Monday 30 April 2018

VARIATIONS IN THE DESIGN OF RC-COUPLED CE AMPLIFIER


EDITOR: B. SOMANATHAN NAIR

NOTE: This article is an extension of the previous blog “RC­-COUPLED  COMMON-EMITTER AMPLIFIER–A SYSTEMATIC APPROACH TO DESIGN”.


1. TO FIX THE Q-POINT EXACTLY IN THE MIDDLE OF THE ACTIVE REGION

The design explained in the previous blog on RC-coupled amplifier design fixes the Q-point only approximately. Because of the large variations existing in the characteristics of transistors, data manuals cannot give us the complete range of the active region accurately. For determining this, at first, we have to plot the characteristics of the transistor experimentally. This is a tedious and time-consuming job. So, normally, this is not resorted to. Instead, we use a potentiometer in series with the R1/R2 biasing network, as shown in Fig. 1 for adjusting VBEQ accurately. The choice of the potentiometer to be added in series with R1 and R2 is based on the criterion that

                                                            Rx + Ry + Rz = R1 + R2             (1)

where Rx and Rz are fixed resistors, and Ry is a potentiometer. In our design problem, R1 + R2 = 80 kΩ.  Therefore, we want

                                                            Rx + Ry+ Rz = 80 kΩ               (2)

Since we are not sure about the actual position of the Q-point, we may choose Rx = Rz = 4.7 kΩ each, and the potentiometer Ry = 68 kΩ. Note that a potentiometer of large value is chosen purposefully. This is to accommodate for the wide range of variation required in fixing the Q-point in the middle of the active region. The fixed resistors R1 and R2 are used to avoid the shorting that may take place due to the wiper arm of the potentiometer accidentally touching its lower or upper terminal if these resistors are not used.
Now, by carefully adjusting the pot, we look for equal voltages across VCE and VRC. It can be seen that when the Q-point is exactly in the middle of the active region, VRC = VCE. In our design, this is exactly 4.5 V. It can now be seen that if VRC = VCE = 4.5 V, then VRE will be 1 V, and IC will be 1 mA exactly, proving the validity of our design. In all practical electronic circuits, precise adjustments of this kind using pots are required to fix operating points. These finer adjustments are protected against tampering by applying wax or lacquer that prevents the wiper arm of the potentiometer from physically moving.




2.  DESIGN FOR DIFFERENT CURRENT SWINGS

The Standard Amplifier design given above was based on a current swing of 1 mA and a voltage swing of 4.5 V. In practice, these requirements may be suitably modified to get the design for any current and voltage swings. Table 1 gives the values of RC, RE, R1 and R2 for various current swings and the same voltage swing of 4.5 V. The values in respect of the Standard Amplifier are shown using bold letters and numbers.

                                                                      Table 1
IC
RC Ω

RE Ω

R1 Ω

R2 Ω


Current-multiplying factor

Resistance-multiplying factor

Wattage rating
1 mA
4.7 M
1 M
12 M
68 M
0.0001
1000
1/8 W
10 mA
470 k
100 k
1.2 M
6.8 M
0.01
100
1/8 W
100 mA
47 k
10 k
120 k
680 k
0.1
100
1/8 W
1 mA
4.7 k
1 k
12 k
68 k
1
1
1/8 W
10 mA
470
100
1.2 k
6.8 k
10
0.1
½ W
100 mA
47
10
120
680
100
0.01
10 W
1 A
4.7
1
12
68
1000
0.001
40 W


From Table 1, when the current IC is multiplied by a factor of 10, a division using the same factor gives the corresponding resistor. But, if the current is divided by a factor of 10, then the corresponding resistor will be obtained through a multiplication by 10.
In fact, if IC gets multiplied or divided by a certain constant a, then the resistors get divided or multiplied by the same constant a, provided bmin, VOP, and VRE remain constant. We now make the following observations from Table 1.

·         Wattage rating of resistors for values of collector current less than 1 mA is very low. Usually, in all such cases, we use (1/8)-W resistors, as they are freely available in the market. However, the picture changes quickly as IC increases above 1 mA. For high currents, the power rating becomes very high.

·         When IC = 1 A, we find that the emitter resistor RE = 1 Ω only. However its power rating is about 4 W. Usually it is very difficult to get low-value resistors, such as 1 Ω. Also, it can be seen that a lot of power is wasted in this resistor. Therefore RE is usually avoided in power amplifiers. When IC = 1 A, we have R1 = 12 ohms. This is a very low value, and hence, will load the input heavily. Therefore, R1 is usually replaced with diode biasing across the base and emitter.

·         Amplifiers with as low a collector current as 1 mA have been designed and tested by the author. There is a little amount of practical difficulty in constructing and testing an amplifier with collector current less than 1 mA. One of the major problems encountered in this case is the non-availability of resistors of values greater than 10 MΩ in the market. Another problem is regarding the stability of the amplifier at very low collector currents. At very low values of IC, collector-to-base leakage current ICO may become prominent and comparable with IC.  In such cases, special care should be taken to ensure the stability of the amplifier.

Caution: Usually, for power transistors, b is very small (typically, less than 20). Therefore, the values of resistors given in Table 5.1 for IC = 1 A are only of theoretical interest. Practically, since the power requirement is very high, it is quite common to use the push-pull configuration for power amplifiers. In such situations, we normally avoid using high-wattage resistors as they unnecessarily waste power.





Sunday 29 April 2018

HYSTERESIS A SIMPLE EXPLANATION


EDITOR: B. SOMANATHAN NAIR

The word hysteresis has been derived from Greek and means lagging behind. The term was coined by Ewing around 1881 to describe the relation between magnetic flux density B and magnetic field intensity H. It was found that when H is zero, B is not zero. Similarly, when B is zero, H is not zero. Thus one lags behind the other. The B-H curve obtained by plotting the corresponding values of B against values of H is called as the magnetic hysteresis loop. Figure 1a shows a typical B-H curve (magnetic hysteresis loop).
       


        Even though hysteresis was described as a phenomenon seen in magnetic domain, it has been found that it exists in many areas of engineering and science. For example, we find hysteresis in Schmitt trigger circuits (Fig. 1b). A Schmitt trigger circuit gets triggered when its input voltage Vin is greater than or equal to its upper trigger (trip) point (UTP). Then the output voltage will be +V volts. The circuit will switch back to ‒V, when the input Vin is less than UTP. This means that UTP will always be greater than LTP and they can never be equal. Hence Schmitt trigger circuit will always exhibit hysteresis.




To illustrate this further, consider a semiconductor silicon diode being forward biased. The diode will start conducting when its anode-cathode voltage is typically 0.5 V (Fig. 2). We find that as long as the anode voltage is greater than or equal to 0.5 V, the diode is in its ON state. It can be seen that it can be turned off only when its anode voltage is less than 0.5 V (Fig. 3) because at 0.5 V, it is still in the ON state. It further means that the ON and OFF voltages can never be the same and are different. This gives rise to hysteresis. 
     


    










RC-COUPLED COMMON-EMITTER AMPLIFIER - A SYSTEMATIC APPROACH TO DESIGN


EDITOR: B. SOMANATHAN NAIR

In electronic circuit theory books, electronic circuit designs are not given in a proper manner. Hence a student finds it difficult to design and construct a circuit so that he/she can test it for proper functioning. A proper engineering design requires step-by-step procedure so that the final design can be implemented directly without any confusion. In this article, we give a standard procedure for the design of a common-emitter voltage amplifier in the class-A mode of operation.
It may be noted that the designs given here are for circuits using discrete devices which are part of first or second year curriculum in electronics. 


1. GIVEN SPECIFICATIONS
An engineering design is done based on the requirements of a given practical problem. These requirements are called specifications. Our first task is therefore to design a single-stage RC-coupled common-emitter class-A voltage amplifier to meet the following specifications:


·         Voltage gain                 :          100 (minimum)
·         Voltage swing               :          9 V (peak-to-peak)
·         Current swing               :          1 mA (peak)
·         Bandwidth                    :          100 kHz (maximum)

2. DC AND AC LOAD LINES
An RC-coupled class-A voltage amplifier can be designed with the help of the DC and ac load lines. Figure 1 shows theoretical DC and ac load lines and Fig. 2 shows them using practical numerical values.  DC load line is the straight line connecting the supply voltage VCC and the maximum DC current ICMAX. VCC also represents the maximum value of collector-to-emitter DC voltage VCEMAX, when the DC collector IC = 0. Similarly, the maximum value of collector current ICMAX occurs when VCE = 0. From Fig. 1, we find

ICMAX = VCC /(RC +RE)

In the above equation, let VCC = 10 V, RC = 4.5 kΩ, and RE = 1 kΩ, then

            ICMAX = 10 V /5.5 kΩ = 1.82 mA

Now we draw the DC load line in this case by joining the VCC (=10-V) point on the voltage (x) axis and the ICMAX (1.82-mA) point on the current (y) axis, as shown in Fig. 2. It may be noted that in computing ICMAX, we have included emitter resistor RE as it forms part of the DC collector-emitter path.
            To draw the ac load line, we have to compute the maximum value of ac collector-emitter voltage Vcemax and the maximum value of ac collector current Icmax. Since emitter resistor RE is usually bypassed with capacitor CE, it can be neglected from computations. Similarly, since the effect of RE does not come under signaling (or, ac) condition, the DC voltage drop across it can be neglected. Therefore the effective peak-to-peak value of the collector-emitter voltage Vce(p-p) under signaling conditions is given by

Vce(p-p) = 2Vcemax = VCC VRE


In our example, if VCC = 10 V, and VRE = 1 V, then

Vcemax = (10 ‒1)/2 = 4.5 V


This represents the point of the ac load line on the voltage axis. To find the current point of the ac load line, we compute the maximum value of ac current using the expression

Icmax  =  Vce(p-p) /RC = 2Vcemax / RC

In our example, since Vcemax = 4.5 V and RC = 4.5 kΩ, we obtain

Icmax  = 4.5 V /4.5 kΩ = 1 mA

By joining the 9-V point on the x axis and the 2-mA point on the y axis, we draw the ac load line, as shown in Fig. 2.
            We have seen that Vcemax is the maximum possible value of the swing in the signal output voltage. For symmetrical swing in the output voltage, therefore, we must have

Vcemax = VCEQ = Vce(p-p)/2

where VCEQ is the steady-state, quiescent, or DC collector-to-emitter voltage. It is also the voltage across the collector and the emitter of the transistor immediately after the circuit has been switched on and with no input signal being applied. ICQ is the DC collector current corresponding to VCEQ.
             It is interesting to note that the ac and the DC load lines cross each other at the VCEQ-ICQ point, as shown in Fig. 1. It can be seen that the amplified signal voltage Vce varies symmetrically with respect to VCEQ to produce the maximum output voltage swing Vcemax, as shown. Similarly, collector signal current Ic varies symmetrically about ICQ to produce the maximum output current swing Icmax, as shown. We thus find that choosing the correct operating (Q) point VCEQ-ICQ helps in achieving the maximum swing in the output voltage and current. Thus it is clear that using the load lines to design an amplifier is an optimum procedure.










Figure 3 shows the voltage drops across various terminals of the BJT and Fig. 4 shows the numerical values of the voltage drops used in the design example.




We now make use of Figs. 2 and 4 in our design example, which is described below.

3. DESIGN PROCEDURE

Step 1: Fixing VCC
As explained above, from Fig. 3, we have

                                                        VCC =VRC + VCE + VRE                (1)

where VRC is the voltage across collector resistor RC, VCE is the voltage across the collector and the emitter, and VRE is the voltage across emitter resistor RE. For maximum symmetrical swing, from load-line analysis, we find the requirement

                                                               VRC = VCE                              (2)

And, for temperature stability, we take (as a rule-of-thumb)

                                                            VRE = 0.1 VCC                                      (3)

It may be noted that the larger the value of VRE, the better it is for temperature stability. However, a larger value of VRE means a lower output voltage swing. To compromise between the two contradictory requirements, we choose the condition given in Eq. (3). Now, using Eqs. (1), (2), and (3), we get the peak-to-peak output voltage Vce(p-p) as

                                                     Vce(p-p) = 2VRC = 2VCE = 0.9VCC          (4)

From this equation, we get the supply voltage

                                                    VCC = Vce(p-p)/0.9 = 2Vcemax/0.9             (5)

where Vcemax (maximum value of the output voltage) = Vce(p-p)/2. In the given problem, Vcepp = 9 V. Therefore we fix the supply voltage

VCC = 9/0.9 = 10 V

Step 2: Fixing IC
The collector current IC is fixed according to the current swing required. From Fig. 3, IC=ICQ=Ic max, where ICQ = value of collector current under quiescent, steady-state, or zero-signal condition. It can be seen from Fig. 3 that IC (=Icmax) also represents the peak value of the ac collector current under signaling condition. Hence IC is fixed at this value. In our problem, the desired current swing is 1 mA. Therefore, we choose

ICQ = 1 mA


Step 3: Selection of the Transistor

Selection of a suitable bipolar junction transistor is based on the following facts:

·      VCEMAX : This is the maximum voltage across the collector and emitter terminals, which the transistor can withstand without breakdown. Assuming VCEMAX = VCC (when IC = 0), we may choose a transistor whose VCEMAX is at least 2VCC. In this case, we have chosen a factor-of-safety of 2.

·      ICPP : This is the peak (maximum) value of collector current that can safely flow through the transistor at any instant. This is equal to 2ICMAX. The transistor must be able to withstand this also. Usually, for safety, we choose a transistor that can withstand at least twice this current.

·      Av : This represents the voltage gain of the transistor = bRL/Ri, where, b or hFE is the current amplification factor of the transistor, RL is the load resistance, and Ri is the input impedance of the amplifier. Assuming that RL= Ri (usually, RL >> Ri), we find that Av = b. This expression puts us in some difficulty. We know that b  varies widely even for the same type of transistor. For example b of BC 107 varies roughly from 125 to 500. Which value of b is to be chosen for designing the amplifier?

                   To find a solution to the above question, consider the value of b of BC 107 itself. As stated above, this varies from125 to 500. This means that all BC 107 transistors will have a             minimum guaranteed value of b (hereafter referred to as bm­in) equal to 125. Current gain b of values greater than this (up to 500) can surely exist; but the manufacturer does not guarantee any value other than the minimum value of 125. This forces us to design the amplifier by using the minimum value of b. In fact, in all our future designs, we shall be using only the minimum value of a parameter, if that parameter has a wide range of variation in its value.

·         Voltage gain given by the expression Av=βR­L/R­I is only approximate because it is dependent on the highly variable parameters of b, Ri, and RL. Assuming RL= Ri, it yields a gain lying between bmin and bmax. To get the required gain to the exact value desired, we must use negative feedback techniques. Taking into account of the above facts, from the data manual, we choose

BC 107

 which has ICMAX=10 mA, VCE MAX= 45 V, and bmin=125, as the amplifier device.

Step 4: Design of RC
Collector resistor RC may be designed by using the expression

                                                 RC = (VCCVRE)/2IC    (6)

Combining the equations given above, we get

                                                   RC = 0.9VCC/2IC           (7)

Substituting,
                       
  RC = 0.45´ 10/10−3 = 4.5 kW

Resistors have finite power rating beyond which, if operated, they are sure to fail. To avoid this, we must also specify the power rating of resistors. Power rating P of a resistor
                                                        P = 2(Imax)2R            (8)                    

where Imax = maximum current through the resistor. The factor 2 is used as a factor-of-safety. To find the power rating of collector resistor RC, we use the modified equation

                                                        P = 2(2ICQ)2RC            (9)                 

  where 2ICQ is the maximum current that can flow through the collector when VCE = 0 V. The maximum value of collector current is also the collector-current swing. So we use Imax= 2 mA. Substituting values,

P = (2 × 10−3)2 ´ 4.7 ´ 103 = 0.0376 W

The nearest value of standard carbon resistor (with ± 10% tolerance) is 4.7 kW with (1/8)-W rating. Therefore, we select

RC = 4.7 kW / 0.125 W

Step 5: Design of RE
As stated earlier, to choose RE, the following conditions must be satisfied:

i.   RE should be as high as possible so that Q point is stabilized. It can be proved from stability analysis that the collector current is stabilized against variations in ICO and VBE, if RE is high. The higher the value of RE the better the stabilization we get.

 ii. However, if RE is high, the swing in the output voltage will be reduced.

Conditions (i) and (ii) are contradictory to each other. In order to compromise between these two conflicting requirements, we choose, as a rule-of-thumb

                                                            VRE = 0.1 VCC             (10)

            In our design problem, VRE = 0.1 ´ 10 = 1.0 V. Since RE = VRE/IC, we find RE = 1/1´l0−3 =1 kW. We select

RE = 1 kW / 0.125 W

Step 6: Design of the Biasing Network consisting of Resistors R1 and R2
The bias network consists of resistors R1 and R2. The design of these resistors is based on the stability analysis of the amplifier since the stability of the operating point is mainly dependent on these resistors. From stability analysis, we have
                                   
                                                             bmin/10 = RB/RE            (11)

where
                       
                                                               RB = R1R2 /(R1 + R2)    (12)

Also, from Fig. 4, we get

VR1 = VCC R1 /(R1 + R2)    (13)

But

                  VR1=VBEQ+VRE                 (14)

where VBEQ is the quiescent or steady-state base-emitter voltage of the transistor. This is the operating or Q-point of the transistor. By steady-state operation, we mean the operation of the amplifier when no signal is present. The base-emitter active-region voltage of silicon transistors usually vary exponentially from 0.5 V to 0.8 V. In such a case, VBEQ is usually taken as 0.7 V. However, with Indian transistors, the authors have found that VBEQ has the best value between 0.6 V to 0.65 V. Choosing 0.65 V, we haveVR1= 0.65+1.0 = 1.65 V. Solving

                        R1 = RB/(1‒a)                  (15)

and

           R1 = RB/a                            (16)
                                                                                                

where
                                                               a =    VR1 / V­CC                      (17)               

Substituting values, we have

R1 = 15 kW and R2 = 82 kW.

The above values are obtained by considering that bmin of BC 107 is 125. However, it is found that bmin of the Indian-made BC 107 transistor is only around 100 in majority of the cases. Therefore, recalculating for R1 and R2 yields the values of R1=12 kW (1/8 W), and R2 = 68 kW (1/8 W).

Step 7: Design of Bypass Capacitor CE
CE is used to bypass AC signal across RE. For designing this, we use the expression

                                                                    fL = β/2πR1C                (18)
               

where fL (lower half-power frequency) = 30 Hz for AF amplifiers. Substituting values, we find

                                                                    CE = 50 mF/10 V

Step 8: Design of Coupling Capacitors CC1 and CC2
CC1 and CC2 are designed using the expression
  
                                                                      fL = β/2πR1CC                      (19)
                                                                 

Substituting values, we get             

                                                             CC1 = CC2 = 5.6 mF
    
   Choose a higher value, say, 10 mF/10 V

Step 9: Design for a Bandwidth of 100 kHz
The bandwidth of an amplifier is determined by the high-frequency response of the amplifier. This, in turn, is determined by the load capacitance C'L. If fH denotes the upper half-power frequency, then

                                                            fH = 1/2πRLCL                        (20)

Given fH = 100 kHz, and assuming R’L = RC we get

CL= 0.4 nF

The completely designed am plifier is shown in Fig. 5. This amplifier is called as the Standard Amplifier in this text.







.







DISCRETE SIGNAL OPERATIONS

EDITOR: B. SOMANATHAN NAIR 1. INTRODUCTION In the previous two blogs, we had discussed operations of scaling and shifting on conti...