# Formule Timer 555 timing ( t = 1.1 x r1 x c1 ) + timer 555 problem solved

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#### Maarten

##### New Member
Hey there,

I am new to this forum so if i place this in the wrong place tell me..

So I started few days ago on exploring the timer 555.. I watched the tutorials and so and then i moved an made a simple flashing led circuit ( a stable right? ). And moved on and made my own flashing circuit.. And as you can guess it didn't work.. it didn't work for 2 days. I tried a lot but nope.. But today i solved the problem. Well I didn't solve it but I found it maybe you guys know why.. The problem was in the DC Supply I was using some school thing DC supply very basic with only like 2/4/6/8/10/12 V DC/AC so when I connected it to the circuit the LED only stays ON and didn't flash.. Today I went to try it again and solve my problem and I had this 9V bat next to me and the dc supply was half meter far from me so i was kinda lazy and used the small 9V bat.. I am glad i was lazy..

I just shared this so maybe other people with the same problem.. try using a 9v bat. But what I don't understand why didn't it work with that power Supply? I also tried it afterwards with a good DC Supply and it worked.

But now back to my question... I just typed this all because I lost my hope and was opening a topic on forum for help.. but while opening the topic, i was just reading one last topic on solving the problem.. and it worked.

Now I have one question about the timing.. in lots of videos i saw this formule on the timing t = 1.1 x r1 x c1 and this should give you the exact time of the filling of the capicator/flashing led. But what I don't understand of the formule it doesn't indicate the Volt, because the volt is a very important thing in the timing right? the 1/3v and 2/3v.. so it would have a lot difference when t = 1.1 x 100k x 1uf and its 5v and when the volt is maybe 15v? Right.. Or did i not understand it good.. Or did i not do enough research yet?

i just feel like sharing this all with you, because im pretty in joy after like 2/3days sturggling on this timer 555.. I am glad it was not a fault from the circuit but just the DC supply it was really driving me mad but patience wins

#### JLNY

##### Active Member
Without going into too much detail, the RC time constant of a 555 timer is independent of supply voltage. Basically, the higher the voltage, the higher the voltage threshold, but at the same time, higher voltage will also increase the charging current through the resistors, so the net RC time constant is the same.

I'm sure in the real world the timing of a 555 probably does vary a tiny amount with supply voltage, but the 555 timer isn't a precision device anyway.

#### MikeMl

##### Well-Known Member
I think you are using the wrong formula...

The usable operating voltage range of a standard bipolar 555 is about 5V to about 15V. It will not run below ~4.5 and more than 18V might damage it.

Look at the attached simulation. Notice that when running as an astable (oscillating), the timing capacitor charges through R1 and R2 together, but discharges only through R2 (when the "discharge" pin is turned on). See the green trace V(rc). Note how it varies between 1/3 of 9V and 2/3 of 9V. Ignore the first charge cycle because it is starting from zero, not 3V like all the subsequent ones.

This makes the period proportional to (R1+R2)*C1 + R2*C1 = C1(R1 + 2*R2) = 1e-6(20K +2*10K) = 4e4*1e-6 = 4e-2 = 40ms. The actual period measured from the simulation is 27.7ms (cursors), so the proportionality constant to calculate the period would have to be ~0.7, or Per=0.7*C1(R1+2R2).

This makes the Frequency = 1/Per = ~1.4/(C1(R1+2*R2)). I will let you calculate the time it takes to charge, and the time it takes to discharge the capacitor. This determines the duty cycle. Note that V(out) red trace is high longer than it is low.

It is tough to make an oscillator with a duty cycle shorter than 50% with this 555 circuit.

Did you have the two bypass capacitors C2 and C3 in your circuit?

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#### Maarten

##### New Member
Without going into too much detail, the RC time constant of a 555 timer is independent of supply voltage. Basically, the higher the voltage, the higher the voltage threshold, but at the same time, higher voltage will also increase the charging current through the resistors, so the net RC time constant is the same.

I'm sure in the real world the timing of a 555 probably does vary a tiny amount with supply voltage, but the 555 timer isn't a precision device anyway.
Thanks for your help. I kind of understand what you are saying. I think I need just to try it out a little more. but as you said in the real world it does differ.. Because i went on playing turning the Dc supply till even under 3V and it kept working flashing, and then i raised it up to even 15v and the timing did differ a bit.

#### Maarten

##### New Member
It is tough to make an oscillator with a duty cycle shorter than 50% with this 555 circuit.

Did you have the two bypass capacitors C2 and C3 in your circuit?

View attachment 106381

#### Maarten

##### New Member
Thanks for you reaction. As you said for the a stable it the wrong formule, I just mixed it up because i kept trying all circuits over and over because it didn't work as i Said the LED kept on... you might also have a opinion of why that supply doesn't work to gether with the 555?

The reason for the Capacator to only discharge from R2 is because the transistor goes on after the 2/3V is reached right ?

As for your question it is a little bit confusing bit i think i get it.. i am new to reading formules on the internet when i write it down it comes easier... But the point which iI don't get is what did you measure from the simulation the 27.7ms? is that like a constant? because in the notes that i took while searching is the fellowing:

T1 (charing) = 0,693(R1+R2) x C1
T2 (discharging) = 0,693 x R2 x C1
T= T1 + T2 = 0,693 (R1 + 2 x R2 ) x C1
f = 1/T = 1.44 / ( R1 + 2 x R2 ) x C1

And the point is in the beginning of your formule you don't have the 0,693 and you add later the 27.7ms and after that it becomes the same formule.. so i that 27.7ms always the same?

as for this formule the charging time will be = 0,693 ( 20000 + 10000 ) x 0,00001 = 0.02ms
and the discharing should be = 0,693 x 10000 x 0,000001 = 0,06 ms ..

1,4/(1e-6(20K + 2 x 10K)) = 35 so what did this 35 means exactly? it's frequeancy but what makes it do exactly?

as for my circuit i didn't use a c2 because from what i understood is that it is only used to regulate some thing such as noices and so on, actually i did want to use it but i only have the + - capacitors ( electrolitic?) which the smalles is 1uf i don't think that i can use that?

as for the C3 it is the first time i see someone using the C3 so I don't really know the function of it and which amount i must use for it?

#### cowboybob

##### Well-Known Member
... as for the C3 it is the first time i see someone using the C3 so I don't really know the function of it and which amount i must use for it?
C3 is simply there to help stabilize the supply voltage swings (due to any possible 555 pin 3 load variations) for Vcc and Reset (and to a much lesser degree, the 2/3Vcc value of the comparator circuit of the 555). For any given 555 circuit, 10μF is generally sufficient.
as for my circuit i didn't use a c2 because from what i understood is that it is only used to regulate some thing such as noices and so on, actually i did want to use it but i only have the + - capacitors ( electrolitic?) which the smalles is 1uf i don't think that i can use that?
Pin 5 (CV) provides "control" access to the internal voltage divider (by default, 2/3 VCC). C2 prevents pin 5 from "floating", i.e., inadvertently altering the 2/3 Vcc value, by noise or whatever.
C2 should never be an electrolytic and can be in the range of 10pF to 100nF.

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#### ci139

##### Active Member
don't have the 0,693
yet another formula for 7555 CMOS v. -- i guess the ∂ stands for duty ratio //// however
• if you don't grasp the following -- leave it alone
• in the 555's std. a-stable cfg. discharge is Vcc (Supply voltage) dependent as R.CE.ON (an ON resistance) of discharge transistor is (also) V.C (Collector voltage) dependent
• there are slight differences in between the same manufacturer's individual ic-s of one specific prefix e.g. NE
• there are slight differences in between the same manufacturer's individual types SA,SE,...
• there may be larger differences in between different manufacturers OR the version updates of the same v. of the same manufacturer
• there may be differences due the same chip age and operating history e.g. at different time ...
• ... (also temp. , parameters of timing elements , also grid noise , parameters of supply e.c.)
what you need is good/sufficient lab equippement to set/tune it work as required

#### Maarten

##### New Member
C3 is simply there to help stabilize the supply voltage swings (due to any possible 555 pin 3 load variations) for Vcc and Reset (and to a much lesser degree, the 2/3Vcc value of the comparator circuit of the 555). For any given 555 circuit, 10μF is generally sufficient.

Pin 5 (CV) provides "control" access to the internal voltage divider (by default, 2/3 VCC). C2 prevents pin 5 from "floating", i.e., inadvertently altering the 2/3 Vcc value, by noise or whatever.
C2 should never be an electrolytic and can be in the range of 10pF to 100nF.
Okay I get it. So I guess it is better to use it then. Kinda weird that on the online videos they don't use it.
I'll try to find one for pin 5.

yet another formula for 7555 CMOS v. -- i guess the ∂ stands for duty ratio //// however
• if you don't grasp the following -- leave it alone
• in the 555's std. a-stable cfg. discharge is Vcc (Supply voltage) dependent as R.CE.ON (an ON resistance) of discharge transistor is (also) V.C (Collector voltage) dependent
• there are slight differences in between the same manufacturer's individual ic-s of one specific prefix e.g. NE
• there are slight differences in between the same manufacturer's individual types SA,SE,...
• there may be larger differences in between different manufacturers OR the version updates of the same v. of the same manufacturer
• there may be differences due the same chip age and operating history e.g. at different time ...
• ... (also temp. , parameters of timing elements , also grid noise , parameters of supply e.c.)
what you need is good/sufficient lab equippement to set/tune it work as required
I didn't get everything but i understand it a bit thanks for helping.

I am thinking now one the charging and discharging time, i guess you can play a lot with these and even make the charging time way more longer than the discharging time right.

Thanks for the information I think now i need just go back to reading and experimenting on these circuits.

By the way does anyone know a good programma where i can write out on circuits such as the one in the earlier posts, and see freqnecies etc? All the programs i found costs money which i can't pay at this moment..

#### cowboybob

##### Well-Known Member
... I am thinking now one the charging and discharging time, i guess you can play a lot with these and even make the charging time way more longer than the discharging time right. ...
Generally speaking, yes.
... By the way does anyone know a good programma where i can write out on circuits such as the one in the earlier posts, and see freqnecies etc? All the programs i found costs money which i can't pay at this moment..
Check out LTSpice. It has a bit of a learning curve, though. I use TINA TI. It's not free (well, there is a free version, but it is severely limited). I find a lot easier to use than LTSpice.

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#### AnalogKid

##### Well-Known Member
The genius of the 555 design is that it is based on ratios of components, not their absolute values. When charging up a capacitor through a resistor, it takes one time constant (resistor value times capacitor value) to reach 63% of the power supply voltage. Thus, with a 15 V supply it takes one time constant to reach 63% of 15 V. with a 6 V supply, it takes one TC to reach 63% of 6 V. So, if the internal comparators also are working with percentages of Vcc for their trip points, they are comparing one varying percentage of Vcc (the capacitor voltage) to one fixed percentage of Vcc (either 33% or 67%, depending on the comparator). Because everything is based on percentages of Vcc, everything tracks even when Vcc changes.

Overall, the 555 acts as a relaxation oscillator with one huge difference. The standard relaxation oscillator usually changes state when the capacitor voltage reaches a specific value based on the device physics. Examples of this are a neon bulb or a unijunction transistor. With the 555, the trip points are based on three resistors, not the forward or reverse voltage of a semiconductor. Accurate resistor values within an IC require laser trimming, a very expensive process step. But accurate resistor *ratios* are easy and cheap, because they fall out of the standard IC fabrication process. While the absolute values of the 555 resistors are not very accurate, their ratios track to within 1%, including temperature drift. Working within the limitations of early 1970's IC fabrication, this was pure genius.

ak

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