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how to convert the signal received by Tsop1738 ir receiver to steady output?

Thread starter #1
Hi All ,,

I want to convert the signal received by IR Receiver Tsop1738 to steady output level for example 1 if there is ir signal 0 if no Ir signal received(an object cross Ir beam).


the signal received by the receiver as the following:


hhh.jpg

>>as you see in figure above the received signal is square wave 32hz(2ms off/29ms on).
>> if there is no signal received due to object cross the ir beam the output of the tsop1738 goes high and still high while the object cross the beam.
>>I see i can do what i need using 555 timer and other ic's like 74123 but i don't want to use them
>>minimum time for crossing the beam 20ms.



>>The transmitter circuit

jjjjjjjjjjj.jpg
as you see the circuit send 38khz for 2ms and off for 29ms.

Regards
 
Last edited by a moderator:

alec_t

Well-Known Member
Most Helpful Member
#6
Will this do?
MissingPulse.PNG
 

crutschow

Well-Known Member
Most Helpful Member
#11
Here are a couple missing pulse detector circuits as alternatives to a 555 circuit.
 

gophert

Active Member
#12
Here is 0-5V out. No 555 but a NPN and PNP to keep from inverting and avoiding a diode voltage drop at the output.

The envelope filter will react to a loss of signal in 0.5 mSec (5 time constants).



7C6BD455-4597-439D-97DF-48BC6AFD8469.png
 
Thread starter #15
Here are a couple missing pulse detector circuits as alternatives to a 555 circuit.

Screenshot_20180611-002757.png

*Can I use lm358 instead of lm393 ?
**For my received signal should I change the value of capacitors and resistors ?
***what is the purpose of c2 and D1?
****which the two circuit better using lm393 or using 4093 ?

Regards:)
 

crutschow

Well-Known Member
Most Helpful Member
#17
Can I use lm358 instead of lm393 ?
No.
The circuit requires the open-collector output of the comparator to operate properly.
For my received signal should I change the value of capacitors and resistors ?
As stated in my writeup, you select the R1C1 value to be somewhat longer than the pulse period.
what is the purpose of c2 and D1?
C2 is to differentiate the pulse edge, giving a short pulse trigger the circuit, and also block the DC bias from the input.
D1 clamps the differentiated pulse from going significantly below ground.
which the two circuit better using lm393 or using 4093 ?
The LM393 circuit's trigger levels, determined by resistors, are more accurate than the 4093's. which depend upon the internal hysteresis value of the 4093 Schmitt trigger input.
But either circuit should be more than adequate to do the job.
 
Thread starter #19
No.
The circuit requires the open-collector output of the comparator to operate properly.
As stated in my writeup, you select the R1C1 value to be somewhat longer than the pulse period.
C2 is to differentiate the pulse edge, giving a short pulse trigger the circuit, and also block the DC bias from the input.
D1 clamps the differentiated pulse from going significantly below ground.
The LM393 circuit's trigger levels, determined by resistors, are more accurate than the 4093's. which depend upon the internal hysteresis value of the 4093 Schmitt trigger input.
But either circuit should be more than adequate to do the job.

Thank you :)

The received signal period about 31 ms .
Is this mean Any object cross the ir signal in time below 31ms can't detect it ?
 

crutschow

Well-Known Member
Most Helpful Member
#20
The received signal period about 31 ms .
Is this mean Any object cross the ir signal in time below 31ms can't detect it ?
Not 100%.
If one pulse is interrupted, then the output will go high for one pulse.
If the object passes so fast that it doesn't block at least one pulse, then obviously it can't be detected.

A block of one pulse is shown in the simulation below for an R1C1 of 31ms.
The output goes high about 33ms after the loss of one input pulse and stays high until the next pulse is received.

upload_2018-6-11_20-2-41.png
 

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