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Question about vu meter

Discussion in 'Electronic Basics' started by [email protected], Apr 10, 2007.

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  1. Guest

    I have a question about the circuit

    http://www.velleman.be/downloads/0/manual_mk115.pdf

    This circuit was as follows,

    when we make noise near the microphone the leds will
    light up. The louder the noise the more leds will come
    on.

    I am trying to figure out the schematic, but my logic seems
    to arrive at an opposite out come, i.e. the louder the noise
    the less number of leds will come on,

    any noise near the mic will produce a voltage which will
    make T1 conduct more current. This will result in a voltage
    becoming lower at base of T2. This will lead to a voltage
    drop at emitter of T2 as T2 is configured as an emitter
    follower. This leads to less transistors in the chain coming
    on i.e T3 to T7

    What am I doing wrong?

    Thanks in advance
     
  2. john jardine

    john jardine Guest

    Yes. The noise will make T1 conduct more current but ... it will then go on
    to make it conduct -less- current and it this that lights the LEDS.
    Remember, noise or sound tones go positive and then go negative and then
    positive and so on.
     
  3. Lord Garth

    Lord Garth Guest

    The ground appears to be missing at the bottom of the schematic as well.
     
  4. The first transistor is an AC amplifier. There will be a DC level
    on the collector, but it will go up and down as the signal into
    microphone (and hence the base of the transistor) varies. That
    AC voltage on the output not only reflects the tone of your voice
    (a single tone would go up and down at the rate of the tone),
    but the amplitude (whistle softly and the AC voltage would be small
    compared to whistling louder). So the signal at the collector is
    the same as the signal at the base, except it's a bigger variation because
    the transistor is amplifying the signal.

    The second transistor acts as a detector, stripping off the variations
    of the tone but following the overall amplitude of the signal. Note
    the capacitor on the emitter, it is a large value so the higher
    frequency variations will not appear there, but it will still follow
    the relatively slow amplitude variations.

    So at the emitter of the second transistor, and hence the base of
    the driver transistors, there is a DC voltage that varies with
    the amplitude of the audio signal into the microphone.

    Then the driver transistors turn on at a somewhat higher voltage the
    further right you go, to turn that varying DC voltage into light.

    Look at the transistors driving the LEDs. The transistor won't conduct
    until the base is more positive than the emitter by about 0.7V (assuming
    it's a silicon transistor, that's the voltage drop of the base to emitter
    junction). Since the emitter of the lowest transistor is connected
    to ground, then the base voltage just has to go above 0.7v for the
    LED in the transistor's collector to turn on (since the collector
    will be low at that point).

    The next transistor does the same, but there is a diode between its
    emitter and ground. Again assuming a silicon diode (I'm too lazy
    to look at the fine print), the transistor's base needs to go an
    extra 0.7v above ground (ie 0.7v for the base-emitter junction and
    then another 0.7v for the diode from the emitter to ground), so that
    second transistor won't conduct until the base is more positive than
    1.4volts.

    Each of the subsequent transistors have such a diode going from
    their emitter. But instead of the cathode going to ground, they
    go to the anode of the previous diode. SO by the time you get to
    the rightmost transistor, its base has to be about 3.5v above ground
    in order to turn on.

    Michael
     
  5. The microphone produces an AC voltage. The louder the noise, the higher the
    voltage swing. This voltage is amplified by T1. T2 acts as a half wave
    rectifier. When the voltage on the collector of T1 is high, T2 conducts and
    loads C4. When the voltage on the collector of T1 falls, T2 blocks and C4
    slowly discharges via the 10k resistors and the base-emitter junctions of
    the transitors T3-T7.

    petrus bitbyter
     
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