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I want to calculate the amplification of the following amplifier. enter image description here

I used the re model.

enter image description here

I do the calculation this way..

enter image description here

So, in my case: enter image description here

So according to my calculation the amplification factor is 39.4.

Here's the problem. When the input signal has amplitude of 0.01v, the output wave swings from 3.54v to 2.01v Thus an amplification of (3.54 - 2.01)/0.01 = 153 (not even near what I have calculated.)

What am I doing wrong?

Ps:re value is multiplied by 2, because, I think with an Ie of only 0.48mA, the junction is in the knee region, where the resistance is more.

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  • \$\begingroup\$ Xce represents C2? \$\endgroup\$ – user28910 Jan 24 '14 at 19:30
  • \$\begingroup\$ You can't talk about gain without having two signals to take a ratio between. Your schematic shows no output, so we can't tell whether this is a common emitter or emitter follower amp. \$\endgroup\$ – Olin Lathrop Jan 24 '14 at 19:32
  • \$\begingroup\$ its common emitter \$\endgroup\$ – Arjob Mukherjee Jan 24 '14 at 19:39
  • \$\begingroup\$ @user28910 I have updated the diagram. \$\endgroup\$ – Arjob Mukherjee Jan 24 '14 at 19:56
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What am I doing wrong?

(1) You're dividing the output peak-to-peak voltage by the input peak voltage. When you say the input has an amplitude of 0.01V, that is the peak value. So, your actual voltage gain is half of what you came up with - 76.5 versus 153

(2) I don't believe the factor of 2 for your calculation of \$r_e\$ is appropriate. If you use the typical

$$r_e = \frac{26mV}{0.48mA} = 54.2 \Omega$$

instead, the calculated gain, ignoring the capacitor impedance, becomes

$$|A_v| \approx \frac{4400}{54.2} = 81.2 $$

which is close to the 76.5 value I believe your actual gain is.

Including the capacitor impedance, the gain is

$$|A_v| = |\frac{4400}{54.2 - j3.18}| = 81.0 $$

so the capacitor impedance is almost insignificant.

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  • \$\begingroup\$ thanks, so much. I sure was calculating the amplification wrong. \$\endgroup\$ – Arjob Mukherjee Jan 24 '14 at 19:50
  • \$\begingroup\$ I multiplied the re by 2, because I thought that the junction is in the knee region, as the Ie is so low. Could u please tell, if this concept is wrong, or does it occur sometime. \$\endgroup\$ – Arjob Mukherjee Jan 24 '14 at 19:54
  • \$\begingroup\$ @Arjob, I don't think that \$I_E = 480\mu A\$ is remarkably low. You should be able to look at the simulator Q-point output and determine the value of \$r_e\$ that was used. \$\endgroup\$ – Alfred Centauri Jan 24 '14 at 21:15

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