Hello spidermanIIII

"Eaten" in that sentence.
It is an idiom used in some countries.
figuratively is like saying that disappears. because it eats the thing disappears. anything.

We're going to do an experiment.

We know from their 556 data sheets:
That setting it as a mono-stable, the IC changes the state of its outputs on the negative transition of a pulse applied to any of its TR(2 or 8) inputs. -TRigger.
Of course if we apply a pulse to the TR(6); Its output PIN 5 will change state.
And if we apply a pulse to the TR(8); Its output PIN 9 will change state.

Any of its outputs: PIN 5 or 9:
The time remaining in that state, is determined by: RC time constant.
This time constant is formed by a capacitor and a resistor connected to its input TH(2) -Threshold.
The resistor goes to the Vcc, the capacitor goes to ground.

In our particular case we have:
R3 = 33000 Ohms.
C3 = 0.0000001 Farads.
So the T (tau) is 33 milliseconds .. Remember I changed the value of C3.

Let's see if this is true.
Familiarize yourself with the picture Testing NE556.PNG attached.
Or the image that appears in Document Testing NE556.PDF attached.

Now look at the picture Testing NE556 (2).PNG attached.
As we can see, the oscilloscope is set as follows:
Horizontally, 5 milliseconds per division.
Vertically, all channels 1 volt per division.

Look at the upper white arrows in the image:
Each negative transition (yellow line)
The Q on the 556 changes state. (Red line).
A transition is jump because it is already in its second state, High.
When tau time (blue line) Ends, the Q 556 returns to its stable state, Low.

Taking the blue line, could you please measure the time and amplitude as the green arrows are pointing ?
With the information you get. calculates the percentage of voltage to the blue line reaches the time the Q 556 returns to its stable state.
Remember that the applied voltage is 6 Volts.
Considering the first white arrow near the blue line.
How much time does take the blue line to reach the voltage you have measured ?

Surely the results you obtained are similar to those we have been discussing.
The inaccuracies are the result of ISIS Proteus Simulator.
Errors in out readings.
Or the error rate of the value of the components used in the circuit.

I hope I have made ​​mistakes in the above.

dear MrCarlos i saw the file you attached and i have some notes i hope you help me with. from this design you choose
R9=1000 ohm
C12=0.00001 F
frequency=50 Hz
so
tau=0.01
and the time period=1/50=0.02
that mean that capacitor C12 will charge and that mean the IR sensor signal which here is 50 Hz will be pure dc but when i use oscilloscope i found it is still pulses what is the wrong

 

Hello spaidermanIIII

You say:
-MrCarlos dear I saw the file you attached and I have some notes I hope you help me with.
Sure I'll help you. but try to follow me. otherwise lose the way.

Forget now the IR sensors and R9, C12 please.
Any way:
I am simulating the IR sensor signals through generators that look in the upper left part of the image that I attached to you, has the name: Testing NE556.PNG.
These generators are set at a frequency of 50Hz.
One has a pulse width of 50%.
The other has a pulse width of 98%.
This part of the circuit is to facilitate the reading of the famous time tau.
There is nothing wrong.

So let me know the results I am requesting in my previous message.

Good luck.

 

Hello spaidermanIIII

You say:
-MrCarlos dear I saw the file you attached and I have some notes I hope you help me with.
Sure I'll help you. but try to follow me. otherwise lose the way.

Forget now the IR sensors and R9, C12 please.
Any way:
I am simulating the IR sensor signals through generators that look in the upper left part of the image that I attached to you, has the name: Testing NE556.PNG.
These generators are set at a frequency of 50Hz.
One has a pulse width of 50%.
The other has a pulse width of 98%.
This part of the circuit is to facilitate the reading of the famous time tau.
There is nothing wrong.

So let me know the results I am requesting in my previous message.

Good luck.

i open testing NE556 and i calculate
tau=35 ms
v=4 volt

 

Hello spidermanIIII

Numbers calculated, message #20:
In our particular case we have:
R3 = 33000 Ohms.
C3 = 0.0000001 Farads.
So the T (tau) is 33 milliseconds .. Remember I changed the value of C3.

Your readings and calculations, Message #23:
i open testing NE556 and i calculate
tau=35 ms
v=4 volt

Well, the difference in time tau is 2 ms.
which represents the 6.06%, I think it's pretty good.

Okay, well:
Take a good study to document that I attached: TSOP1738 IR Sensor Vishay Siliconix.PDF.

On page 1 of document:
In the box titled: Available types for different carrier frequencies, we found that TSOP1738 has, in the column fo, 38KHz.

In the paragraph entitled: Description, we can read that: The demodulated output signal can be ecoded Directly by a microprocessor.
This responds to written by: wayneh in post #15.
And erase, if it could, the third paragraph of my post #3.

Let's go to page 2.

In the box titled: Absolute Maximum Ratings.
What you read in this box, related to PIN3 ?

Study the drawing titled: Application Circuit.
Re-read the quotes *, **.
They seem, even a little, the circuits shown in our old Fig. 1.
I am referring to the devices:
C10, C7, D4, R1 and
C9, C8, D3, R2
In that old figure.

Let's go to page 3.

Studied very, very carefully the paragraph entitled: Suitable Data Format.

In the last pages of this document we can find several graphs.
These graphs, we should study them thoroughly.

Conclusions:

The NE556 IC in its stable state and when it is polarized the Q’s outputs of all NE556 are at low level.
The RS Flip-Flip (N1, N2) is in its reset state, Q = 0.
Thus the pump is off.

Each IC NE556 receive a low or high signal in any input PIN’s 6 or 8.
Each IC NE556/2 responds according to these signals. /2, Because they are two in each pack, IC.

I find some difficulty in explaining, in brief, the full operation of the circuit containing the attached document in its original spidermanIII Post.

Either way, I think in the paragraph entitled: Working of the system on page 24, describes perfectly what happens when the float reaches the points mentioned there.

 

Hello spidermanIIII

Numbers calculated, message #20:
In our particular case we have:
R3 = 33000 Ohms.
C3 = 0.0000001 Farads.
So the T (tau) is 33 milliseconds .. Remember I changed the value of C3.

Your readings and calculations, Message #23:
i open testing NE556 and i calculate
tau=35 ms
v=4 volt

Well, the difference in time tau is 2 ms.
which represents the 6.06%, I think it's pretty good.

Okay, well:
Take a good study to document that I attached: TSOP1738 IR Sensor Vishay Siliconix.PDF.

On page 1 of document:
In the box titled: Available types for different carrier frequencies, we found that TSOP1738 has, in the column fo, 38KHz.

In the paragraph entitled: Description, we can read that: The demodulated output signal can be ecoded Directly by a microprocessor.
This responds to written by: wayneh in post #15.
And erase, if it could, the third paragraph of my post #3.

Let's go to page 2.

In the box titled: Absolute Maximum Ratings.
What you read in this box, related to PIN3 ?

Study the drawing titled: Application Circuit.
Re-read the quotes *, **.
They seem, even a little, the circuits shown in our old Fig. 1.
I am referring to the devices:
C10, C7, D4, R1 and
C9, C8, D3, R2
In that old figure.

Let's go to page 3.

Studied very, very carefully the paragraph entitled: Suitable Data Format.

In the last pages of this document we can find several graphs.
These graphs, we should study them thoroughly.

Conclusions:

The NE556 IC in its stable state and when it is polarized the Q’s outputs of all NE556 are at low level.
The RS Flip-Flip (N1, N2) is in its reset state, Q = 0.
Thus the pump is off.

Each IC NE556 receive a low or high signal in any input PIN’s 6 or 8.
Each IC NE556/2 responds according to these signals. /2, Because they are two in each pack, IC.

I find some difficulty in explaining, in brief, the full operation of the circuit containing the attached document in its original spidermanIII Post.

Either way, I think in the paragraph entitled: Working of the system on page 24, describes perfectly what happens when the float reaches the points mentioned there.

dear Mycarlos i hope to answer my question . when i see your simulation in Proteus i tried to put oscilloscope and see the signal on pin 8 of 556 ic
and i found the signal is pulses at frequency but you told me that signal will be eaten by capacitor 12 . second i hope to answer my questions in post #21 please answer me i am confused

 

! Oh boy. spidermanIIII.
Surely I not explain it as well as I should.

The generators were put into that design, just to test the time called tau.
Only for R3 and C3. Not for R9 and C12.
But say you put the oscilloscope in the NE556 PIN8.
Of course there will be pulses, as we are applying at a rate of 50 Hz. by means of those generators.
In normal operation of the circuit, there will be NO such pulses.
There will be a voltage level change according to the IR sensors.

Put figures 1, 2 and 3 in front of you.
Mentally move the float right through the capillary tube. Up or down Fig. 3.
Carefully analyze what happens with the associated components.

Easily find out: when LED1 light beam is interrupted by the float, We are coming to mind that the tank is full of water. . . is this true ?.
If the pump was on, it should go off.
When the light beam is Not interrupted nothing happens.

But when the float is at the bottom of the water tank indicates that it is empty. So the pump should turn on.
The LED2 light beam is interrupted by the float, the RX2 makes a change in its output level which comes from F to PIN8 NE556 IC. Fig. Our Old Fig 1.PNG. or simply Fig 1.PNG. both were attached.

The components: R9, C12 and R8, C11.
Ensure that the NE556, starting on their steady state when the circuit is biased.
Initially the capacitors are discharged.
When the circuit is biased, the capacitors are charged, right through the associated resistors.
But at that moment the NE556 feel a low level in its inputs TR1 and TR2.
Perform its time tau and return to its stable state.
So and so on.

Forget the word eaten.
Reread paragraph 6 in my post #24. Probably you reach a conclusion.
If I could delete or edit my posts related to that word I would.

Let me tell You a lie: my messages were related to that word just to see if all read what I wrote.
If all read everything, probably would comment something. It is a joke.

REMEMBER: Generators were placed only, at the design, to verify operation of the NE556.

I hope, now, have explained properly.

 

! Oh boy. spidermanIIII.
Surely I not explain it as well as I should.

The generators were put into that design, just to test the time called tau.
Only for R3 and C3. Not for R9 and C12.
But say you put the oscilloscope in the NE556 PIN8.
Of course there will be pulses, as we are applying at a rate of 50 Hz. by means of those generators.
In normal operation of the circuit, there will be NO such pulses.
There will be a voltage level change according to the IR sensors.

Put figures 1, 2 and 3 in front of you.
Mentally move the float right through the capillary tube. Up or down Fig. 3.
Carefully analyze what happens with the associated components.

Easily find out: when LED1 light beam is interrupted by the float, We are coming to mind that the tank is full of water. . . is this true ?.
If the pump was on, it should go off.
When the light beam is Not interrupted nothing happens.

But when the float is at the bottom of the water tank indicates that it is empty. So the pump should turn on.
The LED2 light beam is interrupted by the float, the RX2 makes a change in its output level which comes from F to PIN8 NE556 IC. Fig. Our Old Fig 1.PNG. or simply Fig 1.PNG. both were attached.

The components: R9, C12 and R8, C11.
Ensure that the NE556, starting on their steady state when the circuit is biased.
Initially the capacitors are discharged.
When the circuit is biased, the capacitors are charged, right through the associated resistors.
But at that moment the NE556 feel a low level in its inputs TR1 and TR2.
Perform its time tau and return to its stable state.
So and so on.

Forget the word eaten.
Reread paragraph 6 in my post #24. Probably you reach a conclusion.
If I could delete or edit my posts related to that word I would.

Let me tell You a lie: my messages were related to that word just to see if all read what I wrote.
If all read everything, probably would comment something. It is a joke.

REMEMBER: Generators were placed only, at the design, to verify operation of the NE556.

I hope, now, have explained properly.

Thank you Mrcarlos now i understand how this circuit work because of you. you helped me a lot

 

Remember The RC constant ?.
T = R*C.
The RC constant with values​​:
R8 = 1000000 Ohms.
C11 = 0.000010 Farads.
It is 10 Seconds.
It means that in 10 seconds, the capacitor is charged to 63.3% of the applied voltage.
But:
The period of the signal of 32 KHz is 1/F.
1/32000 = 0,00003125000 seconds, = 31,250 microseconds.
So we're downloading and uploading to C11 at a rate of 31,250 microseconds.
Therefore, the filter will “eat” that 32KHz signal.
As if it came no sign, low level in the NE556 PIN 6.

my question here what happen after capacitor 11 get ,is it eat the train of pulses . second how discharging of C11 happen

 

my question here what happen after capacitor 11 get ,is it eat the train of pulses . second how discharging of C11 happen

my question here what happen after capacitor 11 get charged ,is it eat the train of pulses . second how discharging of C11 happen

 

Hello spidermanIIII

There are No pulses train anywhere around the electronic circuit.
Except, of course, in the circuit seen in the Fig. 2 of the document that you enclose.
But that train of pulses is used by the sensors RX1 (upper limit) and RX2 (lower limit).
For what. . . You should study the data sheets for IR sensor TSOP1738.

In the joints E and F only arrive voltage level changes.
Logic levels between 0 and 1.

0 if the corresponding IR sensor is obscure. It’s IR* beam is interrupted.
1 if the corresponding IR sensor is light. It’s IR* beam is not interrupted.
(* InfraRed beam).

If you have studied the data sheets for the ICs 555 and 556 you've assimilated that 555 is configured as a stable oscillator. Pulse Generator.
While 556 are configured as monostable. Need a change from high to low level at its TR input to generate a pulse at it’s output. This type of configuration is also called one shot.
A noteworthy detail in this configuration is that the stable state (monostable) is 0, low level.
When triggered, it switches to high level, meanwhile time tau, and returns to its stable state.

How discharging C11 ?? -Or C12-
Well, is discharging when the F or E signals reach low level as appropriate.

let me talk (write) just ones and zeros.
Be very present all that has been said here.
Remember that there is a RS Flip-Flip type formed by N1 and N2 gates.
And it changes with a negative pulse, thus: 1, 0, 1 in the 0 state changes its output.
put in view the attached images.

let me talk (write) just logical ones and zeros.
You keep in mind all that has been said here.
Remember that there is a RS Flip-Flip type formed by N1 and N2 gates.
And it changes with a negative pulse, thus: 1, 0, 1 on the 0, the state its output will change.
Put in view the attached images.

If you follow me probably better assimilate My description of the circuit.

Ok, first, to turn on the pump need to change the relay contacts.
T1 & RL1 = 0.
R7 & N1(3) = 1
N1(1) & N3(10) = 0
N3(8 9) & IC2(9) = 1
IC1(8) & RX2(3) = 0

Now, as an exercise that would serve you much, try to turn off the pump.

 

Doesn't this circuit require a schmitt trigger to prevent the O/P of RX1 and RX2 from oscillating due to the water level at D and A ,thus causing the motor to be turned on/off rapidly perhaps burning the motor??

Also NAND gates N2 and N3 are wired together..is that a WIRED AND logic??