**Noobalert** Aaron Chaser

[GermanMafia]

Alright I'm trying to build the AaronCake Chaser LED circuit
https://www.aaroncake.net/circuits/chaser.htm

Now I'm having a few difficulties (i just put it together for the most part tonight, haven't soldered the LED's yet and am trying to figure out which connections I need to solder to on my potentiometer).

So I'm looking at the circuit, there's no -9v. The negative poles on the LEDs drive straight into the ground, and the ground is 0v. Now I am pretty bad at electronic circuitry, so I may be missing something, but something just does not seem right about this. Also for the ground, how do you "make" a ground? I have all my grounds going into the long strip on my Radioshack IC board.

So to sum it up
*Where's the -9v?
*How do I go about creating a ground? (I read this https://en.wikipedia.org/wiki/Virtual_ground but dear god it was over my head).
*Well I guess that's it, look at how big of a windbag I am, all that typing for those two simple questions.

 

[Shax]

Alright I'm trying to build the AaronCake Chaser LED circuit
So to sum it up
*Where's the -9v?
*How do I go about creating a ground? (I read this https://en.wikipedia.org/wiki/Virtual_ground but dear god it was over my head).
*Well I guess that's it, look at how big of a windbag I am, all that typing for those two simple questions.

Why are you concerned about the lack of -9v? There is no need for a negative supply in this circuit!
The "ground" connection is the 0V line in this application....

 

[AllVol]

That little "screw" shaped symbol on pin seven of the 4011 and near the resistor R3 on the 4017 is the ground symbol... it simply means a common attachment. It is the return for the negative side of your battery. All pins adjoining that symbol are connected together. Make sure the short legs of your LEDs are connected there, too.

 

[GermanMafia]

What do you mean "return"? Is my -9v hooked up to that? Without the -9v won't there be no flow in the circuit?

Say I have a metal altoids bin as a case, would that suffice as a ground (would I still have to make a virtual ground?)

 

[Firnagzen]

You have a misconception about -9v.

Now, the battery companies spread this mistake: -9v is suppled from this battery. No. The postive terminal is +9v, the - terminal is actually 0v, NOT -9v. Now, in most cases (exceptions are those high voltages and bipolar power, but forget those) the ground is just the negative terminal of the battery.

Got it?

 

[AllVol]

What do you mean "return"? Is my -9v hooked up to that? Without the -9v won't there be no flow in the circuit?

Say I have a metal altoids bin as a case, would that suffice as a ground (would I still have to make a virtual ground?)

The (-) lead from your battery connects to ground, or the point on your circuit where all the connections come together, so to speak. You are probably thinking in terms of automotive ground, in which the chassis of the vehicle usually serves as ground.

In that case, think of the altoids tin as the chassis. If a wire from the negative (-) post of your battery were soldered to the tin, and a wire run from the connection point on your perf board also soldered to the tin, then yes, you would have ground and return. But it is simplier to just wire direct from the negative battery to the circuit board.

Does this answer the question?

 

[GermanMafia]

Okay, so i made it and it doesn't work. Now I'm screwed. I don't have even the slightest clue as to what it could be. I soldered all the joints (I think correctly, I'm sure many will think otherwise), and when I test it with my 9v battery I get readings of 9.02V on the right pins (and also on the negative input). So I don't know where I should begin troubleshooting.

Also it was made evident to me that I have no idea how a ground should work. So if I wire my negative post to the circuit board, do I still need to ground it to the tin? Is there any common mistakes I can look for? Anybody ever make this and have it not work, only to realize there mistake and then tell me about it? Thanks in advance.

I just wish I was good at this stuff, it's so frustrating looking forward to making it and hoping it will work (whenever I make stuff now I tell myself it won't work just so I won't be disappointed).

Oh yeah another thing, I don't know how fast this thing should rotate through the LED's, but I set the +/- up on the positive and negative ends of a lead and never got anything (should I have?)

 

[House0Fwax]

when I test it with my 9v battery I get readings of 9.02V on the right pins (and also on the negative input).

Please explain what you mean by 'right pins' and 'negative input'.

To start with, take the 4017 out of the circuit by either removing it from it's socket (if it is in one) or by removing the link to pin 14.
Connect the black lead from a meter ( set to DC volts ) to 0v (ground) and the red lead to pin 4 of the 4011 to check that you are getting pulses. If not, the fault is with the 4011 part of the circuit.

Check and check again all connections and polarities.

Post pictures if you can.

 

[audioguru]

Is the 4011 IC backwards?
Is the 4017 IC backwards?
Are the LEDs backwards?

Two connections on the schematic must connect to the + of the 9V battery.
Two connections on the schematic must connect to the - of the 9V battery.

 

[GermanMafia]

2 connections eh? Hmm maybe a picture will do this more justice, but what about if i were to connect all the grounds to one pad (it's a long one) and then I connected the - to that pad, that would do the same thing correct? LED's should be long leg (+) to the outputs correct?

https://en.wikipedia.org/wiki/4000_series
Maybe what's throwing me off (although I think I got it right) is the different numbers, 1-7 pinouts are connected to the LED's, and 9-11 are as well. What's up with the different "output" numbers on the diagram (and probably the schematic)?

So I bought the chips from Electronic Goldmine, they seem to be pretty old, what are the chances they are dead (there's a current going through them fine).

Also one other thing, how vital is it that the capacitor is 16v? I bought the caps from radio shack and they're rated at 50WVDC, but they're 0.1 uF Ceramic Disc Capacitors.

ACk this this is so frustrating.

BTW Thanks for the help so far guys, I have to say this is one of the most embracing first threads I've ever been a part of.

 

[GermanMafia]

Whoops didn't see HouseofWaxes' post. Hmm there's no pulses coming, it's just staying solid at 8.7 volts.

When I said "right" i meant the correct ones (the 9v input pins).

Since it seems as though my 4011 is bad, a 555 could be replaced correct? Is there any 555 and 4017 circuits that would drive 10 LEDs (I've only been able to find ones that drive 2). Thanks in advance.

It seems like every other one of my posts gets lost somehow.

My IC's aren't backwards, and the LEDs should be hooked up with + (long) leg to output correct?

So for the ground, it's still a little hazy to me. If it's connected to the - pole it does not need to be connected to anything correct? And just to make sure, although this is a stupid question, if I were to throw all my grounds onto a single pad and connect the - pole up to that pad that would be fine correct?

Thanks a lot for the help so far guys, this stuff is confusing and I was (and still am) treading water, it's nice to know I now have some helpful and knowledgable people to go to.

 

[House0Fwax]

Is there any 555 and 4017 circuits that would drive 10 LEDs (I've only been able to find ones that drive 2)

LED sequencer
The 10 LEDs are controlled by the 4017. The 555 (or in your case the 4011) only supplies the clock pulse.

Though, since you have made this design, it's best to try and get it working.
+ side of battery goes to 4011 pin 14 and 4017 pin 16
- side of battery goes to 4011 pin 7 and pins 8,13 & 15 of the 4017

Are you sure everything is connected up correctly ?
ICs the right way round ? LEDs ? The resistors and the capacitor ?

 

[GermanMafia]

I'm not really sure of anything anymore, but I'm pretty sure that everything is hooked up correctly. What are the reasons why the 4011 wouldn't be pulsing?


**broken link removed**
Note my inability to solder well (I don't think I crossed any connections I shouldn't have though).

It sized it down a lot, if you'd prefer a higher res one just hit me up.

Also if you'd like an undoctored one tell me

EDIT: better image
**broken link removed**

 

[AllVol]

Hi, Germanmafia,

First, where is the capacitor for your 4011? I can't see it.

EDIT: On second look, I see a cap, but the one I am referring to should be C1, coming from pin 4 of the 4011 to one end of your variable resistor. Without this capacitor, the 4011 will not "pulse".

 

[GermanMafia]

In the first picture it's at the far left, right below the - (black) wire. in the one with all the colors it's the blue line on the lower half right side. It's a light orange color.

Do potentiometers need to be grounded? I have one that has 6 pins on it (It's a 100K pot).

So for a moment (it was like a minute) the thing may have been pulsing. I had one lead from the multimeter on pin 7 of the 4011, and the other on pin 14 on the 4017. The voltages were all over the map, from 8.3 -> 0 and everywhere in between. I wasn't moving at all so I don't think it was the multimeter reading a different pin or anything, but for the life of me I can't reenact that (Also, the LEDs DID not blink during this), but it must have been a fluke. Any reasons why my 4011 doesn't feel like doing anything?

 

[AllVol]

Okay, GM,

I see how you have done it. Sort of the long way 'round, eh? Still, there should be a pot (variable resistor) between the junction of that capacitor and the one meg resistor, and the junction of pins 3 and 5 of the 4011. I don't understand the resistor shown on the very end of your board. It appears to be in the ground line. ???

The attached photo shows the same circuit which I built some time ago, but the LEDs and their resistors are not included.

 

[GermanMafia]

Whenever I test it I put the pot on the 1 meg resistor and the 3/5 jumper (didn't solder it yet cause I don't know which pins are the right ones to use on the potentiometer). The resistor goes into the same pad that the capacitor is on, and then on the other leg of the capacitor pin 4 and the wire to pin 14 on the 4017 are all on the same pad. The lead of the resistor is intercepted by the potentiometer which is connected to the 3/5 (alligator clips). I think I should have just done this on a full board than a 1/2, but I'm a big believer in not wasting valuble PCB.

 

[AllVol]

The attachment below shows you how to check the pot. If yours has six pins, but is similar to the one shown, you have two pots on the same shaft. It makes no difference... use either triplet, just make sure they are for the same resistor. In the original schematic, the pot is used as a rheostat, in that the wiper (center post) is connected to one of the outer posts. The pot is then inserted in the circuit as I believe you are doing, although I'm having trouble following your description.

I believe in future you will find it much easier to use one of the inexpensive plug-in breadboards until you determine the circuit will actually work. Such use saves time and components in the long run, and makes visualizing and understanding the circuit easier.

 

[House0Fwax]

I may be wrong here, but pads 1 & 2 look like they are shorted to pad 3 on the 4011.
Here's an idea, remove the 4011 ( observing anti-static precautions ) and using the multimeter ( set to low ohms ) check for solder bridges. If you find any re-melt the solder and run a knife or similar sharp pointy item through the bridge.

If you have 6 pins on your pot, then it appears to be a stereo one ( I seem to recall a thread not long ago sounding similar ) In which case you will have two rows of three pins. You need to join the middle pin to one of the outer ones ( either one ) and use the two joined pins for one conection and the remaining pin for the other.
Don't worry about the 4017, just concentrate on getting the pulses from the 4011 for now. You can deal with that later.

 

[AllVol]

If you have 6 pins on your pot, then it appears to be a stereo one ( I seem to recall a thread not long ago sounding similar ) In which case you will have two rows of three pins. You need to join the middle pin to one of the outer ones ( either one ) and use the two joined pins for one conection and the remaining pin for the other.

Thanks for the collaboration, pal.