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walid
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Posts posted by walid
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Hi all
you are not respond to my last question, so i'll do more efforts to simplify
the picture to myself, but still need your help.
The table shown below is a cut and eddit from the datasheets of MC14093B.
From this table I understand the following:
When Vcc = 5 vdc, this gate considers (typically) any input voltahe down to
2.9 vdc as logic 1, and any input voltahe up to 1.9 vdc as logic 0
Questions:
(1) Is the above true?
(2) There are three values related to the threshold voltage, min typ and max,
which one to rely on?
(3) the table show three values for Vcc; 5 10 and 15 vdc, what can i do if
my vcc = 7 volts dc?
(4) In the same datasheets, not shown in the table below, i found what is
called Hysteresis voltage, its value is 1.1v at vcc =5v, 1.7 at vcc = 10v
and 2.1 at vcc=15v. what this means and how can be used?
the 2nd step is to use the above info with some exponentioal expressions
related to capacitor to discover how exactly the charging process and the
rate at which voltage developed between its plates.
then I'll see how the R2 (as apart of voltage divider) isfunctioning in the
whole circuit. I'll get the datasheets of 74HC132 and study it carefully
and do all the calculations needed to understand how this circuit works.
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I agree with you that R8 reduce the max resistance between the probes, but R2 is not.
R2 do another job, I think its function is as follows:
Assume R8//soil resistance is very low, say 100 ohm (moisture content is very high)
Our main ocs. changes its state every 250 micro sec (f=2KHz)
If logic 1 is at pin 6 then logic 0 will appear at pin3, this will last for 250 micro sec
Now at this situation, current will flow from pin6 through R8 to:
1) Charge C3
2) Pass through R2 to the virtual ground pin 3
If we omit R2, then C3 will be fully charged in 1.1*100ohm*1n = 0.1 micro sec
let it 1 micro sec, this is very small time compared to the 250 micro.
This means that pin 13 of the NAND gate will be low only for 1 micro sec from the 250 micro sec.
BUT if we let part of pin 6 current to flow through another pass, then we slow the rate of charging C3 and gain more time
I know that the designer said, the led must be dark when the moisture content is high, but what i said will be more clear when the moisture content become low and R8 high.
I can conclude that if we carefully choose R8 and C3 we can then omit R2.
What audioguru say? -
can you please put your circuit for others to look and think abput it.
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Hi audioguru, thank you for everything
(1) If we ignore, for a moment, the formula you gave me later to calculate freq, and concentrate on the statement you told me:((A Cmos Schmitt trigger switches when its input voltage is fairly close to zero volts while an ordinary Cmos oscillator switches when its input voltage is close to half of the supply voltage.))
Looking at the word "half", I notice that this is the key to solve one problem. all the time I calculate the freq and get answer = half yours, so I can conclude that: The freq of an ordinary Cmos oscillator 1/(1.1RC) while for A Cmos Schmitt trigger oscillator, f = 2/(1.1RC), that is the time is half and the freq is double.
Is this approximation is accepted by you?
(2) If C3 charges and discharges through R8 (and the equivalent resistance of the soil in parallel), why we need R2 (=100K)? i think it has no function! -
use google get datasheets and read it
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Hi all
Now I can move to another part of this complex (at least for me) circuit:
In Fig. plants2 shown below I cut this part of the circuit because I think that this is a stand-alone osc.
In his description, Audioguru did not tell us that gate IC1-B is a stand-alone osc, he said that this is a NAND gate.
I'm not sure that this osc. does not affected by neighbor parts connected to it (as shown in Fig. Plants3), but at this moment lets start with Fig.plants2.
Tc = Td = 1.1 R1 C1 = 1.1 * 470K * 1n = 0.517 msec
So, time period = 1.034 msec ~ = 1msec
frequency = 1KHz (if we take guru's last fotmula we get another value)
Now if you look at Fig.plants3, you may notice that I omit the two probes and write on R8 that its value change from few ohms to 47K according to the state of the moisture content.
From this figure we see that the designer connect his osc. (IC1-B) to NOT gate (IC1-A) and this combination works like this:
When there is a 0 at the o/p of the osc. (pin 6), then 1 appears at pin 3, so cap C3 charge through R2 only and the R8 resistor connected in parallel with the C3.
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Hi all my friends, i miss you all, it is good to back
In the Fig.plants1 shown below there is a part of the circuit, this part is the IC1-C work as 2 Hz osc. As the author said.
Pin 10 is connected to 3v Vcc
R3 = 3.9M
R4 = 100K
C4 = 220n
D1 = any small signal diode
The designer wants that the cap C4 charged through R3 and discharged through (R3//R4 = approx R4).
As I understood (may be wrong) the purpose of using this osc. is only to save batt (for more info look at: http://www.electronics-lab.com/projects/science/018/index.html)
If you don't worry about power supply you can omitt this part because it has no real function in the main circuit.
To calculate the charge time Tc = 1.1 R3 C4 = 943.8 msec
The discharge time Td = 1.1 R4 C4 = 24.2 msec
The total period = approx. 1 sec
So, frequency = 1Hz
I know the picture is not so simply, Audioguru did not calculate it like this, his frequency = 2 Hz and he said that: "IC1C is another CMOS Schmitt trigger oscillator at about 2Hz. D1 and R4 discharge C4 quickly so that its output is low for only about 15ms with a 3V battery, and about 25ms with a 2V battery."
From this we conclude that we must take the supply voltage into account; that is the higher this voltage the faster the charging and discharging process.
This looks like filling the same tank, first using 2 hp pump then by using 3 hp. But the question still in my mind is how to calculate these times taking voltage value into account.
After receiving a satisfactory answer I'll discuss more parts of this professional circuit.
Yours,
Walid.
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Hi audioguru
You were taught wrong.
yes this is true, i hope that you always teach me.
I forgot to mention that it wouldn't need a diode for it to have a long capacitor charge time and a short discharge time. -
The 18k capacitor discharges the 10uF capacitor in 180ms, reducing damage to the LED when its voltage is reversed when the power is turned off. The absolute max reverse voltage for most LEDs is only 5V, and this circuit has 9V.
I never operate parts beyond their max ratings so I don't know how soon the LED will fail due to extended time of reversed over-voltage if you remove the resistor.
The LED should have a diode connected in reverse in parallel with the LED to protect it. Then removing the resistor won't make much difference.
this is excellent answer thank you audioguru -
Hi audioguru
First thank you very much for your time with us
Pin 6 never gets to ground
I disagree, long time ago we have taught that caps are approx. short circuit during the charging or discharging process, so it connects pin 6 to ground...
R3 causes positive feedback for a snap-action of the comparator with hysterisis.
What u mean by snap action and hystreisis, i know that snap is a very quick and high voltage pulse.
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Now I have more questions like:
(1) Why the designer did not use another diode in series with R4 to make the discharge process only through R5, since R4//R5 approx = R5
(2) From your point of view, what the modifications that must be done to the circuit to work more efficient with less components?
thank you -
hi audioguru
R3 pulls pin 3 to a more positive voltage than 3V
I think that u mean pin 5 not pin 3
because the output transistor of the comparator doesn't have an emitter-follower loss like opamps.
You may mean the internal (inside the IC) transistor.
When the comparator's output is at +6V, R3 pulls pin 3 to a more positive voltage than 3V. When the output of the comparator is at ground, R3 pulls pin 3 to a more negative voltage than 3V.
This is OK, i calculate and understand it but i have question:
now let the o/p at pin 7 is 6volt then the voltage at pin 5 = 4v, if pin 6 is at ground, the o/p at pin 7 will remain at 6v. to change this o/p, pin 5 voltage must be less than pin 6 (5 more -Ve or 6 more +ve).
can u, please, tell me more about this mechanesim of oscillation.
thank you.
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this is a good and simple answer, i understand it , thank u audioguru
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Hi guru
you deal with us as we a big experts, u must simplify your answers i'm hardly understand u. u must take a low level persons (like me) into your account.
thank you -
Hi guru, thank u very much
You have a negative supply. I didn't notice
no it is 6 volt (4 x 1.5V batt), the +ve terminal is 6V and the -ve terminal is 0 V.
Pin 7 swings almost to +9V and almost to -9V.
From where u get the + and - 9 voltAn NPN transistor with its emitter connected to -9V needs a positive current feed to its base for base current. The output of the comparator doesn't have anything to provide a positive current, so a resistor is needed from the base to the positive supply. It is called a "pull-up" resistor.
u want to say that a "pull-up" resistor connect the point to +batt and the a pull-down resistor connect a point to ground
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Hi audioguru, thank u very much for your kind answer.
Its pin 5 is pulled up to about +2.7V and down to about -2.7V when the output toggles.
(1) pin 5 always connected to the voltage divider formed by R1 and R2 and so it is always at 3v potential, correct me.
(2) How u reach these numbers, +-2.7v
(3) What the values of the o/p at pin 7?
The diode makes C2's negative charging time short
It is the first time for me to hear about negative charging!
An NPN transistor can be used if the comparator's output has a pull-up resistor for its base bias current.
What is a pull-up resistor and what it do?
Than u very much
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I understand the 555 timer IC when used in its astable mode as an oscillator, but the one shown below i need some help in:
(1) how this oscillator work?
(2) the function of the diode?
(3) how to choose the components values to achieve some freq.?
(4) What the formula used to calculate the freq.?
(5) Can I replace Q1 with a NPN one?
Note that bat = 6 v and IC = LM393
Thanks to everybody
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hi
your question is very important for me also, and i wait the answer, but please send the full datasheet to ren_ADD to enable him to answer.
thanks -
Why you enlarge the problem, try to forget the datalogger for a monent, my question is very simple
i ask if you work in TV corporation and you notice that there is some bars in the picture and a hum in the audio and u want to prove to the others that there is a ground loop what you tell them and what measurement you must make to support yourself.
thanks -
Hi indulis,
"Transformers" used in flyback converters ARE NOT transformers at all, but coupled inductors, and almost ALL the energy is stored is in the gap.
I'll concentrate on this air gap, what the distance?
and how the energy stored in this gap?
thank you
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What i mean is that the ferrite core transformers like a flyback or the one used in a switched mode power supply has an air gap to stor energy.
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what about air gap in which they store energy
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Hi
I read this somewhere: "a regular transformer is designed to transfer energy from its primary to secondary and to minimize stored energy."
Can someone please explain how they minimize the stored energy, that is what they do to minimize it?
thanx -
I want to learn
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hi all
u all must know that i am the only person among tens in my work who has the chance of conncting with u and the natural result of this must make me the more understanding person and this is not occur till now
ok i am in advance but slowly.....
now i have a FLUKE 112 TRUE RMS MULTIMETER and a video monitor and a video casitte, there is a hum in the picture, how to do some measurements to fix it?
Greetings...
Merry Christmas.... to all
walid
simple question about batt chargers
in Theory articles
Posted
Hi all my friends
When somebody build a batt charger that stop charging when batt full, he
must use a comparator to compare batt voltage to some referance voltage..
when i conect a 3 volt batt to 5v power supply and wat to measure the
woltage across the connection, what would it be?
i ask this question to know at what level of charge i must stop the process
if i have four 1.2v batts (total = 4.8v), now these batts are not fully
charged, its voltage = 4 v or less.
i want to recharge them using 6v power supply. if i connect a voltmeter
across the connection, what would itsreading at thenbegining and at full
charge
i hope that u uderstand me
yours
walid