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Transistor base-emitter voltage pulled down

Discussion in 'General Electronics Discussion' started by acko1989, Jan 19, 2020.

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  1. acko1989

    acko1989

    5
    0
    Jan 19, 2020
    Hi,

    I have a question related to the circuit in the attachment down below. This is a circuit where LED's glows in music rhythm received from smartphone speaker (or any other source) that fall on electric mic. When there is no sound signal (or any other vibration near mic), Q1 transistor is on, and other transistors are off because there is not enough voltage to the bases of other transistors. With my pocket oscilloscope, I noticed that when there is a sound wave to the mic, voltage between base and emitter of Q1 is pulled down, thus transistor Q1 is opened (no current from collector to emitter) and other transistors are conducting, so LED's glows. My question is: Why is this so? Why is base pulled low when there is sound wave?
     

    Attached Files:

  2. Cannonball

    Cannonball

    191
    53
    May 6, 2017
    When the voltage on the base of Q1 goes down its collector goes high. Q1's collector is directly connected to Q2 ,Q3,and Q4 at their bases pulling them up. The collectors are pulled down. The collectors pull the cathodes of LED 1, 2 and 3 causing the current through them to increase making them to glow brighter.
     
  3. Audioguru

    Audioguru

    3,116
    696
    Sep 24, 2016
    With no sounds, C1 has about +4V at the microphone end and about 0.6V at the base of Q1.
    Sounds cause the voltage at the microphone to wave up and down. The base-emitter diode of Q1 does not allow the base end of C1 to go up so it is charged more by the mic's resistor R1.. When the microphone signal waves down then the base of Q1 is pulled down by the extra charge on C1.
     
    Last edited: Jan 20, 2020
    acko1989 and Cannonball like this.
  4. Nanren888

    Nanren888

    461
    135
    Nov 8, 2015
    I gurss the result depends somewhat on the nature of that microphone but if it is the source of an ac waveform, impressed on the left of the capacitor, c1, then some proportion of that influence results in a current into the base of Q1. If the base emitter junction of Q1 ultimately looks like a diode, I guess the base will ramp downwards for a time, with each positive cycle.
    Have I missed something important?
     
  5. Audioguru

    Audioguru

    3,116
    696
    Sep 24, 2016
    I edited my previous post.
    I think C1 charges more only for very loud sounds.
    Each positive-going half-wave from the mic that has R1 pulling the capacitor up, C1 charges more because the base-emitter diode of the transistor does not allow the base end of C1 to go up, resulting in C1 charging more. Then when a negative-going wave from the mic goes down, C1's extra charge pulls down the base of Q1.
     
    acko1989 and Cannonball like this.
  6. acko1989

    acko1989

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    0
    Jan 19, 2020
    Thank you guys for clear explanation.
     
  7. Cannonball

    Cannonball

    191
    53
    May 6, 2017
    C1 blocks DC and passes the audio signal to modulate the base of Q1 witch amplifies the audio signal
    that drives the bases of Q2,3, and Q4. witch amplifies the signal even more and modulates the cathodes of
    LEDA 1 2 and 3 causing the brightness of the LEDs to varies in sequence with the audio.
     
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