Submission, CMOS 555 Long Duration LED Flyback Flasher

Thread Starter

Wendy

Joined Mar 24, 2008
23,415
That has nothing to do with this circuit. It is verified, it works (well over a month continuous) and it wasn't dim. It also lite up more than one LED.
 
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SgtWookie

Joined Jul 17, 2007
22,230
There are a number of problems with the CMOS 555 circuit as posted.
Firstly it is not efficient as the circuit does not know when the core is saturated.
Secondly it uses a lot of components. And there are other faults.
The circuit I have produced uses half the components and is much more efficient. This
circuit will also drive 2 white LEDs in series without alteration - but the LEDs
will be much dimmer.
Why did you use rod material instead of a toroid?

If your intent was to radiate broadband RF noise, you succeeded.

If you intended for efficiency - well, consider using toroids instead.
Amidon's material 77 would be a good choice for 1kHz to 2MHz; it has an initial permeability of 2,000 and max of 6000. An FT-23-77 has an Al of 390mH/1000 turns.

Even a plain ferrite toroid would be much better than using rod material.
 
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SgtWookie

Joined Jul 17, 2007
22,230
Bill,
Just in case you're interested, here's where you can find Amidon's 77 toroids:
https://www.amidoncorp.com/items/18
Part number FT-23-77 is $0.50 each. 24 evenly spaced turns of wire on that little 0.23" toroid will give you about 225uH. It would be a bit tight unless you're using AWG 30 or smaller.

FT-37-77 is $0.60 each. Al is 880; you'd only need 16 turns for 225uH. AWG 22 magnet wire would work perfectly for one evenly spaced layer.

I'm sure you know this, but for the benefit of others... when I'm talking about turns of wire on a toroid, each time a wire loops through the middle (hole in) the toroid counts as 1 turn.
 
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Thread Starter

Wendy

Joined Mar 24, 2008
23,415
Yeah, I might use them for other projects, but remember for these articles Radio Shack is the preferred provider. :p

They also have toriods too, so it might happen on another article coming up.
 

SgtWookie

Joined Jul 17, 2007
22,230
Yeah, I might use them for other projects, but remember for these articles Radio Shack is the preferred provider. :p

They also have toriods too, so it might happen on another article coming up.
I did a bit of searching and found this fellow's blog:
http://watsonseblog.blogspot.com/2008/05/cat-273-0106-toroid-coils-measurements.html
The blog is about inductors he was making/testing for Joule Thieves.

Looks like he measured some of the lower-current RS toroids at less than 2uH.
The info he gave is far from complete, but I'll guess that the low-current assortment has toroids with a low Al value; meaning a lot of turns would be needed to get the inductance up. I hadn't been interested in buying any; they're overpriced and have no published specifications other than current rating.
 

Audioguru

Joined Dec 20, 2007
11,248
Of the three circuits presented here, only one works because the other two circuits have the LEDs upside-down.

Like Talking says, only a single C555 can be used with its duty-cycle adjusted to quickly pump current into the inductor then release it for a longer time into the LEDs.
 

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talking

Joined Dec 10, 2009
12
Agreed, the CMOS 555 circuit works, but you also have an obligation to produce circuits that use the minimum of components and take note of what other engineers are telling you.
I have sent you a number of personal emails and you have not responded to any of my suggestions.
Some of your circuits DO contain faults and Audioguru has also pointed out that one circuit contains LEDs that are up-side down.
In you explanations you are totally unaware of one of the most important features of an inductor, namely that the voltage produced by the collapsing magnetic field is produced in an opposite direction to the energising voltage – as proven by the LEDs being inserted up-side-down.
On top of this, your explanation of “charging” and inductor is not a term used for the operation of applying a voltage to an inductor. The word “charging” normally applies to the result of a voltage being applied to a capacitor.
There are other points that I will address at a future time.
 

Thread Starter

Wendy

Joined Mar 24, 2008
23,415
Obligation, to whom, you? I think not.

I looked at your website, and talking responses we've established how well you read on the other thread. If you think you can do better feel free to contribute, but while I listen to suggestions I am not obliged to follow them.

Forum crashed mid edit:

My obligation is to write a clear, concise article that is interesting, accurate, and eductional. The circuit must work, but it need not be practical. This is a text book first.
 
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talking

Joined Dec 10, 2009
12
Here is a link to my 50 555 Circuits. It has now grown to more than 61 circuits - many of which I have designed for a specific application:

<snip> Links removed by moderator. The site hosted pop up advertising, had questionable content, and it was quite apparent there was a real gulf in philosophy between the OP and AAC.
 
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talking

Joined Dec 10, 2009
12
1.
When the voltage across pins 2 and 6 reaches 2/3 of the power supply
The sentence should read: When the voltage at pin 6 reaches 2/3 of the power supply. . .

2. The LEDs in this circuit are up-side-down:



3. This is a bad design:



The HIGH on pin 3 of some of the CMOS 555's is up to 1v lower than the supply and this circuit will not work. You may be lucky to get a chip that produces a HIGH that will turn the PNP transistor off.


4.
as well as the inverter made using Q1.
The transistor Q1 is not primarily an inverter, it is a BUFFER.

5.
This circuit uses the exact same principals
Should be:This circuit uses the exact same principles. Remember: principal is your PAL - like your headmaster.

6.
the largest practical resistors are used to minimize current
Should be: the largest practical values of resistance are used to minimize current.

7. You will be lucky to turn off the second CMOS 555 with this circuit, as explained above. Transistors actually start to turn on at 0.52v and since Q1 is in a high impedance stage, it will very quickly bring the reset line of the second CMOS 555 above ground potential and activate the chip.
For many CMOS chips operating at 3v, the high will be 0.6v below rail and the second chip will never be turned off.

 

Thread Starter

Wendy

Joined Mar 24, 2008
23,415
Reality intervines, dispite your wishful thinking. They work as is, and have been fully tested and documented as such.

You really need to study CMOS 555s, with light loading they are rail to rail circuits. What you are talking about is the darlingon transistor drops, which don't apply to CMOS.

Here is a link to my 50 555 Circuits. It has now grown to more than 61 circuits - many of which I have designed for a specific application:

URL Deleted.
Perhaps you are more interested in selling your site than accuracy in your comments? Spam is not allowed on this site.
 
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Thread Starter

Wendy

Joined Mar 24, 2008
23,415
Again, it is obvious you don't know CMOS. I consider anything that gets within 0.1V of the power supply rail, CMOS gets considerably closer with no or little load. Goes with their high input impedance.


It ain’t what folks know that get them into trouble, it’s what they know that ain’t so.

Mark Twain


You keep insisting circuits that have been tested and verified can't work, which makes you look kinda silly. Tearing me down is not going to build you up, though it does let people form their own opinions.

Thing about electronics, it is a lot like science in that it is reproducible. My info in open source, many eyes view it, some will go the next step and duplicate what I've already done. I give complete plans to do exactly what I have already done.

I do this as a labor of love, I get no money from it. How much are you selling your book? Your agenda is showing.

Moderator's note: this reply may seem odd, as the post to which it got addresses has been removed. The post had no merit.
 
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Audioguru

Joined Dec 20, 2007
11,248
The problem is that the schematics have LEDs upside-down but the breadboard circuit has the LEDs with the correct connections. The schematics should be corrected.

Post #9 has text that talks of an "old" circuit and a "new" circuit and that an error was corrected but the "new" circuit is not shown. Maybe the "new" circuit is the corrected schematic.
 

Audioguru

Joined Dec 20, 2007
11,248
The HIGH on pin 3 of some of the CMOS 555's is up to 1v lower than the supply and this circuit will not work. You may be lucky to get a chip that produces a HIGH that will turn the PNP transistor off.
No.
All Cmos 555 ICs have rail-to-rail outputs when the load current is low like when the PNP transistor is turned off.
Intersil's datasheet for their ICM7555 is the most detailed with a graph showing the voltage drop with current. At a load current of 0.1mA the output high voltage is at most 0.1V from the positive supply and the current in the base of the PNP transistor is far less.

It is the ordinary bipolar 555 that has an output high that is at least 1.2V less than the supply voltage.
 

Thread Starter

Wendy

Joined Mar 24, 2008
23,415
I just double checked the layout and the schematic, they agree. I also hadn't got around to tearing the circuit down, so I put some newer batteries in, it works too. It also agrees with the layout. Everything jives, it is confirmed as a working circuit as is.

 

Thread Starter

Wendy

Joined Mar 24, 2008
23,415
The two posts in front are the article, I kept the others as part of the conversations. I never turn down a little help from my friends. You might come off as crusty now and again, but I value your input.

The article has been submitted, BTW. I don't know when it will be merged into the book. At this point it is in Dennis's hands.
 
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