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I'm currently working on this circuit and it appears that my output voltage is not showing the correct values for the positive half wave .

For starters I have these values below - \$U_{CE} = 5 V\$

  • \$R_1 = 6.1kΩ\$
  • \$ R_2 = 1.35kΩ\$
  • \$ R_E = 1kΩ \$
  • \$ R_3 = 1.5kΩ\$
  • \$V_1 = 15 V \$
  • \$ I_C = 2mA \$
  • \$ U_{BE} = 0.4V \$
  • \$ B = 180 \$

Given formula \$r_{BE} \approx \frac {40 mV}{I_B}\$

enter image description here

First I calculated CE with the use of this formula \$τE ≈ (r_{BE}·C_E ) / β = 1/(2 \pi f) \$

These are my calculations:

$$ I_C = B . I_B $$ $$ rBE = 40mV/ I_B = 40mV . B / I_C = 40 . 180 / 2 mA ≈ 4kΩ $$

Let's say the frequency of this amplifier is 1kHz

$$ => τE = 1/(2 \pi f) = 1/(2 \pi 1000) = 1.519 . 10^{-4} s $$

Which makes

$$ => C_E = B.τE / rBE = 1.519.10^{-4} . 180 / 3600 = 7.6 uF $$

And also

$$τE = (r_E / B ). C1 $$ $$ τE = (R_E / B )* C1 $$ $$ r_E = (rBE + B Re) || R1 || R2 = (4kΩ + 180 * 1kΩ) || 6.1kΩ || 1.35kΩ = 1100 Ω = 1.1 kΩ $$

$$ C1 = τE . B / r_E => C1 = ( 1.519 . 10^{-4} s . 180 ) / ( 1.1 * 10^{3} ) = 0.26 uF $$

Then with this question I could calculate C_2.

What is the minimum size of capacitor C2 if the lower limit frequency should be approx.imately 1000 Hz when the output is loaded with 100 kΩ?

$$τE = R * C2 => C2 = τE / R = 1.519 * 10^{-4} / 100kΩ = 1.519 nF $$

$$Ue =√( (1+ω²(R1*C1)²) * (1+ω²(R2*C2)²) . (1+ω²(RE*CE)² )) = 8.5 V $$

Summary :

  • \$C_E = 7.6 uF \$
  • \$C_1 = 0.26 uF \$
  • \$C_2 = 1.519 nF \$

And with these calculations I can simulate my circuit .

Output in red, input in blue: enter image description here

Shouldn't the positive wave of the output voltage be bigger than the positive wave of the input voltage?

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    \$\begingroup\$ You appear to be powering your circuit with AC from V1 -am I reading that correctly? \$\endgroup\$
    – Frog
    Dec 28 '20 at 21:47
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    \$\begingroup\$ Are you really using a sine wave source as the power supply? \$\endgroup\$ Dec 28 '20 at 21:47
  • \$\begingroup\$ Sorry my bad . Forgot to change them . Just wanted to check the simulation with the sin waves \$\endgroup\$
    – Gaston
    Dec 28 '20 at 21:50
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    \$\begingroup\$ Your input voltage is way, way too high. \$\endgroup\$
    – Hearth
    Dec 28 '20 at 21:56
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    \$\begingroup\$ For an amplifier with relatively minimal feedback such as here, start with a small signal (millivolts). Larger signals will distort as g_m varies, even saturating as you see here. \$\endgroup\$
    – nanofarad
    Dec 28 '20 at 22:07
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The emitter capacitor allows the transistor to have a voltage gain more than 100 times but your input level is way too high at 8.5V peak. So I removed the emitter capacitor and reduced the input level to show that the top of the waveform was clipping because the transistor is biased wrong.

Then I re-biased the transistor and reduced the level again. enter image description here

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