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Quick n Dirty answer needed

SparkyCal

Mar 11, 2020
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Failed again:

Here are my steps

Prepared the 555 by shorting pins 2 and 6 and pins 4 and 8
Mounted it on breadboard
Connected Pin 4 to +
Connected + side of Capacitor to Pin 4 and the - side to ground
R1 connected to Pin3 and pin6
Connected other capacitors' + side to pin 3 and the - side to ground
Connected pin 8 to +
Connected Pin 1 to ground
Connected + side of LED to pin 3 and - side to ground
 

AnalogKid

Jun 10, 2015
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The battery is the only power source. Attaching power and GND symbols to it tells the schematic software to connect it to the Vcc and GND pins of any ICs with those pins visually suppressed. This has to do with the way the schematic data is transferred to pc board layout software through a file called a netlist.

ak
 

Harald Kapp

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Prepared the 555 by shorting pins 2 and 6 and pins 4 and 8
...
Connected Pin 4 to +
...
Connected pin 8 to +
No need to short 4 and 8 as you connect them both on the breadboard to "+".
Connected other capacitors' + side to pin 3 and the - side to ground
  1. Please use reference designators, as has been suggested before. With just 2 capacitors one may be able to figure out which one the "other" capacitor is. But generally this makes no sense. A reference designator (short refdes) is the "name" of the component as shown on the schematic, in this case C1.
  2. Connection to pin 3 is wrong, The schematic clearly shows that C1 is connected between pins 2+6 and ground.
Connected + side of LED to pin 3 and - side to ground
The schematic shows the LEDs cathode (-) to the 555, anode (+) to + 9 V. However, the way you swapped polarity of the LED and connection to gnd instead of Vcc will basically also work. The difference being that in the original circuit the LED is ON when the 555's output is low, in your circuit the LED is ON when the 555's output is high.
That is fine, but: You missed the series resistor to limit current through the LED. As you have been told before, this resistor is required, otherwise destruction of the LED (very likely) or the driving output (less likely) may result. Go to our resources setion and read about driving LEDs.
 

SparkyCal

Mar 11, 2020
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Thanks. I am re-doing the circuit on a breadboard, and as i do it, i am typing out my exact steps:

1. I start by placing the 555 on the breadboard. Note, i had already shorted 4 to 8, and 2 to 6 by way of a soldered wire, so I am leaving it as is, despite the comm,ents advising I did not have to short 4 to 8. Does nt make sense to unsolder it, and potentially ruin the 555.

2. I took a jumper wire, and connected Pin 4 to the + rail on the breadboard (which runs to the + terminal of my variable power supply).

3. I connected C2 as follows: the + side of C2. goes to Pin 4 , and the - side of C2 goes to the negative rail on the breadboard (which runs to the - terminal of my variable power supply).

4. R1 (100K Resistor), is connected from Pin 3, to Pin 6

5. The + side of C1 is connected to Pin2 and the - side of C1 is connected to the the negative rail on the breadboard (which runs to the - terminal of my variable power supply).

6. A jumper wire from Pin 3, goes to the + end of the red LED

7. The - side of the LED, is connected to a jumper wire whihc connect it to the negative rail on the breadboard (which runs to the - terminal of my variable power supply).

8. Pin 1 is connected to the negative rail on the breadboard (which runs to the - terminal of my variable power supply).


I only connected one led to see it it would work, Gradually increased volts fro. 1 to 9. no luck
 

Audioguru

Sep 24, 2016
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I guess the LED and/or 555 is/are burned out because of a missing current-limiting resistor in series with the LED.
 

SparkyCal

Mar 11, 2020
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My apologies. I completely did not see R1 to R6. My fault. Let me add that and I will see
 

SparkyCal

Mar 11, 2020
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Still can't get it to work. Used new 555 , new Capacitors and new LED too. Here is a pic
 

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Harald Kapp

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I can’t see any power to the 555.
That is because power is fed to pin 4 which is shorted to pin 8 by a wire jumper:
Note, i had already shorted 4 to 8, and 2 to 6 by way of a soldered wire

Check the LED: is it oriented correctly? LEDs are diodes and require to have the correct polarity of voltage to be applied.
Does the LED work at all? Connect the LED with series resistor to a battery to see whether it works at all. Do not apply power to the LED for testing while the 555 is still connected. This may destroy the 555.

If the LED is o.k., try another 555. This time do not solder any jumper wires to the chip but use jumper wires on the breadboard instead. Makes it much easier to make changes to the circuit.

The photo is not very good (although admittedly much better than many others I have seen here :) ). For your next photo try better uniform lighting and higher zoom, so we can see the wiring and the orientation of components better. Maybe two or more photos with different zoom levels for overview and detail. We have a resource on how to make good shots of pcbs which may help with your breadboard setup, too.
 

Martaine2005

May 12, 2015
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That is because power is fed to pin 4 which is shorted to pin 8 by a wire jumper:
Yes, I saw that earlier but thought I read he’d changed the LED and 555.
Can the IC seat correctly in the breadboard with jumpers under it?
I still can’t see a negative to the 555.

Martin
 
Last edited:

Audioguru

Sep 24, 2016
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Every time somebody makes a circuit on a solderless breadboard, we cannot see which connects to what and it does not work.
 

bertus

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Hello,

I do not see a connection to pin 1 (ground) of the 555:

sparky_555_breadboard.png

Looking further, the circuits seems to be a hysterestic 555 oscillator , as pin 7 is not used:

555-hysteretic.png

(this is an example schematic).

Bertus
 

SparkyCal

Mar 11, 2020
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Ok..I have rebuilt it so that it is easier to look at.

Wires carrying positive voltage are red

Orange wire shorts 8 and 4. 2 and 6 are already shorted with soldered red wire across top of 555

White wires run to ground (-)

Yellow wire goes to + lead of the LED

Both capacitors are correctly set in terms of polarity.

I turned the voltage gradually. At 3 volts, the LED lights. At 4 volts, it lights up very bright. But it does not flash. I did not go past 4 volts. Because it was not flashing, i sense somethign is wrong and did not want to risk going. up in volts until it is figured out.
 

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bertus

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Hello,

What type of 555 are you using?
A bipolair type will work between 4.5 and 16 Volts.

Bertus
 

bertus

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Hello,

On the given page :
  • Model: NE555P
  • Package: DIP-8
  • Frequency: 500KHz of
  • Voltage: 4.5V~16V
You will need to power the 555 with the mentioned voltage.

Bertus
 

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SparkyCal

Mar 11, 2020
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You are right. I moved it to 9 volts and it began flashing. Now, the flash rate is a. little slow. Can I replace something to speed it up? I am guessing that If I reduce the 100 K resistor to a lower value?

P.S. I can't believe it's working. Thanks for all of your patience and help.
 

bertus

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Hello,

When you lower the value of the 100k the flash will go faster.
Here is more info on the 555 circuits:

hysteretic_555.gif

Bertus
 
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