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Questions: 1 2 3 4 5 6 7 8 9 10

PHYS 2420 Transistor Solutions

  1. For the silicon (VBE = 0.7 V) transistor shown in the diagram below determine whether it is in saturation if
    1. βDC = 50, VBB = 6.5 V, RB = 2500 Ω , RC = 500 Ω , VCC = 6.0 V, and VCE(sat) = 0.0 V;
    2. βDC = 75, VBB = 4.0 V, RB = 10000 Ω , RC = 100 Ω , VCC = 4.0 V, and VCE(sat) = 0.3 V;
    3. βDC = 180, VBB = 3.0 V, RB = 16000 Ω , RC = 400 Ω , VCC = 12.0, and VCE(sat) = 0.5 V;
    4. βDC = 125, VBB = 3.9 V, RB = 40 KΩ , RC = 2.0 KΩ , VCC = 9.5, and VCE(sat) = 0.4 V.

    A transistor is in saturation when IC £ βDCIB. To determine the saturation status, we first get Kirchhoff's equations for the transistor.

    Our equations are

    IC + IB = IE
    		 (1)
    VBB - IBRB -VBE = 0
    		 (2)
    VCC - ICRC -VCE = 0
    		 (3)

    From equation (2), we can determine IB = (VBB - VBE)/RB. From equation (3), we can determine Ic(sat) for the given value of VCE (sat), IC(sat) = (VCC - VCE(sat))/RC. If IC(sat) £ βDCIB, then the transistor is in saturation.

    Using EXCEL, we can write a spreadsheet that answers the question automatically.

    Case
    IC(sat) (A)
    IB (A) 
    βDCIB (A)
    In Saturation?
    (a)
    0.0120
    0.00232
    0.116
    Yes
    (b)
    0.0370
    0.00033
    0.025
    No
    (c)
    0.0288
    0.00014
    0.026
    No
    (d)
    0.0046
    0.00008
    0.010
    Yes

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  2. Under the following set of conditions determine IC(sat). What is the minimum of IB that produces saturation. What minimum value of VBB produces saturation. Refer to the diagram above. Take VBE = 0.7 V.
    1. βDC = 50 , RB = 2500 Ω , RC = 500 Ω , VCC = 6.0 V, and VCE(sat) = 0.0 V;
    2. βDC = 75, RB = 10000 Ω , RC = 100 Ω , VCC = 4.0 V, and VCE(sat) = 0.3 V;
    3. βDC = 180, RB = 16000 Ω , RC = 400 Ω , VCC = 12.0 V, and VCE(sat) = 0.5 V;
    4. βDC = 125 , RB = 40 kΩ , RC = 2.0 kΩ , VCC = 9.5 V, and VCE(sat) = 0.4 V.

    As noted in question 1 above, we determine IC(sat) by

    IC(sat) = (VCC - VCE(sat))/RC.            (1)

    From

    IB = (VBB - VBE)/RB,            (2)

    we see that we are free to adjust VBB to get any value of IB. If we increase VBB, we increase IB and the product βDCIB. Saturation occurs when βDCIB ³ IC(sat). Hence the transistor will be in saturation when IB ³ IC/βDC. The minimum value of IB which causes saturation is therefore

    IB(sat) = IC(sat)/βDC.            (3)

    Substituting the value of IB(sat) given by equation (3) into equation (2), we find the minimum value of VBB that causes saturation to be

    VBB(sat) = VBE + RBIC(sat)/βDC .            (4)

    Again using EXCEL, we can write a spreadsheet that answers the question automatically.

    Case
    IC(sat) (A)
    IB(sat) (A) 
    VBB (sat) (V)
    (a)
    0.0120
    0.00240
    1.30
    (b)
    0.0370
    0.00049
    5.63
    (c)
    0.0288
    0.00016
    3.26
    (d)
    0.0046
    0.00004
    2.16

    [Return to Top of Page]

  3. For the following sets of conditions determine the Q-point. Determine the maximum peak value of the base current Ib . If the maximum value is exceeded, will the signal be driven into cutoff, or saturation, or both? Take VBE = 0.7 V and VCE(sat) = 0.0 V.
    1. βDC = 50, VBB = 1.0 V, RB = 2500 Ω , RC = 500 Ω , VCC = 6.0 V;
    2. βDC = 25, VBB = 1.0 V, RB = 2500 Ω , RC = 500 Ω , VCC = 6.0 V;
    3. βDC = 75, VBB = 1.0 V, RB = 2500 Ω , RC = 500 Ω , VCC = 6.0 V.

    The Q-point is a point on a graph of IC versus VCE. Assuming that the transistor is neither in saturation or cutoff, the value ICQ is found from ICQ = βDCIB, where IB is determined by the equation IB = (VBB - VBE)/RB derived in question 1. The value of VCEQ is found by using ICQ and equation (3) from question 1, VCEQ = VCC - ICQRC .

    To answer the rest of the question, we need to examine the "DC Load line". The load line is the line drawn on the IC versus VCE graph that runs from the saturation point (IC(sat), VCE(sat)) to the cutoff point (IC(cutoff), VCE(cutoff)). As we have seen, we get IC(sat) from IC(sat) = (VCC - VCE(sat))/RC. The cutoff occurs when IC = 0. This implies VCE(cutoff) = VCC.

    Using Excel to determine each Q-point, saturation point, and cutoff point yields

    Q-point
    Saturation point
    Cutoff point
    Case
    IC (A)
    VCE (V)
    IC (A)
    VCE (V)
    IC (A)
    VCE (V)
    (a)
    0.0060
    3.00
    0.0120
    0.00
    0.00
    6.00
    (b)
    0.0030
    4.50
    0.0120
    0.00
    0.00
    6.00
    (c)
    0.0090
    1.50
    0.0120
    0.00
    0.00
    6.00
     

    A sketch of the DC Load line is useful

    Note that increasing IB will move a Q-point toward saturation while decreasing it will move the Q-point toward cutoff. Now Ib is the AC signal that is added to IB. The variation in the sum of IB + Ib has the same effect as increasing or decreasing IB; that is if Ib has too large a peak value the transistor will be driven into saturation or cutoff. Examining the graph, we see that the Q-point for part (a) is in the middle of the graph. As the peak value of Ib, Ib(peak), increases, it simultaneously enters both cutoff and saturation. The value of Ib(peak) that this occurs at is

    Ib(peak) = (IC(sat) - ICQ)/βDC ,           (1)

    or

    Ib(peak) = (ICQ - IC(cutoff))/βDC ,           (2)

    whichever is smaller. For part (a), both equations give the same result since ICQ is equidistant to both the saturation and cutoff points. For part (b), ICQ is closer to cutoff so equation (2) is the appropriate one to choose. For part (c), ICQ is closer to saturation so equation (1) is the appropriate one to choose. Calculating Ib(peak) yields

    Case

    Ib(peak) (mA)

    Saturation or Cutoff first?

    (a)

    0.120

    both

    (b)

    0.120

    cutoff

    (c)

    0.040

    Saturation

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  4. The diagram below shows a base-biased transistor. For the following sets of conditions determine the Q-point. Determine the maximum peak value of the base current. If the maximum value is exceeded, will the signal be driven into cutoff, or saturation, or both? Take VBE = 0.7 V and VCE(sat) = 0.0 V.
    1. βDC = 200, RB = 75 kΩ , RC = 200 Ω , and VCC = 15.0 V;
    2. βDC = 100, RB = 75 kΩ , RC = 200 Ω , and VCC = 15.0 V;
    3. βDC = 300, RB = 75 kΩ , RC = 200 Ω , and VCC = 15.0 V.

    First we apply Kirchhoff's rules to the transistor

    and derive the equations

    IC + IB = IE
    		 (1)
    VCC - ICRC - VCE = 0
    		 (2)
    VCC - IBRB - VBE = 0
    		(3)

    We get the Q-point as follows. We find IB from equation (3), IB = (VCC - VBE)/RB. Assuming that we are not in saturation, we get ICQ = βDCIB. Then using equation (2), we find that VCEQ = VCC - ICQRC. As in question 3, the saturation point occurs when IC(sat) = (VCC - VCE(sat))/RC. The cutoff occurs when IC = 0, or VCE(cutoff) = VCC.

    Using Excel to calculate these quantities, we find

     

    Q-point

    Saturation point

    Cutoff point

    Case

    IC (A)

    VCE (V)

    IC (A)

    VCE (V)

    IC (A)

    (a)

    0.0381

    7.37

    0.0750

    0.00

    0.00

    15.0

    (b)

    0.0191

    11.2

    0.0750

    0.00

    0.00

    15.0

    (c)

    0.0572

    3.56

    0.0750

    0.00

    0.00

    15.0

    The graph of the load line for these points looks like

    Note that increasing IB will move a Q-point toward saturation while decreasing it will move the Q-point toward cutoff. Now Ib is the AC signal that is added to IB. The variation in the sum of IB + Ib has the same effect as increasing or decreasing IB; that is if Ib has too large a peak value the transistor will be driven into saturation or cutoff. As shown in question 3, the values of Ib(peak) that this occurs at are Ib(peak) = (IC(sat) - ICQ)/βDC or Ib(peak) = (ICQ - IC(cutoff))/βDC, whichever is smaller.

    For part (a), both equations give the same result since ICQ is slightly closer to saturation. For part (b), ICQ is closer to cutoff . For part (c), ICQ is closer to saturation. Calculating Ib(peak) yields

    Case

    Ib(peak) (mA)

    Saturation or Cutoff first?

    (a)

    0.184

    Saturation

    (b)

    0.190

    cutoff

    (c)

    0.059

    saturation
     

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  5. The diagram below shows an npn emitter-biased transistor. For the following sets of conditions determine the Q-point. Determine the maximum peak value of the base current. If the maximum value is exceeded, will the signal be driven into cutoff, or saturation, or both? Take VCE(sat) = 0.0 V.
    1. βDC = 150, RB = 10 kΩ , RC = 1.0 kΩ , RE = 3.0 kΩ , VCC = 9.0 V, VEE = 9.0 V, and VBE = 0.7 V;
    2. βDC = 300, RB = 10 kΩ , RC = 1.0 kΩ , RE = 3.0 kΩ , VCC = 9.0 V, VEE = 9.0 V, and VBE = 0.6 V;
    3. βDC = 150, RB = 10 kΩ , RC = 1.0 kΩ , RE = 1.5 kΩ , VCC = 9.0 V, VEE = 9.0 V, and VBE = 0.7 V.
    4. βDC = 150, RB = 10 kΩ , RC = 1.0 kΩ , RE = 4.5 kΩ , VCC = 9.0 V, VEE = 9.0 V, and VBE = 0.7 V.

    First we apply Kirchhoff's rules to the transistor

    and derive the equations

    IC + IB = IE (1)
    VCC - ICRC - VCE - IERE + VEE = 0 , (2)
    -IBRB - VBE - IERE + VEE = 0 . (3)

    This set of equations is more complicated than the sets we had for the prior bias circuits. Nonetheless the equations may be solved. First we assume that the transistor is neither in saturation, so that IC = βDCIB and IE = (βDC+1)IB. Equation (3) then yields an expression for IB

    IB = (VEE - VBE)/[RB + (βDC+1)RE] .            (4)

    Replacing IC and IE in equation (2) to get an equation for VCE yields

    VCE = VCC + VEE - IB[βDCRC + (βDC+1)RE] .            (5)

    Note that both equations (4) and (5) depend on VEE, that is that a particular value of VEE determines a value for IB which in turn sets the value of VCE - provided that the system is not in saturation.

    To determine the saturation point, we need to first find the value of VEE(sat). Using equation (4) to eliminate IB from equation (5) yields,

    VCE = VCC + VEE - (VEE - VBE)[βDCRC + (βDC+1)RE] /[RB + (βDC+1)RE] ,

    or, collecting like terms

    .            (6)

    Rearranging the above to find VEE gives

    .            (7)

    Setting VCE = VCE(sat), we get the value of VEE(sat), which we can substitute back into equation (4), do some algebraic manipulation, and get IB(sat),

    IB(sat) = (VCC - VCE(sat) + VBE)/(βDCRC - RB) .            (8)

    Off course, just at saturation IC(sat) = βDCIB(sat),

    IC(sat) = (VCC - VCE(sat) + VBE)/(RC - RB/βDC) .            (9)

    The cutoff occurs when IB = IC = IE = 0. From equation (4), we see that IB = 0 if and only if VEE = VBE. So for cutoff,

    VCE(cutoff) = VCC + VEE = VCC + VBE.

    Using Excel to calculate these quantities, we find

     

    Q-point

    Saturation point

    Cutoff point

    Case

    IC (A)

    VCE (V)

    IC (A)

    VCE (V)

    IC (A)

    VCE (V)

    (a)

    0.00269

    7.19

    0.01039

    0.00

    0.00

    9.7

    (b)

    0.00273

    7.06

    0.01003

    0.00

    0.00

    9.7

    (c)

    0.00526

    4.79

    0.01039

    0.00

    0.00

    9.7

    (d)

    0.00181

    8.01

    0.01039

    0.00

    0.00

    9.7

    There are several important points to draw from these results. The first is that for a given value of RE, the Q-point does not vary very much with βDC. We say that the Q-point is stable. This is important in a circuit since βDC is strongly temperature dependent; in an emitter bias circuit the output will not fluctuate with temperature. The second is that RE can be adjusted to determine the Q-point. The last point is that changing either βDC or RE does not effect the cutoff and has only a small effect on the saturation point. As a result, we can draw one load line for all the points. The graph of the load line for these points looks like

    Note that increasing IB increases IC which will move a Q-point toward saturation while decreasing it will move the Q-point toward cutoff. Now Ib is the AC signal that is added to IB. The variation in the sum of IB + Ib has the same effect as increasing or decreasing IB; that is if Ib has too large a peak value the transistor will be driven into saturation or cutoff. As shown in question 3, the values of Ib(peak) that this occurs at are Ib(peak) = (IC(sat) - ICQ)/βDC or Ib(peak) = (ICQ - IC(cutoff))/βDC, whichever is smaller.

    Examining the graph of the load line, we see that the signal will first go into cutoff for parts (a), (b), and (d) while the signal will go into saturation for (c). Calculating Ib(peak) yields

    Case

    Ib(peak) (μ A)

    Saturation or Cutoff first?

    (a)

    17.9

    cutoff

    (b)

    9.1

    cutoff

    (c)

    34.2

    saturation

    (d)

    12.0

    cutoff

    [Return to Top of Page]

  6. For the npn emitter-biased transistor shown above, what would be the minimum value of RE that keeps the transistor from going into saturation?
    1. βDC = 150, RB = 10 kΩ , RC = 1.0 kΩ , VCC = 9.0 V, VEE = 9.0 V, VBE = 0.7 V, and VCE(sat) = 0.5 V,
    2. βDC = 300, RB = 10 kΩ , RC = 1.0 kΩ , VCC = 9.0 V, VEE = 9.0 V, VBE = 0.6 V, and VCE(sat) = 0.4 V,
    3. βDC = 220, RB = 1200 Ω , RC = 700 Ω , VCC = 6.0 V, VEE = 6.0 V, VBE = 0.7 V, and VCE(sat) = 0.7 V.

    We saw in Question 5, that adjusting RE moves the Q-point around. In this question, we are asking what value of RE will cause VCE to equal VCE(sat). If we take equation (6) form Question 5,

    we need to rearrange this equation to determine RE when VCE = VCE(sat). After a good deal of manipulation, we arrive at

    Using Excel to calculate the results yields

    Case

    RE (sat) (Ω )

    (a)

    770

    (b)

    846

    (c)

    605

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  7. The diagram below shows a transistor with a voltage-divider bias. For the following sets of conditions determine the Q-point. Determine the maximum peak value of the base current. If the maximum value is exceeded, will the signal be driven into cutoff, or saturation, or both? Take VCE(sat) = 0.0 Volts.
    1. βDC = 100, VCC = 8.0 V, VBE = 0.7 V, RC = 1.4 kΩ , RE = 6000 Ω , R1 = 25 kΩ , and R2 = 12 kΩ ;
    2. βDC = 200, VCC = 8.0 V, VBE = 0.7 V, RC = 1.4 kΩ , RE = 6000 Ω , R1 = 25 kΩ , and R2 = 12 kΩ .
    3. βDC = 300, VCC = 8.0 V, VBE = 0.7 V, RC = 1.4 kΩ , RE = 6000 Ω , R1 = 25 kΩ , and R2 = 12 kΩ .

    First we apply Kirchhoff's rules to the transistor

    and derive the equations

    IC + IB = IE (1)
    I1 = IB + I2 (2)
    VCC - ICRC - VCE - IERE = 0 (3)
    VCC - I1R1 - I2R2 = 0 (4)
    I2R2 -VBE - IERE = 0 (5)

    We get the Q-point as follows. First we assume that we are not in saturation, so that IC = βDCIB and IE = (βDC+1)IB and make this substitution into the Kirchhoff equations above. Next we eliminate I2 leaving

    VCC - VCE - IB[βDCRC + (βDC+1)RE] = 0 (1)
    VCC - IB[R1 + (βDC+1)RE(R1+R2)/R2] - VBE(R1+R2)/R2 = 0 (2)

    Then we find IB from equation (7)

    IB = [VCCR2 - VBE(R1+R2)]/[R1R2 + (βDC+1)RE(R1+R2)] .           (8)

    This result may be used with

    VCE = VCC - IB[βDCRC + (βDC+1)RE] ,            (9)

    to uniquely determine the Q-point.

    Cutoff, the point when IB = IC = IE = 0, will occur when the value of R2 is such that equation (7) becomes

    VCC(cutoff) - VBE(R1+R2(cutoff))/R2(cutoff) = 0 ,            (10)

    The value of R2(cutoff) which meets the above criterion is

    R2(cutoff) = R1VBE/(VCC - VBE) .            (11)

    The value of VCE(cutoff) can be seen from equation (6) to be

    VCE(cutoff) = VCC .            (12)

    The saturation point can be found from equation (9)

    IB(sat) = (VCC - VCE(sat))/[βDCRC + (βDC+1)RE] ,            (13)

    and IC(sat) = βDCIB(sat).

    Using Excel to calculate these quantities, we find

     

    Q-point

    Saturation point

    Cutoff point

    Case

    IC (A)

    VCE (V)

    IC (A)

    VCE (V)

    IC (A)

    VCE (V)

    (a)

    0.000309

    5.70

    0.00107

    0.00

    0.00

    8.0

    (b)

    0.000312

    5.68

    0.00108

    0.00

    0.00

    8.0

    (c)

    0.000313

    5.68

    0.00108

    0.00

    0.00

    8.0

    There are several important points to draw from these results. The first is that for a given value of R2, the Q-point is stable and hardly changes with βDC. It changes even less than the two battery bias of Question 6. This is important in a circuit since βDC is strongly temperature dependent; in a voltage divider bias circuit the output will not fluctuate with temperature. The second is that R2 can be adjusted to determine the Q-point. The last point is that changing either βDC does not effect the cutoff and has only a small effect on the saturation point. As a result, we can draw one load line for all the points. The graph of the load line for these points looks like

    Note that increasing IB increases IC which will move a Q-point toward saturation while decreasing it will move the Q-point toward cutoff. Now Ib is the AC signal that is added to IB. The variation in the sum of IB + Ib has the same effect as increasing or decreasing IB; that is if Ib has too large a peak value the transistor will be driven into saturation or cutoff. As shown in question 3, the values of Ib(peak) that this occurs at are Ib(peak) = (IC(sat) - ICQ)/βDC or Ib(peak) = (ICQ - IC(cutoff))/βDC, whichever is smaller.

    Examining the graph of the load line, we see that the signal will first go into cutoff for all three parts (a), (b), and (c). Calculating Ib(peak) yields

    Case

    Ib(peak) (μ A)

    Saturation or Cutoff first?

    (a)

    3.0

    cutoff

    (b)

    1.5

    cutoff

    (c)

    1.0

    cutoff

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  8. The diagram above shows a voltage-divider biased transistor. What is the minimum value of R2 which causes saturation?
    1. βDC = 100, VCC = 8.0 V, VBE = 0.7 V, RC = 1.4 kΩ , RE = 6000 Ω , R1 = 25 kΩ , and VCE(sat) = 0.0 V;
    2. βDC = 200, VCC = 8.0 V, VBE = 0.7 V, RC = 1.4 kΩ , RE = 6000 Ω , R1 = 25 kΩ , and VCE(sat) = 0.0 V.
    3. βDC = 300, VCC = 8.0 V, VBE = 0.7 V, RC = 1.4 kΩ , RE = 6000 Ω , R1 = 25 kΩ , and VCE(sat) = 0.5 V.

    If we examine equations (8) and (9) from Question 7,

    IB = [VCCR2 - VBE(R1+R2)]/[R1R2 + (βDC+1)RE(R1+R2)] ,

    VCE = VCC - IB[βDCRC + (βDC+1)RE] ,

    we see that changing R2 changes IB. Thus we can find a value of R2 that causes VCE to equal VCE(sat). To find R2(sat), we eliminate IB from the second equation using the first. Then we work through the algebra to isolate R2. After a good deal of algebraic manipulation we find

    Using Excel to calculate the results yields

    Case

    RE (sat) (kΩ )

    (a)

    337

    (b)

    267

    (c)

    250

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  9. Determine re, Rin(base) , Rin, Av, A'v for the following amplifier if
    1. Rs = 850 Ω , β = 92, RC = 140 kΩ , RE = 60 kΩ , R1 = 25 kΩ , and R2 = 12 kΩ ;
    2. Rs = 175 Ω , β = 180, RC = 25 kΩ , RE = 5.0 kΩ , R1 = 2.5 kΩ , and R2 = 2.8 kΩ .

    The AC internal resistance of the transistor from Base to Emitter is given by the formula

    re = (25 mV)/IE .            (1)

    To find IE, we use the equations developed in Question 7 to find the Q-point. We can do this because the capacitors block DC signals; in DC the above diagram is simply a Voltage Divider biased transistor. Using Excel to do the calculations, we find

    case

    IB (μ A)

    IE (μ A)

    re (Ω )

    (a)

    0.3390 

    31.53

    792.88

    (b)

    3.8909

    704.26

    35.50

    where we have used βDC = β , i.e. we assumed that the DC and AC current gain are equal.

    For the other quantities we are looking for, we need to consider what the circuit looks like in AC. Some points to note. In AC capacitors act as shorts if XC is small. The DC source VCC act as a ground in AC. The input Vin, or source signal, has a resistance RSource. The output resistance Vout also may have a load resistance, Rload. The transistor may be considered to a combination of two resistors, re and rc. As mentioned re is the base-emitter resistance which is usually small while rc is the base-collector resistance and is huge in the order of megaohms. The resistance of rc is huge because in that direction the transistor is like two reverse-biased diodes in series. The circuit will look like

    The quantity Rin(base), the effective resistance of the emitter branch, is defined as the voltage at the base divided by the base current

    Rin(base) = Vbase/Ib = Ie(re+RE) / Ib = (β +1) (re+RE) ≈ β (re+RE).

    Rin(base) is the effective resistance of the emitter branch because the base-collector junction is a reverse-biased diode and thus can be treated as an open as shown below

    It's not quite that simple since we have Ib entering the emitter branch but the potential drop is much greater than Ib(re+RE) as we have seen. "Looking into the base" the resistance of the arm appears to be Rin(base). In a similar fashion, the total resistance of the transistor circuit as seen by input voltage is the effective resistance of the three parallel branches in the diagram above which is denoted Rin and

    Rin = R1||R2||Rin(base) .

    Rin is connected in series with the source and its resistance. Thus the potential at the base in terms of Vin is

    Vb = Vin [Rin/(Rin + Rs)] .

    The quantity An is the voltage gain, the ratio of the output voltage to the base voltage, defined

    An = Vout/Vbase = IcRC / Ie(re+RE) = β IbRC / (β +1)Ib(re+RE) ≈ RC/(re+RE).

    The quantity An ' is also a voltage gain but is the ratio of the output voltage to the input voltage

    An = Vout/Vin = An Rin / (Rs+Rin).

    Using Excel, we find the following numerical results

    Case

    re (Ω)

    Rin(base) (MΩ)

    Rin (Ω)

    An 

    An '

    (a)

    792.88

    5.65374

    8096.5

    2.28

    2.06

    (b)

    35.50

    0.91143

    1318.8

    4.94

    4.36

    [Return to Top of Page]

  10. Determine Av, A'v, for the amplifier in the problem above if RE is bypassed by a capacitor.

    11. Bypassing a resister by a capacitor means to short it out to AC signal. In DC there is no effect as a charged capacitor act as an open. Thus Q-point would not be affected effected. The effect is to set RE to zero in our equations in Question 9 above. In AC, the first effect of the bypass capacitor would be to change Rin(base) and Rin. Recall that without the bypass Rin(base) = Ib(re+RE) but RE is now zero so

    Rin(base) = Ibre .

    This in turn alters the value of Rin since Rin = R1||R2||Rin(base).

    Following on through the derivations in Question 9, the voltage gain An , the ratio of the output voltage to the base voltage, becomes

    An = Vout/Vbase = IcRC / Ie(re+RE) = [β /(β +1)] RC/re ≈ RC/re.

    The quantity An ', is unchanged except for the value of Rin

    An = Vout/Vin = An Rin / (Rs+Rin).

    Redoing the calculations yields

    Case

    re (Ω )

    Rin(base) (MΩ)

    Rin (Ω)

    An 

    An '

    (a)

    792.88

    5.65374

    8096.5

    2.28

    2.06

    (b)

    35.50

    0.91143

    1318.8

    4.94

    4.36

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