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
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.
--
Posted via a free Usenet account from http://www.teranews.com
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.
schreef in bericht news: snipped-for-privacy@n33g2000cwc.googlegroups.com...
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.
ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.