Why do I need to use two NE555 in this circuit?

• Apr 22, 2006

• #1

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CommanderKeen

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**broken link removed** on the folowing link, we can see 2x NE555 one operating at 5hz, other at 40khz.

But is it necesery to use 2x NE555, can't the job be done with just one?

Can someone please, write a little "resume" over what happens in that circuit, other then what he has written.

 

• Apr 23, 2006

• #2

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Miguel Gaspar

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formulas para el ne555

Well, the project is an intermitent 40kHzand the first 555 is to make the second 555 operate in a intermitent way (5kHz).

Added after 3 minutes:

if you eliminate de first 555.

your circuit transmit a continuous 40kHz signal.

 

• Apr 23, 2006

• #3

ahmad_abdulghany

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ne555 40 khz

You may use NE556, it contains internally two 555 ICs, if you aim to use one IC instead of two...

I hope that helped you,
Regards,
Ahmad,

 

• Apr 23, 2006

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ne 555 sensor

Well i have already build the circuit. Okay thanks, so if i remove the first NE555 (5hz) it would be a stabil 40khz output.

BTW can someone show me a illustration when something is intermitent

 

• Apr 23, 2006

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IanP

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ne556 (in place of or instead of) ne555

Intermitent:

Sometimes you may face intermitent downtime of EDA board ..
From time to time there will be a database error message ..
Another time it simply says "cannot connect to server" ..

Regards,
IanP

 

• Apr 23, 2006

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ne555 40khz

Ok thanks. And the for last, what formulars to calculate the resistors in reciever schematic, if we say that the comparator is a LM339

 

• Apr 23, 2006

• #7

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ne555 40khz circuit

In circuits like that there is no "formula" as a such ..
For example, resistors R4 and R5 are used to set 1/2 of the supply voltage at the inverting input of the comparator .. Almost any values of resistors will work correctly - you can use 1kΩ, 10kΩ, or as in this example 100kΩ ..
Instead of fixed resistors you may consider a 100kΩ potentiometer which will allow you to set the sensitivity of this receiver ..
All other values are not critical, too .. If you select R2=1kΩ you can lower the value of R3 to, say, 200kΩ .. and so on ..

Regards,
IanP

 

• Apr 23, 2006

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Re: NE555 problematic!

Thanks once again. But why such a low value for R1 (100ohm). ?

 

• Apr 23, 2006

• #9

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Re: NE555 problematic!

R1 and C1 build a local supply voltage filter for the sensor ..
If the value of R1 is much bigger than 100Ω the and because the sensor needs (draws) some current the voltage may drop, the sensor will be fed of to low voltage and may not work at all ..
This value is not critical and if you have 47Ω or 82Ω resistor, you can use them ..

Regards,
IanP

 

• Apr 23, 2006

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Re: NE555 problematic!

Thanks again for the fast replies.

Added after 11 minutes:

and the 1N4148 works as a switching diode.?

 

• Apr 23, 2006

• #11

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Re: NE555 problematic!

The reason you have 5Hz switchig 36kHz is that the 36kHz is the carrier and the 5Hz is the modulatig signal ..
The output of the TSOP sensor will be the demodulated 5Hz signal ..
To produce a usable control signal for the comparator (+) input this 5Hz signal has to be converted in a dc voltage and that is done by a negative peak detector build aroud the 1N4148 diode, R2 and C2 ..
So to some extend one can say that the diode's function in this circuit is to work as a switch ..

Regards,
IanP

 

• Apr 23, 2006

• #12

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Re: NE555 problematic!

Well thank you. It all makes some sence now. If I could give you 10.000 point, you would have them

 

• Apr 23, 2006

• #13

ahmad_abdulghany

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Re: NE555 problematic!

CommanderKeen said:

If I could give you 10.000 point, you would have them

Click to expand...

IanP is one of the most active and helpful members on the board, i think even if you gave him 1000,000,000 points, this will be not enough!

This is the case also for too manyothers such as tsb_nph, haadi20, flatulent, ... ando too many others sorry if i didn't mention some one, i don't mean they are only helps!

Thank you all that help people!