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. From eqn.(2) we have (i) When VcE = 0, Ic = Vcc/Rc (ii) 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. 2
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. 3
(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 ---------(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 --------- (1) { Since VBE is small & β+1= β } The collector current is Ic = Ic = ----------- (2) 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 Differentiating w.r.t. Ic, we get ---------(1) 4
As Ic = β IB + (1+ β) IcO Differentiating w.r.t. Ic we get 1= β 1= β.{ } Putting the value of 1= - 1+ from eqn.-(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 ---------- (1){Since VBE is small} 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 (1+β) IB Vcc = IBRB+VBE+(1+β) IB RE 5
Since VBE is very small & 1+ β = β ---------(1) The collector current is given by, -------(2) Ic = β 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 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+ = S = + ).{ = } -----(2) from eqn..(1) 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 ): 6
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 -------------(2) Applying KVL to the input section, VTH = IB RTH + VBE + IERE --------(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 ------------(4) { Sine VBE is small &(1+ β) = β } The collector current is given by, Ic = 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 Differentiating w.r.t. Ic, we get = 0 0 - = - ------------(1) Since Ic = β IB + (1 + β) ICO Differentiating w.r.t. Ic, we get 1 = 1 = 7
Substituting the value of 1 = 1 + S = from eqn.(1). ----------(2) 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 ---------(1) 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 & Ic / = IE /. RB +VBE + IE RE =VEE IE = --------- (2) If we want the operating point to be independent of RE >> IE = ------------ (3) {Since VBE is small} we should have 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 8
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 = (- ) + S = from eqn.(5) -------(6) 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 - 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 2) 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 9
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. 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 RB = = = 40k Ώ. ii) Stability factor = S = 1+ β = 1+ 100 =101 iii) Voltage between collector & ground (emitter) = VCE. Since Ic = β 100*0.3*10-3 A = 30*10-3 A 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*10 3 Ώ RB = 100k Ώ = 100*10 3 Ώ, β dc= 100, VBE = 0.7v We know that, 1)Ic = =. /β / =0.845*10-3 A = 0.845mA 2) Collector to emitter voltage, VEC = Vcc Ic.RC = 10 [(0.845*10-3 ) *(10*10 3 )] = 10-845 = 1.55v 3) D.C. load line- Ic (sat) = = = 1*10-3 A = 1mA 10
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) =. β. ( ) (. ) = 0.0118*10-3 = 0.0118mA (2) IC = β. 90*0.0118 = 1.06mA (3) VCE = VCC IC.RC = 3 (1.06*10-3 ) * (1.8*10 3 ) = 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) = β. ( ) [ ] = 0.04*10-3 A = 0.04mA (2) IC = β. 100*0.04 = 4mA (3) VCE = VCC (RC+RE) Ic = 20 [(2*10 3 ) + (1*10 3 )] 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*10-3 A = 9.9mA 2) VCE = VCC IC.RC = 25 - (9.9*10-3 ) *820Ώ = 25 8.12 = 16.88v (3) S = ( β) (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 VTH = Vcc = ( )* 10 = 3.33v RTH = R1I I R2 = [ β ] = =.. = = 3.33*10 3 Ώ = 3333Ώ. = 4.93*10-5 A = 0.0493mA Ic = β*ib 100 * 4.93*10-5 A = 4.93mA (2) Collector to-emitter voltage: - VCE = Vcc (RC+RE) Ic = 10 [(1*10 3 + 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) Since IE = =. = 1.86*10-3 A = 1.86mA ii) Since Ic = IE = 1.86mA iii) Since VCE = Vcc Ic Rc = 10-1.86*10-3 *1*10 3 = 8.14v 13