Voltboy

Hey,
I was building a circuit, and it didn't worked for me.
When the QSE113 receive light so voltage is higher than the voltage form the pot voltage divider the output must go high. In my circuit the Output is always high, even when I put my finger over the QSE113
My question is why there is R2 because there is an open way to +V always.
I heard something of pull-up resistor, but I don't heard much, if this is the reason please tell me.
And another thing is that instead of 5v I used 3.7V (cellphone battery).
Oh.. and R1 is 10k
Thanks

Attached Images

circuit1.GIF (3.9 KB, 17 views)

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audioguru

R1 should be about 47k ohms or a very bright light is needed to turn on the photo-transistor.
Maybe your photo-transistor is connected backwards.

Look at the datasheet for the LM339 quad comparator or the LM393 dual comparator. Their output is the collector of an NPN transistor and they need a pull-up resistor to the positive supply or the output will never be high.

[quote]And another thing is that instead of 5v I used 3.7V (cellphone battery).[quote]
The minimum supply voltage of the comparator is 2.0V but its inputs don't work when they are higher than 1.5V less than the positive supply voltage.

Attached Images

QSE113.PNG (4.8 KB, 1 views)

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Uncle $crooge

Voltboy

So if my Vcc is 3.7 my inputs most be higher than 2.2 because Vcc-1.5 which is 3.7-1.5=2.2?
So to do this I set the pot to divide to 2.2 or more volts and make a way the the PT produce more than the pot divider voltage?

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audioguru

No.
The inputs must be less than +2.2V.

The pot must be set to +2.2V or less.

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Uncle $crooge