Emitter follower o/p coupling cap question

Discussion in 'General Electronics Chat' started by samy555, Feb 11, 2014.

  1. samy555

    Thread Starter Active Member

    May 24, 2010
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    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.​
    [​IMG]
    [​IMG]
    The input signal was 300mvolt 1kHz, and the problem was that the output signal was distorted as the following picture​

    [​IMG]
    But when I removed the coupling output capacitor C1 I got a clean output signal like this:​
    [​IMG]
    How can I get a clean output signal in the presence of C1??​
    Thanks​
     
  2. crutschow

    Expert

    Mar 14, 2008
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    The problem is that you calculated values for the small signal linear operation of the amp but it is actually operating in the large signal region where the dynamic range of the circuit comes into play. As long as you keep the peak AC signal at less that 20mV or so you stay in the linear region and your calculations work. Above that you need to look at the circuit biasing which affects the large signal operation.

    R3 must supply (sink) the current through the capacitor to the load for the negative portion of the waveform, which limits the negative current. You have biased the base of the transistor at about 2.4V meaning the emitter is biased at about 1.8V. This voltage will be divided by the ratio between the 8 ohm and 470 ohm resistor on the negative half cycle. The maximum negative voltage is thus 1.8V * (8 /(8+470)) = 30mV, which is about what your simulation shows.

    To get a larger negative voltage you need to bias the base at a higher voltage and/or reduce the value of R3. Typically to drive a low impedance load such as a speaker a push-pull complimentary (class B or AB) stage is used where a NPN provides the positive current and a PNP provides the negative current. That way you don't have to waste so much bias current to get a reasonable output level.
     
  3. samy555

    Thread Starter Active Member

    May 24, 2010
    116
    3
    Thank you crutschow very much
    why you divid the VE= 1.8v between RE and the 8 ohm load?
    Why not to divid it between Zo = 8 ohm and the 8 ohm load, ie fifty fifty?
    thank you
     
  4. MrChips

    Moderator

    Oct 2, 2009
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  5. crutschow

    Expert

    Mar 14, 2008
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  6. studiot

    AAC Fanatic!

    Nov 9, 2007
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    Can you explain this further?

    3*(20/26.2) = 2.29 and this will be reduced further by any shunting effect of the transistor base current.

    Also if samy is right about Ic = 3.25mA I make Ve to be around (3.25)*(.47) = 1.53 volts

    Of course 2.29 - 1.53 = 0.76, which is a tad on the high side for a well biased 2N3904
     
  7. crutschow

    Expert

    Mar 14, 2008
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    I slipped a digit. The approximated base voltage should have been 2.3V, not 2.4V giving about 1.6V at the emitter (give or take). This would give a bias current of 1.6V / 470 = 3.4mA. But my calculations were rough just to illustrate my point. A simulation gives a base voltage of 2.24V and an emitter voltage of 1.55V, giving an emitter current of 3.3mA.
     
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