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Quick n Dirty answer needed

SparkyCal

Mar 11, 2020
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lol..thank-you for the link to the Amazon already made heart. Thank-you also for the advice. Leave this with me. I don't think I will buy the heart. Part of this is challenging myself to learn how to do it, so that it has more meaning when I give it to her. I am not sure I will be able to do this in time, and perhaps worth doing even if I use it for another occasion. I've learned:

Don't trust Youtube videos not from accredited sources
LEDs can't be strung endlessly because there are current/voltage considerations.
Colour of LEDs make a difference in terms of forward voltage
Resistors are required for each string of LEDs.
Probably 4 LEDs per string is best
Can use one 555 timer. (I'm not sure what circuit though). I have a few 555 timer circuits that i have built, that flash. Some of these have been from good sites. Maybe I just pick one of those?
Plus, I would not have known about 555 timers, had it not been for this site ;-)
 

SparkyCal

Mar 11, 2020
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Bertus- I may be getting confused due to an abundance of riches. I did look at a web page that someone provided a link to. On that page, it had a number of 555 timer circuits, some entitled Blink. I did build from that page and that is one of the circuits, i assume would be reliable as it was provided by either you or someone else here. Problem is, I need to catalogue what i build, so I know where it came from. I'll have a look at the link you reposted just now as well. I am going nthrouhg what a new person does. I get excited when things work, and csan't wait to move on. to the next. But I need to be more methodical now so that i learn more.
Thanks
 

AnalogKid

Jun 10, 2015
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First pass at a schematic. Flash rate is 1 Hz; adjust R1 to change. C2 is for power source decoupling, and should be as close as possible to 555 pins 1 and 8. It is shown on pin 4 because that is a convenient place to tie it to Vcc. Many schematic programs hide the power and ground pins for clarity. R2-R6 do not need to be 1% parts; that is what's in my design library.

ak
LED-Blinker-1-c.gif
 

SparkyCal

Mar 11, 2020
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Thank-you for taking the time to send this. I will give it a shot.
 

bertus

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Nov 8, 2019
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Hello,

Here is a PDF with some more schematics.
It also contains some tables for the values.

Bertus
 

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SparkyCal

Mar 11, 2020
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First pass at a schematic. Flash rate is 1 Hz; adjust R1 to change. C2 is for power source decoupling, and should be as close as possible to 555 pins 1 and 8. It is shown on pin 4 because that is a convenient place to tie it to Vcc. Many schematic programs hide the power and ground pins for clarity. R2-R6 do not need to be 1% parts; that is what's in my design library.

ak
View attachment 48383
Well AnalogKid. In tried your circuit, bit it turns out that it is beyond my skill level at this point in time. I got confused along the way.

Here's some stuff I was wondering:

Is the 2 and 6 pins of the 55 the only ones shorted?

You said that C2 is for power source decoupling, and should be as close as possible to 555 pins 1 and 8. This cinbfusd me. Should it be connected to pins 1 or 8? Or 4? I connected it to 4, but why would it need to be close t 1 , and 8..As in physically close? Why would that matter?

I may try it again. For now, it is temporarily in my circuit cemetery, where all my failed projects go. Of course, this is not your fault. It's mine for not being ready. But each attempt I make and fail at, I always leanr some things. And that makes it worthwhile.
 

Harald Kapp

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Well AnalogKid. In tried your circuit, bit it turns out that it is beyond my skill level at this point in time.
This circuit can hardly be made any simpler. Have you observed @AnalogKid 's comment about pins 1 and 8 for ground and Vcc not being shown on the schematic? You need to connect those to gnd and 9 V respectively.
The only simplification I can see (just for testing the blinking circuit per se) is to remove the bunch of LEDs and resistors to the right and replace it by a single LED and resistor (220 Ω to 470 Ω, whatever you find in your parts bin).

Why would that matter?
Any length of wire is an inductor at high frequencies. This capacitor is meant to deliver pulse current for switching transients to help keep the operating voltage of the chip stable when the chip draw a high pulse current (e.g. when the output turns from on to off or vice versa). Any inductance, on the contrary, is a high impedance at high frequencies. Therefore a pulse current through the inductor will create a voltage drop. This in turn reduces the operating voltage of the chip and the chip may not work reliable. Therefore this capacitor should be as close as possible to pins 1 an 8.
If interested, read more about this topic e.g. here.

Is the 2 and 6 pins of the 55 the only ones shorted?
No, 4 and 8 are shorted too (both go to Vcc).

Note that C1 and C2 at 22 µF are (I assume) electrolytic or tantalum capacitor. These are polarized, meaning the y have defined (and labeled) "+" and "-" pins. You need to observe the orientation of these capacitors and make sure "-" is connected to ground.

The circuit from @AnalogKid 's post #25 is may seem incomplete as pin 7 is not connected. This is a variant of the standard circuit as shown e.g. here (scroll down to "50% Duty Cycle Astable Oscillator").
Depending on which variant of the 555 chip you use an additional capacitor (10 nF to 100 nF, whatever you find in your parts bin) should be connected from pin 5 to gnd.
 

SparkyCal

Mar 11, 2020
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Harald:

Thanks for explaining. I did not mean to imply that the circuit posted was not properly specified; only that I don't have enough experience to understand all aspects of it.

As a small example, I still don't understand what Vcc means. I assume that is just a way of saying voltage?

From my questions, I think it's clear that I don't ave the experience to know that I needed to connect pins 1 and 8, nor did I know enough to assume that 4 and 8 are also shorted. The only reason I thought to ask, was that is previous 555 circuits I have used, 4 and 8 are usually shorted.

This is not about AnalogKid's schematic or your explanation. I'm just trying to point out that I need more experience to be able to interpret things.

I'm working on. it. I may try the circuit again. this time, i will start with a breadboard rather than soldering it.
 

SparkyCal

Mar 11, 2020
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P.S. I understand that it must be frustrating for people who know what they are doin g to explain simple things to me. But I assure you I am learning. Maybe not as quickly as I'd like though.

I appreciate the help.
 

SparkyCal

Mar 11, 2020
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Well, I took another serious run at this circuit- this time, on a breadboard.,

The good news, is that it worked I think, but with a modification. After getting everything in place, I connected a single red LED and started very low with voltage via the power supply. The Red LED started blinking, albeit at a slow rate when I got to a certain voltage level.

I turned off the voltage, as i concluded that the circuit would work if I saw it through. But then i did the same thing, but added another Red LED. This time, nothing worked, and I could not get the original single LED to work again either. So I don't know if I moved something by accident,. or whether the circuit was not really working, but I thought it was.
 

Harald Kapp

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added another Red LED. This time, nothing worked
How did you connect the 2nd LED? If in series do you realize that this will increase the minimum operating voltage by ~1.6 V before the LEDs are going to blink?
Did you observe correct polarity?
Do you have a current limiting resistor in place?
 

AnalogKid

Jun 10, 2015
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Use the reference designators to tell us which LEDs (and other parts) you have in the circuit. Note that as you add LEDs in series, the minimum operating voltage goes up. Depending on the exact LED part number, each LED adds approx 1.8-2.0 V.

AND - If you have only one LED in a string, you should not run the circuit at the full 9 V. At 9 V and only one LED, the LED current will be around 90 mA. A bipolar 555 can sink this with no problem, but the LED might fail due to overheating.

Suggestion: Start with a 6 V battery and one LED string with one resistor and two LEDs in series in it. R2, D1, D2. This should blink safely. Now you can mess with R1 and C1 to get the blink rate you want. Once everything is happy, add D3 and D4 to the string and change the battery to 9 V. Blink rate should be the same. At this point you are running approx 11 mA through the LEDs; very safe, but the LEDs will be a bit dimmer than before. Adding the other strings will increase the battery current but shouldn't change the brightness of the individual LEDs much.

With the circuit as shown in the schematic, the current in each LED is approx 10-11 mA. If you want more brightness (with shorter battery life), you can reduce R2 - R6 to as low as 39 ohms. This yields around 20 mA current per string, a common "full brightness" value for low cost generic LEDs. At this point the 555 is sinking 100 mA, so its low output voltage will be higher than before. That 100 mA probably is a high estimate, but safe.

ak
 

SparkyCal

Mar 11, 2020
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Thanks for all the thoughts. I plan to rebuild it from scratch on a breadboard, so, at that time, I will red everything carefully before proceedings. I did have the resistors in place the right polarities and I was using a power supply that i can dial in the volts. The LEDS were sytri=ung in parallel, not i. series. I always start very low unti i see the LED light up.

Anyway, i will try it again and let you know. I have all the parts, so i did not stray from the specs in the schematic.


One thing that I am confused about: what is that stuff at the far right of the schematic. It looks like (9 volts and a couple of batteries that also go to ground. Is that a repetition of the voltage requirement, or is it suggesting two different sources of power?
 

Martaine2005

May 12, 2015
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That is a battery symbol. The arrows (triangles) are 9v positive rail. The gnd symbol is the negative rail.
It’s simpler to do it that way or you’ll end up with a confusing spaghetti junction!.
So all + are connected to battery positive. All - to battery negative.

Martin
 
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