[SOLVED] Lm339 comparator with photodiode.........................

[kdg007]

what is the purpose of 1K resistor ? can i use the same circuit without it ?
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its a circuit in my lab,i am trying to figure out the use of the 1k resistor,...

thank you

 

[mister_rf]

The 1k series resistor help protecting the photodiode for adjusting the 1M variable resistor to a very low value… :wink:

 

[kdg007]

thank you..........................

 

[Pjdd]

I'll answer your PM question about the purpose of the 1M pot here instead of by PM. mister_rf has already explained the purpose of the 1k resistor.

The pot is a means of adjusting the current through the LED. If the LED is a red one, the current can be adjusted from about 3.5uA to 3mA. Further explanation is difficult without knowing the rest of the circuit.

 

[kdg007]

what if its a photodiode instead of an LED,what is the purpose the 1M pot ? its the same circuit !! nothing new added.

 

[Pjdd]

It's unlikely to be a photodiode. First, the direction of the arrows should be reversed in a photodiode symbol. Secondly, a photodiode should normally be reversed in polarity; that is, the cathode should point upwards in the diagram.

its the same circuit !! nothing new added.

The circuit as shown is quite meaningless by itself. LED current is adjusted with the 1M pot. The forward drop Vf of the LED changes non-linearly with the LED current. The voltage at pin 6 of the comparator is set by the 5k pot, and that at pin 7 is set by Vf. When Vf is lower than pin 6, the LM339 output goes low. When Vf rises above pin 6, the output goes high.

That's all that can be said without knowing what the circuit is supposed to do. When asking questions about a circuit which is obviously a part of a larger circuit, it's best to explain what the whole thing is supposed to do. Otherwise, we're working in the dark. It places an unnecessary burden on the person who's trying to help. He has to make guesses and assumptions which may turn out to be wrong.

 

[kdg007]

this is the complete design .. the output from the 2.2k pull up resistor is a square wave ...

 

[Pjdd]

Now that's a completely different circuit - different not only in the details, but in the purpose and functioning principle. There's still one error: the output of the LM339 is shorted to ground. The junction of pin 1 and R4 should not be connected to ground.

D1, R2, R1 are in series across the positive supply of 5V. When the D1 is in the dark, only a small leakage current (also called dark current) flows through the series circuit, causing only a small voltage drop across R1+R2 (let's call this V1-2). When light falls on the photodiode, its leakage current increases, raising V1-2. This raises the input voltage of the comparator at pin 7.

V1-2 is compared with the voltage at pin 6 which is set by R3. If R2 and R3 are properly set, the voltage at pin 7 is lower than that at pin 6 when D1 is in the dark. When light falls on D1, pin 7 rises above pin 6. This causes the LM339 output to switch between high and low states, depending on whether or not there's light falling on D1. Thus, shining pulses of light on D1 causes the LM339 to produce a rectangular wave at its output.

Note: It should now be apparent how an incorrect and/or incomplete diagram can cause a lot of confusion. The error or the missing portion may look unimportant to the inexperienced, but it can cause a big difference in the way an experienced person analyses the design.

 

[kdg007]

Wish i could think the way you do i am still learning a lot ...
so,what is the purpose of 1M pot ,does is have to be low as possible ..i mean we can keep a normal resistor of 1M instead of a 1M POT ? and 2.2k is to protect the photodiode ?