Emitter follower o/p coupling cap question

I've designed an Emitter Follower stage working on a 3-volt battery and IC = 3.25 mA
I do the design so that the Output Impedance = 8 ohms because I want to run a 8 ohm load. I do not intend to actually run the circuit, I do it for study and analysis.
Her is the circuit and some of my calculations.




The input signal was 300mvolt 1kHz, and the problem was that the output signal was distorted as the following picture


But when I removed the coupling output capacitor C1 I got a clean output signal like this:


How can I get a clean output signal in the presence of C1??
Thanks

Your amplifier has Ic = 3.25 mA, How will you get significant current in your 8 Ohms load, without distortion? Your 470 Ohms resistor will sink maximum 8 mA (when Q1 is fully off).

You would be able to drive maximum 3 mA into your 8 Ohms load. That equals just 24 mV across the 8 Ohms load.

Do a search on complementary output stage (also called push pull output stage or push pull amplifier). That will help you further.

OK
thank you WimRFP for fast reply.
I'm try to study
I am trying to understand the emitter follower now, I do not want to go into the study of push pull amplifier
I expect that the emitter follower can run the 8 ohm load
I'm wondering, why when C1 is removed I got a clean output signal and when I put it again shows distortion in the signal!!
Is it real or because the software programs give us results far from reality sometimes????

When you remove C1, the collector current will increase significantly and then you have more output current, hence more output voltage across the 8 Ohms load. Of course, you batteries will be flat/discharged very soon.

For a class A stage, the peak output current can't be more then the bias current. In real world it will be somewhat less to have low distortion.

samy555 said:

Is it real or because the software programs give us results far from reality sometimes????

Click to expand...

Yes, it's real. The problem is not the capacitor. The problem is that the load is very low resistance, and the quiescent current is small, so you can only get a small output voltage.

You set quiescent current IC = 3.25mA so maximum undistorted output = 3.25mA * 8Ω = 26mV.

When you remove the capacitor the quiescent current is much higher so you can get higher output voltage.

godfreyl said:

Yes, it's real. The problem is not the capacitor. The problem is that the load is very low resistance, and the quiescent current is small, so you can only get a small output voltage.

You set quiescent current IC = 3.25mA so maximum undistorted output = 3.25mA * 8Ω = 26mV.

When you remove the capacitor the quiescent current is much higher so you can get higher output voltage.

Click to expand...

Yes,,, now I understand, thank you
I hope you give me some information about some of the points that must be taken into account when I want to design an emitter follower that runs on a 9-volt battery, for example, and that the load impedance = 100 ohm
thank you a gain

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WimRFP said:

When you remove C1, the collector current will increase significantly and then you have more output current, hence more output voltage across the 8 Ohms load. Of course, you batteries will be flat/discharged very soon.

For a class A stage, the peak output current can't be more then the bias current. In real world it will be somewhat less to have low distortion.

Click to expand...

I am very sorry not to be aware to your reply
Thank you
Yes, your words correctly and I benefited from it

samy555 said:

...................
Is it real or because the software programs give us results far from reality sometimes????

Click to expand...

Spice simulation seldom give results far from reality, particularly for simple circuits with good part models.