No, no, Im saying using the voltage divider as you did wont work. You shouldnt need it at all.

Take a 1k Ohm resistor and the wall wart. run it over the 1k resistor. What voltage do you read? If you measured the wall wart with no load, you will get quite a fluctuation. but It should calm down with a load.

You want to go from wall wart to 7812 to the rest of the circuit.

Completely skip the voltage divider.

 

It drops .05 volts with the resistor. I just don't have a 7812 on hand so I'll just use the 7805 for now.

Will that fix the voltage drop on the board?

 

the 7805 will drop it to 5v

You need the 7812 to get the 12v

Getting rid of the voltage divider will do it.

 

I just noticed that on Bill's long duration minimum parts diagram, the red LED connected to output at 3 is running from power to 3 (if I am reading correctly). Is the term 'output' on the 555 misleading?

Still, I hooked the led up that way and it made no difference. I have 5.02 volts going in now, and I read 1.82V at pin 3.

 

The long duration circuit uses the extreme low power supply voltage and some oddities of CMOS IC to do something that isn't normally recommended. It is the exception that proves the rule.

Look at the 555 Hysteretic Oscillator instead, though the design is slightly different it is more typical.

I have to admit I'm a bit confused by the voltage divider references, unless you are talking about post #7 and R1/R2. They force the 555 circuit to go to a voltage just below the standard 1/3Vcc (which is common to practically all 555 designs) and only works once on startup.

I don't necessarily approve of the end use, but I'd rather have it done right and you understand how to use the chips. The LEDs, 555s, Flashers, and Light Chasers also covers a lot of this ground too. You strike me as just slightly above beginner, I think you understand Ohm's Law and most of the circuit, but need a little help with the rest of the understanding. AAC is a teaching site as much as anything else. My Cookbook is fairly new, you can see some other configurations for a 555.

 

No, the 555 can Source or Sink current.

See, you can switch a circuit from the positive side or the negative side.

The OUT on the 555 isn't the only the output, it is the switch out.

So if it is a LOW or (0) it can sink current from another V+ source.
If it is a HIGH or (1) it can source current from itself to a GND.

 

Neat. Thanks.

Turned the circuit to sink, replaced one of the 1K resistors with a 1M, and threw in one of the piranha LEDs (perhaps it is clear?), and it flashes like a son of a gun. Very dim though - I would guess because it's flashing rapidly.

I think the circuit is wired correctly it's the math that is tripping me up. I think I am making a simple exponential error, but haven't nailed it down yet. I would like to try a 2HZ cycle.

I can't find a resource that will tell me what my outputs should read given my inputs, or perhaps I'm not knowledgeable enough yet to decipher the information I have yet.

I get these readings -

From 7805 - 5.02V
1 - 0V
2,6,7 - cycling rapidly between 2.27V and 3.34V (I can just barely follow)
4,8 - 4.45V
5 - 2.93V (though 5 is unconnected)


edit: LOL Went and changed out some more stuff. Swapped out the other 1K resistor for a 1M, switched the .1uF capacitor for a 1uF and now I have a period of about 2 seconds. Which means I was off by either a factor of 100 or 10,000.

 

Them zeros will get you every time.

 

Like I said, replace the 1µF with a 10µF.

I keep a calculator next to my keyboard just for such a contingency.

 

the 78xx series of regulators sinks about 5.5mA (give or take) via the GND terminal.
You can increase the output voltage by adding resistance between the GND terminal and ground.

To make a 7805 regulator output 12v (>14v in required) you can add (12v-5v)/5.5mA = 7v/0.0055 = 1,273 Ohm resistor between the GND terminal and ground. Try a 1.2k or 1.3k resistor and see what you get.

[eta]
If you're going to use a regulator in your application, it'll have to be either a low-dropout type, or you'll have to settle for around 10v out of a 78xx series regulator.

To get 10v out of your 7805, try using a 910 Ohm resistor between the GND terminal and GND.

During load dumps, you'll have rather high peak voltages on the wiring harness. You can help reduce those peaks by using a 5 Ohm resistor between your supply and the IN terminal of the regulator, and a large capacitor (1,000uF or more) to ground.

 

Hallo,

Also a 6.8 Volts zener diode in the ground pin of the 7805 would work.
That would give you about 11.8 volts at the output.

Bertus

 

Bertus,
That could work, but typically available Zeners require around 10mA to 20mA to regulate properly. To source the additional current, a resistor would be needed from the IN terminal to the Zener/GND terminal connection.

Just using a simple resistor from the GND terminal to ground would be more reliable and easier to find.

 

Since it is the zeros, can someone tell me the proper units for the formulas, or point me to the place I can find out? The source I was using said they should be ohms and farads. Since I usually trust myself to do basic multiplying and dividing, I wonder if this is correct.

I was at radio shack and found a 7812 regulator. For the LEDs I don't think I will need a regulator in the circuit. The regulation of the arrays between 12V and 14V is not that different. I'd just as soon not use one for the board either, but I haven't gotten that far yet. If I do I may just bring it down to 5V.

If only I had all this stuff laying around...

I hope that trick about the resistors holds in the old memory banks. Looks like a good one to know.

 

You will have more heat dissipating to bring ~16v to 5v then you will from ~16 to 12v.

That means at 5v, shorter life and more possible problems.

 

Since it is the zeros, can someone tell me the proper units for the formulas, or point me to the place I can find out? The source I was using said they should be ohms and farads. Since I usually trust myself to do basic multiplying and dividing, I wonder if this is correct.

Ohms are specified in Ohms, the C is in microFarads (uF).

Generally, if you see your values of resistance going below around 2k or so, you need to use smaller caps. If your resistors are getting pretty large (over 1 Meg) you need to use a larger cap.

I was at radio shack and found a 7812 regulator. For the LEDs I don't think I will need a regulator in the circuit.

Then you are not thinking correctly.
The first load dump that comes along will eat your LEDs.

The regulation of the arrays between 12V and 14V is not that different. I'd just as soon not use one for the board either, but I haven't gotten that far yet. If I do I may just bring it down to 5V.

If you are going to use a 78xx regulator, you need to set your sights for around 10v out.

You might be using compression braking, but if your engine is at idle, you will not be generating power, and your electrical system will be down to 12v. The 78xx series has a minimum 2v dropout from in to out, so 10v is the most you can hope to get out of it.

If only I had all this stuff laying around...

...then you wouldn't be able to find it. Ask me how I know this...

I hope that trick about the resistors holds in the old memory banks. Looks like a good one to know.

There are lots of "tricks", but the resistor thing is all Ohm's Law.

 

OK. Another teachable moment. I saw this before. What is a load dump? And how does it come from my 12V system?

Got a 500K resistor you can borrow me?

(taking a break from trying to figure out the 4094 part of the system)

 

Bertus,
That could work, but typically available Zeners require around 10mA to 20mA to regulate properly. To source the additional current, a resistor would be needed from the IN terminal to the Zener/GND terminal connection.

Just using a simple resistor from the GND terminal to ground would be more reliable and easier to find.

Seem like a resistor from the output to the zener/GND terminal would result in less ripple and more predictable current. A 1k resistor would provide 5mA, with another 5,5mA (nominal) trovided through the GND pin.

 

OK. Another teachable moment. I saw this before. What is a load dump? And how does it come from my 12V system?

A "load dump" is when a load is suddenly turned off, like lamps. The generating system takes a moment to adjust it's output. In the meantime, current is still flowing out of the generator, and you can wind up with some pretty high voltage spikes.

Got a 500K resistor you can borrow me?

Sure. Two 1MEG resistors in parallel = 500K.
What size? 1/4W axial, 1206, 0805 SMT?

(taking a break from trying to figure out the 4094 part of the system)

4094? CD4094B 8-stage shift and store register?

 

Seem like a resistor from the output to the zener/GND terminal would result in less ripple and more predictable current. A 1k resistor would provide 5mA, with another 5,5mA (nominal) trovided through the GND pin.

Good idea - that works for me

A 1N754A 6.8v Zener drops it's rated voltage at 20mA; so you'd need 15mA (approx) from the output. A 330 Ohm resistor to the junction of the GND terminal and the Zener would do it - for that particular Zener.

The trouble with Zeners is that many hobbyists simply don't realize that they have to check the datasheet for their specific Zener to find out what current they need to have flowing through it, in order to get the rated voltage out of it.

 

Sure. Two 1MEG resistors in parallel = 500K.
What size? 1/4W axial, 1206, 0805 SMT?
DOH! I should have known that. It was in there when I was refreshing my memory on series resistors. Thanks.

4094? CD4094B 8-stage shift and storet register?
HEF4094B Which is on my dining table doing odd stuff with a couple of LEDs right now.