Light sensor turns sound off, I want it to turn on

[mettam]

Hi,
I am building a project found in the "Electronics for Dummies" pg 315 titled "When There's light, you hear this noise".
I am building it on a bread board and the idea is that when light hits the sensor a sound is made and when there is no light there is no sound, I have the oposite happening and I dont know what component needs to be switched for the desired result to happen.
The components are:

• IC LM555 Timer
• 2N3906 PNP transistor
• 100k potentiometer
• resistors and capasitors (1 capacitor is polarised)
• and a speaker.

I am not sure what more information I should give to make it easier for someone to give advise.

Thanks in advance
Justin

 

[Russlk]

I don't see any phototransistor or LDR (light dependent resistor) listed ? If you have one, where is it connected?

 

[mettam]

Sorry I left out the Photoresistor.
So the components are:
* IC LM555 Timer
* 2N3906 PNP transistor
* 100k potentiometer
* Photoresistor
* resistors and capasitors (1 capacitor is polarised)
* and a speaker.

The Photoresistor is connected between a 100k potentiometer and the ground (to the neg of the battery, I asume it's the same as the ground).
One lead of the transistor (2N3906 PNP) is connected between the resistor and the Photoresistor the other is connected to the +V and the 3rd lead of the transistor is connected to the 4th Pin of the 555 IC as well as a 3.9Kohm resistor.

Can you give me an idea of how a curcuit can be described in words and I will do my best to decipher my schematic to you. Or I can send you a scan of the schematic.

 

[AllVol]

Attached is a general circuit for what you are doing. I used an LED to signify the output... it could be sound, etc. As shown, the LED would glow in the presence of light. Reversing R1 and the photoresistor would give the opposite result.

 

[ZIGGY_DAN]

I have that book, i don't rate it, iv'e never tried that circuit, iv'e looked it up in the book, and what you need to do is put the photoresistor where the potentiometer is and vice versa, if you wish to learn more about electronics i suggest you get:

ISBN: 0-7195-7205-3

Success In Electroncs 2ND Edition by Tom Duncan, it is really good, lots of information about the operation of components, it's a few years old though, but still very good.

if you need anything more then just ask.

hope this helps.

 

[Roff]

Attached is a general circuit for what you are doing. I used an LED to signify the output... it could be sound, etc. As shown, the LED would glow in the presence of light. Reversing R1 and the photoresistor would give the opposite result.

AllVol, you've got it backwards. Your circuit will turn the transistor off in the presence of light.

 

[AllVol]

If it's (notice, not its) not one thing, it's a mother.

AllVol

 

[Roff]

If it's (notice, not its) not one thing, it's a mother.

AllVol

I've been wondering if anyone understood my stupid signature. I think you do.

 

[bibinjohn]

do this way. Use LDR and use LM324 comparator. Connect output of LM324 to the reset pin of 555. See
**broken link removed**
**broken link removed**

Bibin John
www.bibinjohn.tk

 

[jemilsan]

reverse the position of the photoresistor and potentiometer..

 

[mettam]

Yes reversing the R1 and PhotoResistor did the trick. Now I just have to understand why.
Thanks for all your help.

Justin.

 

[ZIGGY_DAN]

The potentiometer and the photoresistor are acting as a potential divider, the voltage between them is dictated by the ratio of there resistances, since the resistance of the photoresistor is changing with light intensity, the potential diferance across the photoresistor is changing, with the photoresistor connected to the negative rail the voltage will go lower as the resistance of the photoresistor falls, with the photoresistor and potentiometer reversed the voltage across the divider will rise as the resistance of the photoresistor falls.

potential dividers are very common, and have a wide use of applications.

potential diferance means voltage.

hope this helps

 

[AllVol]

The formula for a voltage divider is: Vout = R2/R1+R2 * Vin

 

[Roff]

The formula for a voltage divider is: Vout = R2/R1+R2 * Vin

At the risk of sounding anal, the equation needs parentheses:

Vout=Vin*R2/(R1+R2)

 

[AllVol]

You did. rofl

 

[Roff]

You did. rofl

I thought so.
However, if our OP (or any other troll) has learned how to evaluate equations, but knows squat about electronics, your equation would evaluate as follows:
Vout = (R2/R1)+(R2 * Vin)
Which ain't gonna give the right answer.