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Determine the total input resistance of the emitter-follower in the figure below. Also find the voltage gain, current gain, and power gain in terms of power delivered to the load, RL. Assume \$\beta_{ac} = 175 \$ and that the capacitive reactances are negligible at the frequency of operation.

enter image description here

This is the analysis from the textbook:

$$R_e = R_E \parallel R_L = 470 \ \mathrm{V} \parallel 470 \ \mathrm{V} = 235 \ \mathrm{\Omega}$$

$$R_{in(base)} \approx \beta_{ac} R_e = (175)(235\ \mathrm{V}) = 41.1\ \mathrm{k\Omega}$$

$$R_{in(tot)} = R1 \parallel R2 \parallel R_{in(base)} = 18 \ \mathrm{k\Omega} \parallel 51 \ \mathrm{k\Omega} \parallel 41.1 \ \mathrm{k\Omega} = 10.1 \ \mathrm{k\Omega}$$

$$V_E = (\frac{R_2} {R_1 + R_2}) V_{cc} - 0.7 = \frac{51 \ \mathrm{k\Omega}} {18\ \mathrm{k\Omega} + 51\ \mathrm{k\Omega}} 10\ \mathrm{V} - 0.7\ \mathrm{V} = 6.69\ \mathrm{V}$$

$$I_E = \frac{V_E} {R_{E}} = \frac{6.69\ \mathrm{V}} {235\ \mathrm{\Omega}} = 14.2\ \mathrm{mA}$$


The following is my analysis:

enter image description here

$$R_{th} = R_1 \parallel R_2 = 18\ \mathrm{k\Omega} \parallel 51\ \mathrm{k\Omega} \approx 13.3\ \mathrm{k\Omega}$$ $$V_{th} = V_{cc} \frac{R_2} {R_1 + R_2} = 10 V \frac{51\ \mathrm{k\Omega}} {18\ \mathrm{k\Omega} + 51\ \mathrm{k\Omega}} \approx 7.39\ \mathrm{V}$$

$$ R_e = R_E \parallel R_L = 470\ \mathrm{\Omega} \parallel 470\ \mathrm{\Omega} = 235\ \mathrm{\Omega} $$

$$I_E = \frac{V_{th} - 0.7\ \mathrm{V}} {\cfrac{R_{th}} {\beta_{ac}} + R_E} = \frac{7.39\ \mathrm{V} - 0.7\ \mathrm{V}} {\cfrac{13.3\ \mathrm{k\Omega}} {175} + 470 \ \mathrm{\Omega}} \approx 12.2643\ \mathrm{mA}$$

$$r'_e = \frac{25\ \mathrm{mV}} {I_E} = \frac{25\ \mathrm{mV}} {12.2643\ \mathrm{mA}} \approx 2.0384\ \mathrm{\Omega}$$

$$ \begin{align} R_{in(tot)} &= R1 \parallel R2 \parallel \beta_{ac} (r'_e + R_e) \\ &= 18\ \mathrm{k\Omega} \parallel 51\ \mathrm{k\Omega} \parallel (175 \times (2.0384 +235) \times 10^{-3}\ \mathrm{k\Omega} \\ &\approx 10.087415\ \mathrm{k\Omega} \end{align}$$

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  • \$\begingroup\$ What answer do you expect to get from us? \$\endgroup\$
    – G36
    Commented Dec 28, 2022 at 11:19
  • \$\begingroup\$ @G36 Which analysis is more accurate? Mine or the Textbook's? \$\endgroup\$
    – kile
    Commented Dec 28, 2022 at 11:20
  • \$\begingroup\$ In that case, your analysis is more accurate than the book version. \$\endgroup\$
    – G36
    Commented Dec 28, 2022 at 11:30
  • \$\begingroup\$ @G36 Thank you for your feedback! BTW, could you check the total input resistance of the emitter-follower? Is my analysis for total input resistance correct? \$\endgroup\$
    – kile
    Commented Dec 28, 2022 at 11:32
  • \$\begingroup\$ Yep, sure. Rin will be around 10k. so your answer looks good. But why 1.16 instead of 2? \$\endgroup\$
    – G36
    Commented Dec 28, 2022 at 11:39

1 Answer 1

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Here is what I get for variables ... (Simulated for checking).
If the capacitor's impedance is "negligible", read data where curves are "horizontal" (frequency high enough).

Output impedance is calculated as usual ( Vopen (load open) / Icc (load shorted).

enter image description here

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  • \$\begingroup\$ Could you explain what step data mean? \$\endgroup\$
    – kile
    Commented Dec 28, 2022 at 11:36
  • \$\begingroup\$ The beta of BJT (beta is very variable) is stepped (from 163 to 652) for checking the dispersion of answers depending on that parameter. One can see that results are quasi "independent" of beta. \$\endgroup\$
    – Antonio51
    Commented Dec 28, 2022 at 11:47
  • \$\begingroup\$ Is beta of BJT changing all the time in AC circuit? \$\endgroup\$
    – kile
    Commented Dec 28, 2022 at 12:08
  • \$\begingroup\$ No. Change with type of BJT, and "only" with the temperature. \$\endgroup\$
    – Antonio51
    Commented Dec 28, 2022 at 14:04

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