Download 7 Transistor Biasing and Thermal Stabilization

UNIT :7
Transistor Biasing and Thermal Stabilization: - 9
The operating point, The DC and AC load line, Need of transistor biasing and stability of Q point, Thermal
instability. Method of Biasing: - Fixed bias, Collector to base bias, Self bias or Emitter bias, potential Divider
bias, Stabilization factors Definition of stability factors S. temperature compensation using single diode and two
diodes, Transistor Rating and specifications for typical transistor SL 100 and BC 148 or BC 548. 10
The operating point & DC load Line: Consider a CE amplifier circuit without any a.c. input signal. This
condition is called ‘quiescent condition’. The battery VCC sends current
Ic through the load resistance RC & the transistor.
Applying kirchhoff’s voltage law to the collector circuit ,we get
VCC = Ic Rc + VCE --------(1)
Rearranging and solving for Ic,
---------(2)
This eqn. is similar to y = mx +c which is the equation of straight line.
Thus plotting eqn.(2) on the transistor output characteristic we get a
straight line, whose slope is ( & intercept on the Ic axis is (Vcc / Rc). The slop of this line depends on the
d.c. load resistance Rc hence this line is called as d.c. load line.
(i)
(ii)
From eqn.(2) we have
When VcE = 0, Ic = Vcc/Rc
When Ic = 0, VcE = Vcc
Joining these two points gives the dc load line as shown in graph.
The dc load line intersects the output curves. The point of intersection of
load line for specified base current IB is called as ‘quiescent point’ or
‘operating point’ or ‘Q-point’. The exact location of Q-point is decided by
Vcc,Rc,RB,VBE and VBB. For given values of Vcc & Rc, the Q-point
depends upon the base current IB, which is calculated as follows –
Applying Kirchhoff’s voltage law to the base circuit, we get,
VBB = IB. RB + VBE
Since VBE is small (VBE is 0.7 v for Si & 0.3 v for Ge)
Analysis of Amplifier using d.c. load line :Consider a CE amplifier circuit as shown in fig.(a). A transistor can amplify ac signals only after its dc
operating point is suitably fixed. Generally the operating point is selected at the centre of the load line.
Under quiescent condition, the base current has a constant dc value. It is determined from Q-point.
Now when a.c. input signal is applied, the base current varies as per the input signal. As base current varies, the
operating point moves along the dc load line ( Q1-Q2).
Thus the instantaneous values of collector current & voltage also vary according to input signal as shown in the
graph. The variation in the collector voltage is many time larger than the variation of the input signal. The
collector voltage variation reaches the output terminal through capacitor Cc2. The output is therefore many time
larger than the input.
The current gain, voltage gain & power gain of amplifier is given by,
1) Current gain, Ai =
2) Voltage gain, Av=
1
3) Power gain ,Ap =
CE amplifier has large current gain, voltage gain and power gain.
A.C. load line :-
Consider CE amplifier circuit with external load RL as shown in fig. For ac signals the coupling capacitor Cc2
acts as short circuit. Hence the external load resistance RL comes in parallel with the d.c. load resistance Rc.
Thus the effective value of load resistsance for ac signal is R’L = Rc II RL. The R’L is always less than Rc. Thus
as far as a.c. signals are concerned the load line has the slope
, which is greater than the slope of
d.c.load line. A line passing through the same operating point Q and having slope
is called as a.c.
load line as shown in graph.The a.c. load line is steeper than d.c.load line.
Selection of the operating point :
Fig.(1) shows the operating point near the saturation region. In this case even though the base current
varies sinusoidally, the output current is clipped at the positive peak. This results in distortion of the signal. Thus
operating point near the saturation region is not suitable.
Fig.(2) shows the operating point near the cut-off region. In this case the output signal is now clipped at
negative peak, resulting in a distortion. Thus operating point near the cut-off region is not suitable.
Fig.(3) shows the operating point at the centre of load line. In this case the output signal is not clipped
even though the input signal is sufficiently large. Thus the centre of the load line is most suitable position for the
operating point.
Need for BIAS STABILIZATION: After fixing the operating point suitably, it should remains there only. But there are two reasons for the
operating point to shift.
1) The transistor parameters such as VBE & β changes from device to device.
2) Transistor parameters are also temp. dependent.
Since the collector current is
Ic = β IB + (1+ β) Ico
Here β, IB and Ico are temp. dependent.
Fig.1 Fig.2
β of a transistor is strongly dependent on temperature. As temp. increases, β increases as shown in fig.1.
The junction voltage VBE is also a function of temp. As temp. increases, VBE decreases. But IB increases
as shown in fig.2.
The Ico is also a strong function of temp. Ico doubles for every 100c rise in temp.
The net result of these increase in β, IB and Ico is that, Ic increases, shifting the operating point in the
upward direction. This causes distortion in the output i.e. Operating point is not stable against temperature
variations.
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Thermal Runway: - As temp.increases, the leakage current Ico increases. This increases the collector current.
The power dissipated at the collector junction also increases. This further rises the temperature of the collector
junction. This increases the leakage current & thus in turn increases the Ic. Thus the whole cycle repeats again.
Such cumulative increase in Ic will ultimately shift the operating point into the saturation region. This situation
is called ‘Thermal Runway.’
Stability factors:The extent to which the stabilization of IC is successful is measured by stability factor.As Ic is a function of
three variables Ico,VBE, & β , there are three stability factors.
1)Stability factor(s) :- It is defined as the ratio of change of collector current IC w.r.t. leakage current Ico,
holding β & IB constant.
…………. β & IB constant
2)Stability factor (s’) :- It is defined as the ratio of change of collector current Ic w.r.t. VBE keeping β & Ico
constant.
…………. β & Ico constant
3)stability factor(s’’) :- It is defined as the ratio of change of collector current Ic w.r.t β keeping Ico & VBE
constant.
…………….. Ico & VBE constant
For better stability the value of stability factor must be very small
Method of biasing:The requirements of biasing circuit are 1) It must establish the operating point at the centre of the load line.
2) It must stabilize the operating point against temp variations.
3) It must provide the operating point independent of β of transistor.
(1) Fixed Bias circuit:- Applying kirchhoff’s voltage law to the input
section, we get
Since VBE <<Vcc
------(1)
In this case the supply voltage Vcc is
fixed and once RB is selected, IB is also
fixed.
Hence the name ‘fixed bias circuit’.
Since Ic = β IB + ICE0
As ICE0 << Ic
Ic
=
β
IB
………..2
Applying kirchhoff’s voltage law to the
output section
VCC = IcRc + VCE
VCE = VCC -IcRc
………3
The voltage drop IcRc can never be more than Vcc. If it is more, the
operating point is in the saturation region & the Collector saturation
current is given by
Ic (sat) = Vcc/Rc …………4
Thus for transistor to be in the active region, the collector current must be
always less than Ic(sat).
Ic < VCC/Rc ----------(5)
Drawbacks: 1) With rise in temp., a cumulative action takes place and the collector current goes on
increasing. This leads to thermal runway.
Sine Ic = β IB + ( 1+ β) IcO
Differentiating w.r.t. IcO,
=0+(1+ β)
S =(1+ β)
The value of s is large indicating that circuit is less stable against temp variation.
(2) Since Ic = β IB & IB is already fixed, Therefore Ic is solely dependent on β.
When transistor is replaced by another transistor with different value of β, the operating point will shift.
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(2) Collector to Base bias circuit: In this circuit IB is reduced with increase in temp, which is achieved by obtaining the bias voltage from the
collector of the transistor instead of from collector supply as shown in fig.
Applying kirchhoff’s voltage law to input section,
Vcc = ( Ic + IB) Rc + IB.RB + VBE
Vcc = Ic. Rc + ( Rc + RB) IB + VBE
IB =
---------(1)
Applying kirchhoff’s voltage law to the output section,
Vcc = ( Ic + IB) Rc + VcE
VcE = Vcc - ( Ic + IB) Rc
VcE = Vcc - Ic Rc
--------(2){Since
}
Putting this value in eqn. (1) we get
IB= VcE-VBE/(Rc+RB)
---------(3)
If temp. increases then leakage current increases. This increases the collector current. But as collector current
increases voltage VcE decreases. This decreases the base current
The lowered base current in turn reduces
the original increase in collector current. Thus there exist a mechanism in the circuit because of which collector
current is not allowed to increase.i.e.the circuit stabilizes the operating point.
Here the input base current depends on the output collector voltage. That is there exist the feedback. Hence this
circuit is called as voltage feedback bias circuit.
The operating point can be determined as follows.
From input section,
Vcc = ( Ic + IB) Rc + IB. RB + VBE
= Ic Rc + (Rc + RB) IB + VBE
Since Ic = β IB
Vcc = β IB. Rc + ( Rc + RB) IB + VBE
= [β Rc+ Rc+ RB] IB + VBE
= [(β +1) Rc + RB] IB + VBE
IB =
--------- (1) { Since VBE is small & β+1= β }
IB =
The collector current is
Ic =
Ic =
Applying KVL to output section,
Vcc = ( Ic + IB) Rc+ VcE
VcE = Vcc-( Ic + IB) Rc
--------- (3) { Since IB is small }
VcE= Vcc - Ic Rc
Stability factor: Since Vcc = Ic Rc+( Rc+ RB) IB + VBE
IB =
Differentiating w.r.t. Ic, we get
---------(1)
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----------- (2)
As Ic = β IB + (1+ β) IcO
Differentiating w.r.t. Ic we get
1= β
1= β
…….{
Putting the value of
}
from eqn.-(1)
1= 1+
--------- (2)
S=
Here the stability factor is smaller than (1+ β). Hence there is improvement in the stability.
Drawback: (1) The resistor RB not only provides d.c. feedback for stabilization of operating point, but it also
causes an a.c. feedback. This a.c. feedback reduces the voltage gain of the amplifier.
(3) Bias circuit with emitter Resistor: This is a modified circuit of fixed bias. A resistor RE is connected in the emitter circuit as shown.
Applying KVL to input section,
Vcc = IB.RB + VBE + IERE
IB =
---------- (1){Since VBE is small}
IB =
As temp increases, the leakage current increases. This increases the collector current as well as the emitter
current. As a result the voltage drop IERE also increases. This reduces the numerator of eqn. (1) & hence the
current IB also reduces. This reduces the collector current. Thus there exist a feedback mechanism which does
not allow the collector current to increase as shown below.
Above circuit also provides the stabilization against β
variation. If the transistor is replaced by another transistor with
different value of β, then it stabilizes the operating point as shown
below.
The operating point can be determined as follows:
From the input section,
Vcc = IB RB + VBE + IE RE
But IE = IB + IC = IB+ β IB = (1+β) IB
Vcc = IBRB+VBE+(1+β) IB RE
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IB =
Since VBE is very small & 1+ β = β
IB =
---------(1)
The collector current is given by,
-------(2)
Ic = β IB =
Applying KVL to output section.
Vcc = IC RC + VCE + IE RE
Vcc = IC RC + VCE + IC RE
{IC=IE}
Vcc = IC (RC + RE) + VCE
VCE = Vcc-(RC + RE) Ic ---------(3)
Stability factor: Vcc = IB RB + VBE + IE RE
Vcc = IB RB + VBE + (IB + Ic) RE
Vcc = IB (RB + RE) + VBE + IC RE
IB =
Differentiating w.r.t. Ic we get
=
---------- (1)
Since Ic = β IB + (1+ β) ICO
Differentiating w.r.t. Ic, we get
+ (1 + β)
+
)……….{
Substituting the value of
1= 1+
=
}
from eqn..(1)
+
=
-----(2)
S=
Thus the stability factor is smaller than
. Hence there is an improvement in the stability factor.
Drawback of the circuit: - This Resistor RE provides negative feedback for a.c. signals. This reduces the
voltage gain of amplifier. This drawback can be overcome by putting a capacitor CE across the resistor RE. CE
acts as a short circuit for a.c. signal.
(4) Voltage Divider bias ( Potential Divider bias ):
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The name ‘voltage divider’ comes from the voltage divider formed by the resistors R1 and R2 as
shown in fig.(1). The resistor RE provides the negative feedback which stabilizes the operating point.
The Thevenin equivalent circuit is as shown in Fig.(2).
The Thevenin voltage
is given as,
---------(1)
The Thevenin resistance R
is given as R
Applying KVL to the input section,
VTH = IB RTH + VBE + IERE
-------------(2)
--------(3)
As the temp. increases, the leakage current increases. This increases the collector current as well as
emitter current. As s result the voltage drop IE RE also increases. This reduces the numerator of eqn. (3) &
hence the current IB also reduces. This reduces the collector current. Thus there exist a feedback mechanism
which does not allow the collector current to increase as shown below. This action stabilizes the operating point.
The operating point can be determined as follows: Applying KVL for input circuit,
VTH = IB RTH + VBE + IERE
= IB RTH + VBE + (1+ β) IB * RE …….. {IE (1+ β) IB}
= IB [RTH + (1+ β) RE] +VBE
IB =
IB =
------------(4) { Sine VBE is small &(1+ β) = β }
The collector current is given by,
Ic = IB =
Ic =
--------(5)
Applying KVL to output circuit,
Vcc = Ic. Rc + VcE + IE RE
VcE = Vcc – IcRc – IE RE
--------------(6) { Since Ic = IE }
VCE = Vcc – ( Rc + RE ) Ic
Stability factor:Since VTH =IB RTH + VBE + IERE
VTH = IB RTH + VBE + (IB + Ic ) RE
= IB (RTH + RE) + VBE + Ic RE
–
IB =
Differentiating w.r.t. Ic, we get
=0–0=Since Ic = β IB + (1 + β) ICO
Differentiating w.r.t. Ic, we get
1=
------------(1)
1=
7
Substituting the value of
from eqn.(1).
1=
1+
----------(2)
S=
Stability factor is less then (
Advantage of circuit: -Since
indicating improvement in the thermal stability.
, the collector current is almost indepenclent of . This stabilizes the
circuit agains vatiation.
(5)Emitter Bias circuit: The circuit gets this name because the negative supply VEE is used to forward-bias the emitter junction through
resistor RE. As usual, the Vcc supply reverse biases the collector junction.
Applying KVL to base-emitter loop,
-VEE = - IE RE – VBE –IB RB
VEE = IERE +VBE + IBRB
IB =
---------(1)
IB =
This circuit also gives stabilization to the operating point as shown.
The operating points can be determined as follows:
Applying KVL to base-emitter loop,
IB RB + VBE + IE RE =VEE
Since Ic ≈ IE & IB = Ic / = IE /
. RB +VBE + IE RE =VEE
IE =
--------- (2)
If we want the operating point to be independent of
RE >>
we should have
------------ (3) {Since VBE is small}
IE =
The above eqn. show that the emitter is virtually at ground
potential. All the VEE supply voltage appears across RE.
If the emitter is at ground point, the collector to emitter voltage
VCE is given by,
VCE = Vcc – IcRc
---------- (4)
Stability factor: Applying KVL to base-emitter loop,
IBRB + VBE + IE RE = VEE
Substituting IE = IB +IC,
IBRB + VBE + (IB + IC) RE =VEE
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IB =
Differentiating w.r.t. Ic,we get
=0-0=------------ (5)
Sine Ic = β IB + (1+β) ICO
Differentiating w.r.t. Ic
1=
1=
Substituting the value of
1 = (-
from eqn.(5)
)+
-------(6)
S=
Stability factor is less than ( 1+ ) indicating improvement in the thermal stability.
Advantage of the circuit: Since IE =
The collector current is independent of & VBE, therefore Q-point is not affected by variation in these
parameters.
Bias Compensation: - Signal diode compensation:A diode of the same material as that of transistor is placed across the emitter base junction of the
transistor. The diode is reverse biased by the base emitter junction voltage. It allows the reverse saturation
current to flow through it .When the temp. increases, the reverse saturation current ICO through the transistor
also increases. This in turn increases the reversers saturation current Io (leakage current) through the diode,
which will reduce the base current IB by keeping the current I constant. This action keeps constant the value of
collector current.
(1) Compensation using Two diodes: Here the voltage across the compensating diodes produces
the bias voltage for the emitter diodes of transistors. When
temp. increases, the compensating diodes produces less
voltage, which in turn reduces the base current & hence the
collector current.
Transistor Ratings and specifications :
1) SL-100 / CL-100
- NPN
- General purpose medium power transistor
- Package- TO-39
2)
Maximum collector current = 0.5 A
DC current gain = HFE = 25 to 300
VCE = 50 V, VCB = 60 V, VBE = 5 V
Power dissipation = 800 mW
BC-148 or BC 548
-NPN
-General purpose small signal amplifier
-Package- TO-92
-Maximum collector current = 100 mA
-DC current gain = HFE = 110 to 800
-VCE = 30 V, VCB = 30 V, VBE = 0.6 V
-Power dissipation = 500 mW
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Ex. (1) Fig. shows a fixed bias circuit using NPN transistor. Determine the values of base current, collector
current and collector to emitter voltage
Given:- Vcc = 25v, RC = 820 Ώ, RB = 180k Ώ β = 80
i)base current, IB
=
= 0.14mA
∗
ii) Collector current, Ic = β.IB = 80*0.14mA = 11.2mA
iii) Collector to emitter voltage,
VcE = Vcc - Ic.Rc = 25 – (11.2*10-3 ) * 820 Ώ = 25 – 9.2 = 15.8v
Ex.(2) Determine the value of bias resistor RB & the stability factor for the fixed bias circuit. Also determine
the voltage between collector and ground.
Given:- Vcc = 12v, RC = 330 Ώ, β = 100,
IB 0.3mA = 0.3*10-3 A
i)Value of bias resistor RB,
Since IB =
RB =
= .∗
= 40k Ώ
ii) Stability factor = S = 1+ β = 1+ 100 =101
iii) Voltage between collector & ground (emitter) = VCE.
Since Ic = β IB = 100*0.3*10-3A = 30*10-3A
VCE = VCC-IC.RC = 12 – (30*10-3) * 330 Ώ = 2.1v
Ex :(3) Calculate the Q. point values of collector current & collector to emitter voltage for the d.c. bias circuit
shown in fig. Also draw the load line and locate Q. point on it. Assume VBE = 0.7v .
Given: - Vcc = 10v, RC = 10k Ώ = 10*103 Ώ
RB = 100k Ώ = 100*103 Ώ, β dc= 100, VBE = 0.7v
We know that,
.
1)Ic =
= ∗
=0.845*10-3A = 0.845mA
/β
∗
/
2) Collector to –emitter voltage,
VEC = Vcc –Ic.RC = 10 –[(0.845*10-3) *(10*103)] = 10 - 845 = 1.55v
3) D.C. load line- Ic (sat) =
=
= 1*10-3A = 1mA
∗
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Ex: (4) Determine the d.c. bias current and voltage for the circuit shown. Also determine the stability factor .
Assume VBE = 0.7
Given Vcc = 3v, RC =1.8k Ώ = 1.8*10-3 Ώ
RB = 33k Ώ =33*103 Ώ, β = 90, VBE = 0.7v
.
(1) IB =
= (
)
∗
β.
( . ∗
)
= 0.0118*10-3= 0.0118mA
(2) IC = β. IB = 90*0.0118 = 1.06mA
(3) VCE = VCC – IC.RC = 3 – (1.06*10-3) * (1.8*103)
= 3 -1.9 = 1.1v
( β)
(4) S =
= 16.1
(β
/
)
Ex: (5) calculate the operating point for given circuit. Neglect VBE of Transistor.
Soln. (1) IB =
β.
=
(
∗
) [
∗ ∗
]
= 0.04*10-3A = 0.04mA
(2) IC = β. IB = 100*0.04 = 4mA
(3) VCE = VCC – (RC+RE) Ic
= 20 – [(2*103) + (1*103)] 4*10-3
= 20-12 = 8v
Ex: (6) Determine the value of collector current, collector-to-emitter voltage & stability factor. Assume VBE =
0.7v
11
1) Ic =
.
=
/β
∗
/
= 9.9*10-3A = 9.9mA
2) VCE = VCC – IC.RC
= 25 - (9.9*10-3) *820Ώ
= 25 – 8.12 = 16.88v
(
β)
(3) S =
=
(β
/
)
=
.
= 74.44
.
Ex: (7) Determine the value of collector current & collector-to-emitter voltage for voltage Divider circuit
shown.
Assume VBE = 0.7v and β = 100.
(1) Collector current: -Since
RTH = R1I I R2 =
IB =
[
β
]
=
∗ Vcc
VTH =
=
.
.
-5
∗
∗
∗ ∗
∗
∗ ∗
=
∗
=(
∗
∗
)* 10 = 3.33v
= 3.33*103Ώ = 3333Ώ
.
= 4.93*10-5A = 0.0493mA
Ic = β*IB – 100 * 4.93*10 A = 4.93mA
(2) Collector –to-emitter voltage: VCE = Vcc – (RC+RE) Ic
= 10 – [(1*103 + 500)]*4.93*10-3
= 10 – [1000 + 500]*4.93*10-3
= 10 -7.395 = 2.605v
Ex: (8) Find the values of emitter current, collector & collector-to-emitter voltage for the emitter bias circuit.
Assume VBE = 0.7v
12
i)
ii)
iii)
.
=
∗
= 1.86*10-3A = 1.86mA
Since Ic = IE = 1.86mA
Since VCE = Vcc – Ic Rc = 10-1.86*10-3*1*103
Since IE =
= 8.14v
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