E
electricked
Guest
Hi all,
I've been reading some electronics theory and I've finally decided to do a
simple project. My project consists of having a microphone and 4 red LEDs.
The 4 LEDs act as a volume meter. So when someone talks in the mic, the leds
light up depending on the volume. If the volume is low, then no LEDs light
up. If it's very high, then all 4 LEDs light up. If it's somewhere in
between only 2 LEDs light up.
I'm thinking on loading the circuit from a 9V battery. So my initial
thoughts were to use a transistor and build the circuit as follows:
R=105ohms
___
-------------|___|----------------------
| | | | |
--- | | | |
- VCC 9V | | | |
| | | | |
| | | | |
| V V V V
| - - - -
| | | | |
|/ | | | |
-----------| | | | |
| _ |> | | | |
|-/ \ | | | | | |
(Mic)| | | | | |
|-\_/ | | | | | |
-----------------------------------------------------
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
First of all, how does a microphone work? Do I need to apply voltage to it
in order for it to work, or does it produce current on it's own when noise
is detected?
What I was thinking is having the microphone switch a transistor if someone
is talking and then depending on the loudness of the volume then the
transistor will let current flow from C to E accordingly. At least that's my
understanding of transistor so far. I'm thinking, if the base current is
lower, then the current in main circuit will be less, if current at base is
higher, then current in main circuit will be higher. Having this figured
out, I was thinking the transistor will drop 0.6 volts so now I have
9-0.6=8.4V when the transistor is fully open. So next I was thinking I'd let
the max current be 80mA since 4 LEDs times 20mA each in parallel will be
80mA max current required for them to light up fully when the transistor is
fully open. So I'll need a resistance value to limit the current in the main
circuit. So I have R=V/I=8.4/0.08=105OHMs. So that's my resistance. Next
thing, I figured say the transistor is half way open it will allow 40mA to
pass and only two LEDs will need to be lighted up. But if I connect all 4
LEDs in parallel then isn't the 40mA going to split into 4 10mA for each LED
connected in parallel? So this is not the effect I want. This will light up
all LEDs but with less brightness. What if I connect the LEDs in series?
Then I'll have to change the resistance since only 20mA max have to be
allowed at fully open transistor. But then if the LEDs are in series and I
have 20mA then the brightness depends on the voltage, right? So each LED
will drop about 1.6V is it?
So I'm stuck here. If I put the LEDs in parallel will that get me the effect
I want? Basically, if the transistor is half open only two LEDs will light
up at their full brightness and the last two in series won't light up at
all. Is that going to work? How do I figure the value for the resistor in
that case?
All help is welcome. Sorry for the long-winded post. I'm just trying to put
my thoughts into writing so it makes sense.
Thanks!
--Viktor
I've been reading some electronics theory and I've finally decided to do a
simple project. My project consists of having a microphone and 4 red LEDs.
The 4 LEDs act as a volume meter. So when someone talks in the mic, the leds
light up depending on the volume. If the volume is low, then no LEDs light
up. If it's very high, then all 4 LEDs light up. If it's somewhere in
between only 2 LEDs light up.
I'm thinking on loading the circuit from a 9V battery. So my initial
thoughts were to use a transistor and build the circuit as follows:
R=105ohms
___
-------------|___|----------------------
| | | | |
--- | | | |
- VCC 9V | | | |
| | | | |
| | | | |
| V V V V
| - - - -
| | | | |
|/ | | | |
-----------| | | | |
| _ |> | | | |
|-/ \ | | | | | |
(Mic)| | | | | |
|-\_/ | | | | | |
-----------------------------------------------------
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de
First of all, how does a microphone work? Do I need to apply voltage to it
in order for it to work, or does it produce current on it's own when noise
is detected?
What I was thinking is having the microphone switch a transistor if someone
is talking and then depending on the loudness of the volume then the
transistor will let current flow from C to E accordingly. At least that's my
understanding of transistor so far. I'm thinking, if the base current is
lower, then the current in main circuit will be less, if current at base is
higher, then current in main circuit will be higher. Having this figured
out, I was thinking the transistor will drop 0.6 volts so now I have
9-0.6=8.4V when the transistor is fully open. So next I was thinking I'd let
the max current be 80mA since 4 LEDs times 20mA each in parallel will be
80mA max current required for them to light up fully when the transistor is
fully open. So I'll need a resistance value to limit the current in the main
circuit. So I have R=V/I=8.4/0.08=105OHMs. So that's my resistance. Next
thing, I figured say the transistor is half way open it will allow 40mA to
pass and only two LEDs will need to be lighted up. But if I connect all 4
LEDs in parallel then isn't the 40mA going to split into 4 10mA for each LED
connected in parallel? So this is not the effect I want. This will light up
all LEDs but with less brightness. What if I connect the LEDs in series?
Then I'll have to change the resistance since only 20mA max have to be
allowed at fully open transistor. But then if the LEDs are in series and I
have 20mA then the brightness depends on the voltage, right? So each LED
will drop about 1.6V is it?
So I'm stuck here. If I put the LEDs in parallel will that get me the effect
I want? Basically, if the transistor is half open only two LEDs will light
up at their full brightness and the last two in series won't light up at
all. Is that going to work? How do I figure the value for the resistor in
that case?
All help is welcome. Sorry for the long-winded post. I'm just trying to put
my thoughts into writing so it makes sense.
Thanks!
--Viktor