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I'm designing a common emitter - voltage divider bias amplifier circuit like below:

enter image description here

Requirements:

  • \$|A_v| = 5\$
  • \$Z_{in} \geqslant 10k\Omega\$
  • \$P_{consumption} \leqslant 0.2W\$
  • \$V_{CC} = 15V\$

Procedure:

Assuming \$I_C \approx I_E\$

We have:

\$|A_v| = \frac{R_C}{r_e} = \frac{R_C}{\frac{26mV}{I_E}} = 5\$

\$\Leftrightarrow R_C = \frac{130mV}{I_C}\$

On the other hand, we have the DC loadline equation:

\$I_C = \frac{V_{CC} - V_{CE}}{R_E+R_C} (1)\$

For optimal operation, we assume that the Q-point will be in the middle of the load line, or:

\$V_{CE} = \frac{1}{2}V_{CEmax} = \frac{1}{2}V_{CC} = \frac{1}{2} 15V = 7.5V(2)\$

\$I_C = \frac{1}{2}V_{I_{Cmax}} = \frac{1}{2}\frac{V_{CC}}{R_C+R_E} = \frac{7.5}{R_C+R_E}(3)\$

From \$(1)\$, \$(2)\$, and \$(3)\$:

\$\Rightarrow \frac{130mV}{R_C} = \frac{7.5}{R_E+R_C}\$

\$\Leftrightarrow \frac{R_E}{R_C} = \frac{737}{13}\$

This is an unrealistic ratio for \$R_E\$ and \$R_C\$ as I always see \$R_E\$ smaller than \$R_C\$. Was my calculation wrong, or was it my approach?

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    \$\begingroup\$ Try another design with signal feedback (only a part of RE bridged with C ). Gain: A=Rc/(re+RE)) \$\endgroup\$
    – LvW
    Nov 21, 2023 at 15:27

3 Answers 3

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There's nothing wrong with your calculations and approach.

The issue is with the design because of a relatively low voltage gain without any emitter degeneration. If we were talking about higher (e.g. 3-digit) voltage gains or higher current-demanding loads (relatively smaller load resistors) then we'd consider shorting emitter degeneration resistor.

The emitter degeneration resistor (RE) appears at DC only to bring a local feedback and therefore set the quiescent current \$I_C\$ (As you can see, the entire circuit forms a constant-current sink at DC).

So if RE wasn't shorted (presumably) with C2 then the gain equation would be

$$ A_V\approx -\frac{R_C}{R_E} $$

and therefore the collector resistance would be greater than emitter resistance. Since the emitter resistance is shorted at AC, it has to be much greater than collector resistance, ~57 times for your circuit.

I'd remove C2.

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  • \$\begingroup\$ A more exact gain expression (as mentioned in my comment to Halfys question) is Av=-Rc/(re + RE) with re=1/gm=Vt/Ic. \$\endgroup\$
    – LvW
    Nov 26, 2023 at 14:35
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You can pretty much determine collector current from the power requirement and the supply voltage. So, 0.2 watts divided by 15 volts = 13.333 mA so, broadly assume the collector current is 10 mA. That makes RC equal to 7.5 volts divided by 10 mA = 750 Ω or maybe a bit greater if you wanted. 1 kΩ seems a nice round value (7.5 mA quiescent collector current).

For an Av of 5 I would ditch the nearly-always-problematic emitter capacitor and make RE 200 ohm. I might be tempted to lower it to maybe 180 Ω if the gain didn't work out at exactly 5.

Can you take it from here?

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A study that gives much more satisfactory results than the previous one is the following (from an idea taken from an old electronics book "Principles of Electronics" by Paul E. Gray, Campbell L. Searle published by John Wiley & Sons 1969):

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    \$\begingroup\$ Question: Do you really think that the above could be considered as an answer to the TOs question? \$\endgroup\$
    – LvW
    Nov 27, 2023 at 14:23
  • \$\begingroup\$ The simulation is OK! \$\endgroup\$
    – Franc
    Nov 27, 2023 at 14:26

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