555 flashing led

Thread Starter

Senz_90

Joined Jul 11, 2013
70
Hi everyone. I am beginner and I just understand a little bit about this circuit. I am simulate it on Livewire because I don't know how to using other simulation tool and lack of knowledge how this circuit work.

I hesitate about the results because Led D5, D10 and D3, D6 each series connected looks more brighter than others led on livewire simulation. I have try measure voltage across the led. and get like this picture result, that is maybe okay for white led but I want to use a red led where the voltage around 2.5 as I know.

please could someone look at this and give some advice and what component should I change to get same brighter for all leds? I don't makes this pcb yet because I don't want get a bad result or my led get burns before make sure this circuit will working.

should I change current limiting resistor value for this series led?

sorry for my bad english.
 

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pwdixon

Joined Oct 11, 2012
488
Where did you get that circuit?
The 4017 is a dead duck straight away with it's outputs joined together. And a 4017 can only have at most a 7V supply so that's going to help do in the transistors too.
 

Jony130

Joined Feb 17, 2009
5,487
First try to remove all resistor starting from R1, R8, R9, R12, R14.
Next try to choose LED resistor values and set LED's current around 10mA or so.
Add diodes between Q7 and Q1 ; Q6 and Q2; Q5 and Q3.
Also kept in mind that real LED's will behave completely differently than LED's from Livewire.
 

djsfantasi

Joined Apr 11, 2010
9,155
First, you should not connect the outputs of the 4017 together as you've done. You'll need diodes on each pin, to prevent short circuits when one output is high and the other low. Add a diode to output 4 as well so that the voltage drop is the same for all LED strings.

Second, your current limiting resistors will need to change if you change the LEDs to a different color. The resistor value depends on the forward voltage drop of the LED (Vf). Remember that LEDs are current devices. So to calculate the resistor value, use this formula:
R=(Vss-Vf)/Iled
where Vss is your supply voltage, Vf and Iled are the forward voltage drop and the current required by the LED respectively.
 

Thread Starter

Senz_90

Joined Jul 11, 2013
70
Where did you get that circuit?
The 4017 is a dead duck straight away with it's outputs joined together. And a 4017 can only have at most a 7V supply so that's going to help do in the transistors too.
what is mean "dead duck straight away with it's outputs joined together".

I have to lower the battery voltage? I don't see datasheet for 4017, maybe I need to see it.

I get this from EFY(electronics for you) file, but i am modify it a little bit so it can move back and forth. I forget the site address where I get.

So what will need to do to correct this? please help
 

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eetech00

Joined Jun 8, 2013
3,847
Where did you get that circuit?
The 4017 is a dead duck straight away with it's outputs joined together. And a 4017 can only have at most a 7V supply so that's going to help do in the transistors too.
Not true...4017 can be operated with a 15vdc supply.


eT
 

Thread Starter

Senz_90

Joined Jul 11, 2013
70
First try to remove all resistor starting from R1, R8, R9, R12, R14.
Next try to choose LED resistor values and set LED's current around 10mA or so.
Add diodes between Q7 and Q1 ; Q6 and Q2; Q5 and Q3.
Also kept in mind that real LED's will behave completely differently than LED's from Livewire.
btw what diode should I use? 1N4148 or 1N4004? where is the cathode anode should go? I am little confused to choose resistor values because the led are 2 series connected, R=V/I. voltage is Vf red led 2.5x2 / 10mA? ah nevermind djsfantasi give the ideas. I will try that and report the result. thank you.


First, you should not connect the outputs of the 4017 together as you've done. You'll need diodes on each pin, to prevent short circuits when one output is high and the other low. Add a diode to output 4 as well so that the voltage drop is the same for all LED strings.

Second, your current limiting resistors will need to change if you change the LEDs to a different color. The resistor value depends on the forward voltage drop of the LED (Vf). Remember that LEDs are current devices. So to calculate the resistor value, use this formula:
R=(Vss-Vf)/Iled
where Vss is your supply voltage, Vf and Iled are the forward voltage drop and the current required by the LED respectively.
thank you sir. I will try this.
 

Thread Starter

Senz_90

Joined Jul 11, 2013
70
Exact calculation are not so easy for this circuit. Too many unknowns.
But check this version
ah. my livewire is absolete. I couldn't open it because the version didn't same. would you mind give another format file like pdf or jpeg sir? btw thanks for your time.
 

Thread Starter

Senz_90

Joined Jul 11, 2013
70
I'm using Circuit Wizard.
pop up window said that my livewire pcb wizard didn't support your new version :D. If I want to make it could move back and forth, just attach 1N4148 diode towards 4017 like my file on latest edit?

I am usually using Eagle for make a PCB, I don't used to use PCB Wizard.
 

Jony130

Joined Feb 17, 2009
5,487
OK I see. So you need to add those diodes, but this time we don't want any short circuit in our design. So we need to add more diodes.
And notice that Q3 is PNP.
 

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Thread Starter

Senz_90

Joined Jul 11, 2013
70
OK I see. So you need to add those diodes, but this time we don't want any short circuit in our design. So we need to add more diodes.
And notice that Q3 is PNP.
wow. Note !! thank you sir :). I want design my pcb so it could attach on my motorcycle :D
 
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