Re: Question about vu meter
- From: et472@xxxxxxxxxxxxxxxxxxx (Michael Black)
- Date: 10 Apr 2007 22:45:34 GMT
(hc04@xxxxxxxxxxxxxx) writes:
I have a question about the circuitThe first transistor is an AC amplifier. There will be a DC level
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
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
.
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