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This link feature a video starting at a specific time: https://youtu.be/t0UOSIUve9E?t=16m24s

  1. For a common collector configuration, The teacher chooses the emitter voltage (output voltage) to be the half way between Vcc and gournd, Why?

  2. Also, He does not explain why he choosed 5mA as a quiescent current?
    What considerations should I take care of while choosing those values?

  3. If I'm using PN2222A transistor and I want to drive 8-ohm speaker as a load (audio amplifier). What would be the values of emitter voltage and quiescent current ?

Datasheet:https://www.fairchildsemi.com/datasheets/PN/PN2222A.pdf

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  1. By setting the output voltage of the circuit to about half the power supply voltage, the output can swing up about equally as much as it can swing down. For many simple input signals (eg. audio), this usually what you want. In practice you want to keep some headroom to prevent distortion.
  2. The reason for choosing 5mA is in some sense similar to #1. It means in "worst case scenario" the input signal can reduce the current through the transistor by 5mA, resulting in 0mA. Again with the audio signal as an example this would mean a maximum current of 5+5=10mA. This is somewhat theoretical, in practice you will want some headroom to prevent distortion. The choice highly depends on the load you want to drive and of course the capabilities of the transistor.
  3. Hard to say how to size the circuit for an 8 ohm speaker, it largely depends on input amplitude of the signal. I would personally use a different architecture for such a low impedance load. You could try your luck with an output series capacitor in the order of 220µF (+ goes at emitter, speaker goes to ground). When driving an el-cheapo speaker and being tied to this circuit, I would personally probably just try my own luck by putting the speaker in the collector lead.
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  • \$\begingroup\$ "Hard to say how to size the circuit for an 8 ohm speaker". Would you suggest me a suitable load? and What would be the input? I need to practice what I've learned on a breadboard. \$\endgroup\$ – Michael George Sep 26 '15 at 6:34
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    \$\begingroup\$ @MichaelGeorge What signal source will you be using? You really want the AC load current to be smaller than the quiescent current. Ohms law tells us that 5mA × 8 ohm = 40mV. Your input voltage would be 40mV max (voltage gain is approximately 1), but you probably shouldn't go much further than half of that because you don't want too much distortion. It may sound weird, but for experimenting you may want to consider a PC's active loudspeaker for starters. Then you can change the load by adding extra resistors in parallel to that input and see how it changes the result. \$\endgroup\$ – jippie Sep 26 '15 at 11:58
  • \$\begingroup\$ The signal source will be the output of a cell phone (earpiece socket). "You really want the AC load current to be smaller than the quiescent current." What if I need to drive large current load (50mA)? Should I choose a large quiescent current more than (50mA)? or Can I remove the emitter resistor out of the circuit and put my load instead? \$\endgroup\$ – Michael George Sep 26 '15 at 18:24
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    \$\begingroup\$ If you need a larger current, you'll have to redesign the circuit, taking into account everything mentioned in the video and on top of that component selection and also take thermal limitations into consideration. It depends on your load whether you can drive it with a DC bias or not (which is essentially what you do when you replace the emitter resistor with the load). On the other side it doesn't hurt to experiment a little, you'll learn a lot from that. Just be careful not to overload your phone's output. \$\endgroup\$ – jippie Sep 26 '15 at 20:14

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