LM324N problem!

[kpsg25690]

Hey all!
Since this is my first post i should introduce myself..
My name Karan and i am an engineering student.

I have been working on a project on a project that requires a comparator.
I am using LM324n as a comparator and the circuit is working correctly.

I am using a +5V supply power to the ic but when the output is high i get around 2.1V on the output pin,which is enough to drive an LED but i want to give this output to a PIC18F4520 which will not identify 2.1V as high.

What do i do?
I have tried using a 12V supply for the IC also without any change.

I've been at it for some time and need some help...

 

[carbonzit]

It would help us a lot if you would post a schematic of your project, showing what you have connected to what. (Hand-drawn is fine if nothing else).

 

[Gobbledok]

Do you have the LED connected to the output without a resistor? That will probably give you about that voltage.

 

[Reloadron]

Do you have the LED connected to the output without a resistor? That will probably give you about that voltage.

I agree, if the output of the LM324 is directly driving the LED sans a series resistor with the LED it sounds to me like you are seeing the forward voltage drop across the LED. Place a resistor in series with the LED and measure the voltage at the output pin. Has it increased? Start with about a 100 Ohm resistor.

Ron

 

[kpsg25690]

Thanks for the reply guys!

and yeah i was driving the LED without a resistor,i followed your advice and connected a resistor in series,Now i get 3.6 V with a 2.2K resistor with o/p pin of lm324 as source and 3.7 V with the o/p pin as sink...

Which configuration should i use?

and if i connect the output pin to the PIC18F4520 as input would the uC identify it as logic 1? (i think it should)

 

[Gobbledok]

According to the datasheet, the LM324's output can swing within 1.5 v of VDD. That's why you're only seeing 3.6v out.

According to the PIC datasheet, you need 0.8 x VDD on a schmitt trigger input for it to be considered a logic 1.

0.8 x 5v = 4v so your output will not be high enough.

You need to either use a different device (a rail-to-rail op amp or a comparator) or place something on the output (a transistor) to level shift.

 

[Reloadron]

That sounds about right. The LM324 is not a rail to rail operational amplifier so the output voltage swing will be 0 volts to VCC -1.5 or in your case 5 - 1.5 = about 3.5 volts.

Large output voltage swing:0VDC to VCC-1.5VDC

Given a choice if you need a LED in there between output and PIC I would let it source and see how that goes. That short of adding a transistor or anything else.

<EDIT> Gobbledok is way ahead of me. </EDIT>

Ron

 

[kpsg25690]

I considered using LM339 but i would have to use pull ups.
Could you suggest a rail to rail amplifier IC or a comparator IC that is easily available and meets the requirements?

 

[Reloadron]

I would have used the LM339 since what you want is a comparator. You are only looking at a few resistors added to the circuit and then you are using a chip designed more for what you are after, that being a comparator. As to single supply rail to rail operational amplifiers you may want to give this a read. For a single comparator there is the LM311, for a dual the LM393 and for a quad the LM339 all of which are very common and inexpensive choices.

Ron

 

[Gobbledok]

I considered using LM339 but i would have to use pull ups.
Could you suggest a rail to rail amplifier IC or a comparator IC that is easily available and meets the requirements?

MC33201 would suit nicely if you only need one op amp.

MC33202 if you need 2.

 

[kpsg25690]

thank you for your reply guys.
you really helped me out.

 

[MikeMl]

Why not reverse the inputs to the LM324, and wire the LED/resistor between 5V and the LM324 output?
The LM324 can drive it's output within mV of ground when it is sinking current.