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audio amplifier design |
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| Oct14-12, 09:18 AM | #1 |
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audio amplifier design
I am studying basics of analog circuit design from microelectronics circuit analysis by donald neamen
And in that i encountered a problem in design of audio amplifier. The question and solution is given in the attachment. And i want to know that in the output emitter follower why there are no biasing resistors apart from emitter resistor ? Hope my question is clear and if not then please do tell me. I want to clarify this Thank you |
| Oct14-12, 12:00 PM | #2 |
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Good question, but it's hard to answer it without seeing the circuit for the complete amp.
You are right in thinking that SOMETHING should be controlling the base voltage of the output transistor. Maybe the output stage is direct coupled to the stage before, so the previous stage has fixed the base voltage? |
| Oct14-12, 12:50 PM | #3 |
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In the calculations in your pdf Vce for the output stage is stated to be 6 volts.
Since Vcc is 12 volts the base voltage is calculated to be (12 - 6) = 6 volts. thus the base is held at 6 volts (midway betwwwn Vcc and ground) There must be some circuitry to cause this, it just isn't shown or relevent to the section. Was the chapter on emitter followers (common collector) amps? Note it is more normal to use other circuitry than resistors to achieve this bias in CC output stages, maybe that is why it was not shown, but will be covered in a later chapter? |
| Oct15-12, 10:06 AM | #4 |
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audio amplifier design
chapter is based on bjt amplifiers................
i agree to both of you , but can you tell me " Why it is more normal to use other circuitry than resistors to achieve this bias in CC output stages ?? " I think gain stage and output buffer are dc coupled. So operating point of output buffer is fixed by gain stage. Any advantage of this ?? |
| Oct15-12, 11:55 AM | #5 |
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The impedance of any biasing network needs to be 1/10 of this for stability say 200Ω. This is very low and needs to be driven by the previous stage which can be difficult. So much of the effect of using the emitter follower is lost. |
| Oct15-12, 11:56 AM | #6 |
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The disadvantage is that the operating point may not be so well controlled when the operating temperature changes, or because of the variation of semiconductor parameters. But for your circuit, it doesn't need to be accurately controlled, because the supply voltage is a lot bigger than the peak-to-peak output voltage you need. A shift of say 1V in the operating point of the output stage isn't going to make much practical difference to anything. |
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