJT saturation
npn and pnp BJT
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npn
pnp
Symbols
An ohmmeter’s
view of transistor
terminals
There are
interactions
between the 2
junctions.
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The terminals of a BJT are labelled:
B – base (similar to G of MOSFET);
C – collector (similar to D of MOSFET);
E – emitter (similar to S of MOSFET).
The arrow on the emitter terminal points in the
direction of the positive current
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Simplified
structure of
a npn BJT
The transistor effect consists in a current flowing through
a reverse biased junction (base-collector) due to its interaction with a
forward biased junction (base-emitter) placed in its close vicinity.
For the transistor effect
• very thin base region; considerably thinner than the diffusion length
of the minority carriers in the base region;
• the emitter region more heavily doped than the base region;
• the emitter and collector regions wider than the diffusion length of
the minority carriers in the these regions.
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Terminal
characteristics
of npn BJT
iB =
IS
β
e
v BE
VT
iC =βiB
Valid in active region
Input characteristic
iC = I S e
Transfer characteristic
v BE
VT
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Output characteristics, npn BJT
IS=2·10-15A and β=100
Active region:
iC=βiB
Saturation region:
iC <βiB
VCEsat≈0,2V
Off region:
iC=iB =0
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Operating regions of npn BJT
(off)
vBE<0.6V ; vBC<0.6V
(aF)
vBE>0.6V; vBC<0.6V
(exc)
vBE>0.6V; vBC>0.6V
(aR)
rarely used
Observation: the transistor shouldn’t be biased very close
to the origin of the axes or to one axis
switching mode (off)
Ways to use
(exc)
as an amplifier (aF), eventually (aR)
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The currents
through BJT
iE =iC+iB
Always valid
In the active region (aF)
iC=βiB
iE = iC +
1
β
iC = iC (1 +
1
β
)
iE=(β+1)iB ≈βiB
This relations does not hold in the saturation region(exc)
where
iC <βiB
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Limiting the control current for a BJT
difference BJT –MOSFET : junction in the control circuit B-E.
one have to use a series resistance in order to establish
(limit) the base current.
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BJT Saturation
The values of the resistors and voltages should be chosen
such that the BJT would operate in the desired region
BJT can be seen also as a current-controled current
source iC=βiB for the active region (aF).
i Bsat =
iCex
β
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Exemplification
Operating regions, RB=50K
i) vCo=0,4V;
ii) vCo=1,7V;
iii) vCo=5V
vCo=2,7V; β∈(25…200)
RB domain so that T is in:
i) (aF);
ii) (exc).
i) because vCo=0,4V < VTh=0,6V T-(off)
ii) vCo >VTh ⇒ T in (aF) or in (exc).
Consider vBE=0,7V for conduction and β=100.
Assume T in (aF) so that iC =β·iB.
We should compare iB with iCex /β.
If iB >iCex /β, T - (exc), if iB <iCex /β, T - (aF)
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VPS − vCEsat 12 − 0,2
=
= 5,9mA
iCex =
2
RC
vCo − v BE 1,7 − 0,7
iB =
=
= 0,02 mA
RB
50
iCex
5,9
=
= 0 ,059 mA
β 100
Because iB=20µA <
iCex
β
= 59 µA, ⇒ T is in (aF);
iC = βiB = 100⋅ 0,02 = 2mA
OP
vCE = VAl − RCiC = 12 − 2 ⋅ 2 = 8V
vBC = vBE − vCE = 0,7 − 8 = −7,3V < 0,6V
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iii)
vCo − vBE 5 − 0,7
iB =
=
= 0,086mA
RB
50
Because iB =86µA>iCex /β=59mA, results that T is in (exc);
vBE ≈0,8V;
vBC=vBEsat -vCEsat ≈0,8V-0,2V=0,6V=VTh
Alternatively we can solve supposing T in (aF) ⇒ iC=β·iB=8,6mA
vCE=VPS - RC·iC =12-2·8,6=-5,2V
Obviously an impossible value (vCE can be only positive) so our
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supposition is false. Thus T is in (exc)
b) i) For T in (aF) we must be sure
that iB<iCex /β regardless the β
value in the specified range; the
worst case: β=βmax= 200.
iB <
vCo − v BE i Cex
<
β max
RB
iCex
β max
vCo − vBE
2,7 − 0,7
RB > βmax ⋅
= 200⋅
= 67,8KΩ
iCex
5,9
ii) For the saturation the following condition must be fullfield:
iB >
iCex
β min
vCo − v BEsat
iCex
>
RB
β min
vCo − vBEsat
2,7 − 0,8
= 75 ⋅
= 24KΩ
RB > β min ⋅
5,9
iCex
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