Re: International standards (2022 Update)

rest

Well, only in your bible-centric world.

The most disturbing part of that particular fable is the "it's okay to shag your daughters in a cave" bit.

Reply to
John F. Miller
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I am trying to get a small gas powered engine running, (Rockwell) and I need some help electrically. I have the unit running on the bench, with a simple on/off switch, and a 12 volt marine battery. I can get it started absolutely ok, but what I'm wondering, is it possible to have the starter motor charging the battery while its running, so it can power lights, just like a car does?. If I can do it, is it simply a matter of using a relay, that will take the electrical power generated from the the starter, once the engine is running. In other words a relay that will switch between running (starting the motor), and generating (running from the motor). Do I need to place a voltage regulator between the starter and the battery?. What do I need to keep the battery from overcharging? Thanks Kim

Reply to
Neil

Generous readers of this forum walked me through the building of a Pulse Width Modulator. This was earlier in the year in January.

The project was to reduce the speed of a small

12v DC motor which turns a 4 foot disk, in this case a painting on canvas mounted on a round laminated wooden stretcher, so fairly light. The point of the PWModulator is to conserve the batteries which power the motor whereas a voltage regulator would discharge the batteries faster.

Other things intervened but having returned to the project I was pleased to see that the electronics work and that the speed of the motor can be adjusted to very slow.

I now need to hardwire the components. As a total newbie, I am not even sure which side of the board the components are mounted on.

The schematic of the PWModulator and pictures of the breadboard with components and of two sides of boards for hardwiring is at

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I am writing to ask if there is an online FAQ on basic electronic wiring or failing which whether there is a useful book you can recommend to learn how to do this.

Thanks in advance.

Mike Eisenstadt

Reply to
Michael Eisenstadt

The hybrid alternator/starter assembly used on marine engines isn't regulated well at all. The voltage can go up as much as 16V and this is bad for the battery.

Use a small automotive alternator. They have a regulator installed already and it will maintain steady 13.8V to 14V even with varying RPM. You can tweak the pulley size to change the maximum output.

Reply to
AC/DCdude17

Older small marine engines are often fitted with a thing called a dyno start, which is a starter motor dynamo and regulator all in one. The system is not terribly well liked and a lot of people remove the dyno starter and fit separate alternator and starter motor.

The trouble with automotive starter motors is that usualy they are very crude and inefficient, because of the abundance of power available in a typical application where a starter motor is used, there is no point developing a more efficient/expensive solution.

If your stater motor is not typical of automotive gear, and is of a reasonable quality, then it may be practical but you ned to regulate the output such that it will be suitable for your battery charging.

You might look at fitting a separate alternator, you get a known working solution with regulator built in and they are very cheap from the local breakers yard.

KevinR

Reply to
KevinR

The component side, of course.

;-) Rich

Reply to
Rich Grise

Hi, Mike. You've already got it working on a protoboard. Why not buy a PC board that's set up like your perfboard, and just transfer it over? If you know how to solder, Radio Shack has a Matching Printed Circuit Board for $3.29 USD (Catalog #: 276-170) which will do the job.

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While you're there look at the newbie books to get you started.

Good luck Chris

Reply to
CFoley1064

When using an LED from a power source whose voltage is higher than the rating of the LED, a resistor is typically used. If more than one LED is to be used (let's say three), each one would have its own respective resistor, and the schematic would look something like this ...

+----------+-----------+-----------+-------------- | | | | | | | | | LED-1 LED-2 LED-3 power | | | source | | | | resistor-1 resistor-2 resistor-3 | | | | | | | | +----------+-----------+-----------+--------------

But if all the LEDs were of the same voltage and amperage (e.g., if they were all of the typical 3.6V 20mA), isn't there a simpler way to do this by using only ONE resistor? In other words, can't it be done like the following ?

+--------------+-----------+-----------+--- | | | | | | | | power LED-1 LED-2 LED-3 source | | | | | | | | | | | +---resistor---+-----------+-----------+---

And what would be the proper value of the resistor?

(I'm guessing it would be one third of what it was in the first diagram because the required voltage reduction would be the same but the effective current draw of the 3-LED assembly would be three times what it was before. So if the power source were 6V and the LEDs were 3.6V and 20mA each, then the required resistance would be 120 ohms in the first case and 40 ohms in the second case. Correct?)

Reply to
wylbur37

Most LEDs don't come with a built-in resistor. Ones that do are more expensive.

The LED driver chip makers get around this problem by connecting up to maybe 6 LEDs in series. All the LEDs then get exactly the same current.

Also, if you are parelleling 3 or more LEDs to get more light, then instead, use a single more powerful LED such as the Luxeon Star. There are also other brighter LEDs such as the 'spider' LEDs.

With blue or white LEDs, if you do parallel LEDs without a separate resistor for each, be prepared to have one or more LEDs that overheat with a greatly shortened lifetime and/or much dimmer light output. Also, one or more LEDs may overheat and its voltage drop will go up, which then causes that LED to get dim, and the other LEDs will then end up with more current. They then may overheat,, etc.

Murphy's Law always applies. :-P

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Reply to
Watson A.Name - "Watt Sun, th

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One factor needs further explanation. LEDs have a negative temperature coefficient. So if one LED gets a bit more current, it gets hotter, which then causes its voltage drop to be lower, which then causes it to hog even more current. This vicious circle of current hogging usually causes the LED to overheat and have a much shorter lifetime. This may take hundreds or thousands of hours, but the closer the LEDs are run to their maximum, the less time it will take.

One way to reduce this tendency is to have all the LEDs closely coupled thermally. But then this usually means they will get hotter, because there's a lot of heat in a small space. I put a dozen white LEDs on a predrilled 0.1" hole spacing PCB spaced every other hole, and they got too hot. So I spaced them out every third hole and they stayed cooler. (View with Courier font)

Originally: 0 . 0 . 0 . 0 . 0 . 0 0 . 0 . 0 . 0 . 0 . 0

Much Better: 0 .. 0 .. 0 .. 0 .. 0 .. 0 0 .. 0 .. 0 .. 0 .. 0 .. 0

I've been leaving these cheap white LEDs from Hong Kong on 24/7 at their maximum current, and they just get real dim after a few months and a thousand or so hours. Just remember that whatever you do, heat is the enemy of LEDs, and the hotter they get, the shorter they will live. It's just a law of physics.

Reply to
Watson A.Name - "Watt Sun, th

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Reply to
Watson A.Name - "Watt Sun, th

The LED's will not normally be accurate enough to each other. You will most likely have a situation where one would light normally, and the rest would be dim, or not lit at all.

You need a separate resistor for each one, or you should look for LED's with built in resistors. If they have built in ones, then you will be restricted to the voltage range indicated for the LED's.

--
Jerry G.
==========================
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Reply to
Jerry G.

Put the LEDs in series, not parallel. That is:

  • -----/\/\/\/-----LED-----LED----LED----- -

They will all carry the same current but need not have the same voltage (e.g., you can have a 3.2-volt blue LED, a 2.1-volt green, and a 1.8-volt red, all in series).

The power supply must be higher than the sum of the LED voltages, with the resistor calculated to make up the difference.

Example: With the 3 LEDs I just mentioned, 3.2 + 2.1 + 1.8 = 7.1 volts.

Suppose the supply is 12 volts and you want 20 mA.

Then the resistor is (12 - 7.1)/0.020 = 245 ohms (use 270, which is a standard value).

As someone pointed out, if you put the LEDs in parallel, they will not share current equally unless their voltages are *exactly* equal, which depends on temperature, manufacturing lot, etc., even if they're all the same color.

Reply to
Michael A. Covington

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Yes; contrary to what I just said, they probably *do* have enough internal resistance to even things out if you make a good faith attempt to use the same kind. Obviously one red and one green will not do... even one bright red and one dark red might not do.

Reply to
Michael A. Covington

I think this circuit should be fairly simple, but I have no idea how to find the correct chip for this application. Here is what I'm trying to do:

Basically I have a variable resistor (10k pot). When the resistance is < 2.5k I want an output1 to be set to GND and output2 open. If the resistance is >7.5k then I want Output 1 open and Output 2 set to GND. At all other times I want Output 1 and 2 to be open (so from 2.5k to

7.5k).

Is there a chip or a couple of chips and some resistors that I can put together that will accomplish this? I have access to 3.3v and 5v.

Thanks for any help,

-Zinfari

Reply to
Zinfari

parallel

That is NOT parallel, that is SERIES.

[snip]
Reply to
Watson A.Name - "Watt Sun, th

Actually you still need the resistor even when the voltage is at the rating of the LED albeit a small one. It serves the purpose of current limitation in addition to dropping voltage.

Right. It's the correct way to get uniform brightness among the set.

It serves the safety purpose, but then you get variability on the brightness because each LED will not draw exactly the same current. So some will be dimmer than others.

The second thing is that the total current through the resistor will be a factor of the current through each LED. So for your example the resistor above would need to allow 60ma of current to flow through to feed the LEDs downline. The problem is that if one LED blows out then that current will have to be pushed through the other working LEDs, which can cause the others to fail also.

Single resistor value divided by the number of LEDs across it.

Right. But see the caveat above.

A better way to do it is to bump the voltage up and put the LEDs in parallel like this:

+V -> resistor -> LED1 -> LED2 -> LED3 -> GND

You treat the equation the same as a single LED except that you add the Vf voltages of the LEDs. So for your example the total voltage of the LEDs would be 10.8V. So the resistor you'd use with a 12V +V would be:

(12V - 10.8V) / .02A = 60 ohms.

But there are significant differences:

1) Each LED will draw the same current, so you'll get uniform brightness. 2) If a single LED (or the resistor) fails, then the whole string goes out with only the single failed component, instead of going into cascade failure due to increasing overload.

When doing a single resistor, that's the better way to handle it.

BTW, thanks for crossposting this message. There are many who don't understand the concept. But I think I'll limit my replies to basics and misc, which are appropriate.

BAJ

Reply to
Byron A Jeff

Thanks Chris for your prompt reply. I already have the board that you recommend. There's a picture of the 2 boards already in inventory reproduced on the charlesumlauf.com/wiring.htm I referenced above.

I have already soldered the 3 terminals on the pot.

It seems sorta likely that you mount the components on the bare side of the board at the same time putting an additional wire in that hole to connect to some component somewhere else on the board. Then on the other side of the board you trim the wires that poke through the various holes and spot solder them to the copper ring around the holes.

Would the official basic soldering FAQ ISO DIN 9000 manual confirm this supposition?

Thanks to sci.electronics.basics folks I already got it past Proof of Concept on a breadboard. Do I have to buy a book at this point? I hope not.

TIA

Michael Eisenstadt

Reply to
Michael Eisenstadt

My last post proves once again that the fingers are faster than the brain.

The holes in the board are too small for both the MOSFET's legs and an additional connecting wire to go through together. And most likely the CMOS NAND gate legs as well.

(schematic etc. is reproduced at

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So my original guess about how to put and connect components on a board was wrong.

I am overlooking something very obvious. Thanks for your help.

Mike Eisenstadt

Reply to
Michael Eisenstadt

Or if the supply voltage is higher than the sum of the LED voltage drops AND they have similar current and light output ratings, put the LEDs in series with a single resistor.

With the example you gave though, there is no choice but to have individual resistors for each LED.

--- sam | Sci.Electronics.Repair FAQ Home Page:

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Reply to
Sam Goldwasser

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