because i want to understand common collector so that i can use it in my circuit. where i am getting 8v(RMS) and 4mA at the output of my circuit. i want to amplify it to atleast 100mA with same voltage.
because i want to understand common collector so that i can use it in my circuit. where i am getting 8v(RMS) and 4mA at the output of my circuit. i want to amplify it to atleast 100mA with same voltage.
I did not read your link but will offer some info. First off an emitter follower (common collector) will give less than unity voltage. So you will have a slight voltage loss. Now, it sounds like you want allot of current. So there is a modification of an emitter follower known as the ever popular Darlington pair. The Darlington amplifier will give you beta times beta of the two transistors. I think this is the way to go for you. Here is an arrangement I use quite often using simple 2n3904 transistors. They make nice buffer amplifiers as well so you will setup your base current of the first transistor with the voltage divider. Most people use a high value to keep from loading the input and then the emitter resistor sets up the output Z which is a low impedance making it a stiff current source.
now look at my circuits. i have designed a phase shift Oscillator which is supplying me a 4.4mA with 7.945v(RMS) at the output. then i designed a common collector Amplifier (using this link given down) to amplify the current but all the things got down. i don't why. That's why i was asking about the calculations on the made on the link, I've posted in post#1.
i am attaching my circuits with and without the common collector Amplifier.
Look at the output from your oscillator. It is only 1.62V RMS, not 8V RMS because it is shorted by the extremely low values for R7 and R8. They are only 72 ohms in parallel but an opamp has trouble driving less than 1000 ohms.
Look at the reactance of the 2.3uF capacitor C4 at 977Hz. it reduces the input to the transistor to 1.14V RMS which is the same as the output.
The transistor and its emitter resistor will be extremely hot with such a high current. Why not use a class-AB push-pull amplifier instead?