Beginner circuit help.

I'm working on a circuit to help learn all this jazz, and wonder how this could be accomplished:
Two devices to be used on the circuit are a speaker and an LED.
Source: 12V battery (Two 1.5V + 9V batteries)
LED: 3V / 20mA Speaker: 12V / 15mA
How could I design a circuit that houses both these devices at their rated amps? Right now I'm putting a parallel circuit near the beginning to cut up the amperes, but then I seem to get stuck when trying to use the lower voltage device at 20mA.
Any help?
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In article <3423d99b-7740-479f-bf38-
says...

Not a good idea.

???Where did you get this information???

Maybe it would be better to start with what you're trying to accomplish.
--
Keith

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My goal is to understand how to design simple circuits, as purposeless as they may be, and my posting here was for aid, not for snide response.
It may not be a good idea to string two batteries together, but I did it anyway to see if the voltage output was 12V, and it was. Insight to this would be helpful, not just tossing the idea back at me in dismissal.
The voltage/amp ratings were on the package for the components. Regardless, the purpose was not to create a meaningful device, but the goal of creating a circuit with certain restrictions for my education.
Again, my purpose for this is general circuit design knowledge, and your response uninformative to no end.
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In article <968b028f-0092-4d2c-bb2f-2a8bb43e8d41

It wasn't a snide response. You clearly have no idea what you're doing so thought it was a good idea to try to figure out what you wanted to do before starting. OTOH, you only have three devices, each with two leads. There aren't that many ways to make smoke.

Batteries aren't something to take lightly. In this particular case, it's not the fact that you're stringing two batteries together, rather that they are completely different batteries. A mistake can make one explode. You can hurt yourself doing things that aren't smart. Generally people try not to give you tips on how to accomplish this fear.

"Speakers" aren't rated in volts and milliamps and certainly aren't useful in a DC circuit. I don't know what you have, but it's not a "speaker".

Hurt yourself. But if you want help you'd better be able to ask a meaningful question (e.g. "how do I make the LED light"). BTW, sci.electronics.basics is a better place to ask beginner's questions.
-- Keith
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For jazz, first get a saxophone. <g>

First hint: the speaker needs to see an AC audio waveform.
Second hint: the LED will have a voltage rating, probably around 2 volts. It will also have a current rating. if it 20 mA you need to drop a volt @ 20 mA calculate the resistance needed and the power needed to size the dropping resistor.
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Dan Schuman wrote:

I assume that the "12v speaker" is a 12 volt buzzer of some kind. In which case you can put it right across your "12v battery" - observing proper polarity as required (many of these "buzzers" use electronics to generate the noise and thus have a + and - terminal that must be connected to the appropriate + and - terminals of the battery).
The LED needs a series resistor of 470 ohms before it can be connected across your "12v battery". If it doesn't light, swap the wires over that are going to either the LED or to the battery.
Having got each of the above circuits to work independently, you can connect both of them to the battery, simultaneously, to get both noise and light.
How you could work this out for yourself is to go on a course, either via a correspondence school or local college. There you can learn stuff like how to work out resistor values needed to limit LED currents to 20mA. There are some really good distance learning courses these days, many that come with a box of electrical and electronic components and mix a lot of practical work, with the theory behind it all.
Playing with electricity can be great fun but learning by trial and error is probably not the best route to a good understanding. A structured course is usually a much better approach. Be warned that even a lowly AA cell can cause a fire, intentionally or by accident. Once you start using mains, or car batteries, that remote possibility of something going catastrophically wrong becomes a near certainty - unless you know what you are doing.
-- Sue
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Thanks for your help. I'll test it out and (wearily) post back again with other basic questions. Gauging the responses thus far, it seems not to be too friendly of a place, this Internet of ours.
Think there'd be a better place to go for questions on basic things other than this forum? I don't have the time for courses right now which is why I'm resorting to a few books on my own.
Again, thanks Sue.
Dan
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Dan Schuman wrote:

The only difference between "a few books" and a distance learning course, of the type I mentioned, is that the latter is *designed* to make it easy to learn. So it builds up knowledge in a structured way, and reinforces it with practical exercises that work. Plus many of them come with a tutor at the end of the phone or (e)mail - to help with anything that you aren't sure about.
I don't know about the US, but here in the UK there is a range of excellent distance learning courses that can take anyone with the will, the interest and a bit of ability from zero to PhD - or any point in between that they want to achieve.
This place, apart from a few intolerant wierdos, is friendly and helpful. However, it is full of engineers/technicians who are used to and respond better to questions framed like, "I want to achieve this, how do I do it?". Or, "My book says to use Litz wire, but how do I solder it?"
It is also full of people who have seen (or know) of horrific burns caused when electricity "escapes". They really don't want to see or hear of any more.. Electricity is *incredibly* dangerous - it only takes a tiny mistake and, literally in less than a millionth of a second, it can reach out and kill you. There are no old, bold, electrical engineers. Safety is always uppermost in everything they say, do and write.
-- Sue
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Actually, the problem I'm presenting to myself is that the source ideally would be 12V from one point. I'm not trying to divvy up the voltage, but work with one source of 12V. Circuit I've been thinking would be more like this:
----|+ 12V -| ----- | ___|__ | | | | 3V.LED Buzzer.12V | |___ __| |____________|
Sorry if the ASCII came out badly. I'm guessing it could work something like that (resistors in needed places but not labeled), but I was trying for the whole circuit to draw as few amperes as possible, ideally 20mA. So far I can't think of a way to have that work with two points that respectively will have 12V @ 15mA and 3V @ 20mA. For it to work on above circuit I suppose it'll draw more amperes than 20mA overall..
Sorry again if I'm wasting time on something that's really off the mark. Just trying to attain a solid grasp of fundamentals and find myself hard-pressed in drumming up talks with anyone knowledgeable.
The 9V / 2x1.5V batteries are linked in series. You said this is OK, but if there's really a chance of these things exploding that's something I'd like to know.
Dan
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In article <d65a870b-298d-4c4b-b941-868818bdcde3

When drawing ASCII art, always use a monospace font, like Courier. That way everyone can view the diagram.

Ok, it's a buzzer, not a "speaker". This makes more sense. Your buzzer needs 12V and all you have is 12V. It will work supplied directly from your batteries (though I'd rather see you use a string of the same batteries in a suitable holder - be careful not to short them) without a resistor. It will draw 20mA.
The LED will likely light with 5-10mA. 20mA is likely the maximum. LEDs are current controlled devices and must always have current limiting. 3V sounds high (is it a white LED?), but assuming that's right, you have 12V and need 3V for the LED. That means the resistor needs to have 9V across it. Let's choose 10mA, just to be well within the LED's spec. 9V at 10mA is (R=V/I) 9/.01 = 90 ohms. Next you need to calculate the power dissipated by the resistor. P = V*I = 9*.01 = .9W. To be safe, you should use a 2W resistor.

Like I said, sci.electronics.basics is meant for these sorts of questions.

You're much better off with 8x1.5V batteries in a holder or 2x6V lantern batteries. 9V batteries are easy to abuse. They have very little current/energy available and aren't really suited to your needs anyway.
-- Keith
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krw wrote: <snip>

> 9*.01 = .9
-- Sue
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[This followup was posted to alt.engineering.electrical and a copy was sent to the cited author.]
@privacy.net says...

Oh, good grief! 9/.01 = 900. Make that 900 ohms.

...and .09W. a 1/4W resistor should work.
Thanks Sue. I shouldn't do arithmetic just after waking up.
--
Keith

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krw wrote:

YVW. I just thought I would point it out for you to correct before some idiot tried to make a big deal of it.
-- Sue
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krw wrote:

Or before! ;-)
--
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snipped-for-privacy@att.bizzzzzzzzzz says...

9/.01 = 900 *not* 90.

9*.01 = .09 *not* .9 You can safely use a 1/4W resistor.
Sorry for the confusion. Thanks again Sue.
--
Keith

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----------------------------

Actually, the problem I'm presenting to myself is that the source ideally would be 12V from one point. I'm not trying to divvy up the voltage, but work with one source of 12V. Circuit I've been thinking would be more like this:
----|+ 12V -| ----- | ___|__ | | | | 3V.LED Buzzer.12V | |___ __| |____________|
Sorry if the ASCII came out badly. I'm guessing it could work something like that (resistors in needed places but not labeled), but I was trying for the whole circuit to draw as few amperes as possible, ideally 20mA. So far I can't think of a way to have that work with two points that respectively will have 12V @ 15mA and 3V @ 20mA. For it to work on above circuit I suppose it'll draw more amperes than 20mA overall..
Sorry again if I'm wasting time on something that's really off the mark. Just trying to attain a solid grasp of fundamentals and find myself hard-pressed in drumming up talks with anyone knowledgeable.
The 9V / 2x1.5V batteries are linked in series. You said this is OK, but if there's really a chance of these things exploding that's something I'd like to know.
Dan ---------------- In the diagram that I sent- somehow the LED has slipped to the left - it should be across the 3V side. Your parallel circuit would work only if you put enough resistance in series with the led to drop the voltage across the LED to 3V- as someone has already suggested- about 450 ohms (9V drop across the resistor-going to heat) . You will draw 35ma from all the batteries rather than just the 1.5V batteries in my model (corrected as below)(which doesn't have any extra resistors).
In this case there is no battery problem as long as the 1.5 V batteries can handle 35ma-which they should. The problem with batteries in parallel is that if they are mismatched (an extreme example would be a 12V battery and a 9V battery in parallel- then you would have a 3V difference driving current limited only by the internal resistance of the batteries). This could be excessive and could cause explosions with some kinds of batteries and won't improve the lifetime of others.
-->15ma -->35ma |---9V+-----o----3V+---|

| <--20ma |

<--15ma Think Kirchoff's Laws as that is all that you need at this stage. KCL: What goes in comes out. KVL: If you go around the block you end up where you started and haven't gone anywhere.
--

Don Kelly snipped-for-privacy@shawcross.ca
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<snipped stuff I think was a bad (confusing) quotation>

If the 3V LED is connected to the 3V side it will likely burn up or perhaps not light, depending on the actual voltages. There is no series resistance to limit the current. LEDs must be driven from a current source, which the 12V in series with the 900ohm resistor does a fair job of approximating. 3V isn't going to cut it since there is no head room for a current limiting device. If you're *LUCKY* it might work using the internal resistance of the battery to limit current, but it isn't a good idea to try this for a number of reasons.
<snip>
--
Keith

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----------------------------
says...

-------------------------- Yes, you are right, current limiting is needed, whether 900 ohms at 12V or 200-250 ohms at 3V (10-15ma range rather than 20ma). With such limiting, both circuits would work.
Unfortunately Dan's original problem should have specified two Ohmic loads, avoiding the non-linear characteristics of a LED at his stage of knowledge. His "ratings" were 3V,20ma and 12V, 15ma and the type of batteies (and internal resistances) were not specified.
I recognized that in my original response to his query.
"assuming that the batteries are ideal and that the devices actually draw their rated current at rated voltage, there is a simple circuit by putting the batteries in series. ..."
On the basis of that, I have said nothing wrong. (neither have you on the basis of your experience with LEDs) .
--

Don Kelly snipped-for-privacy@shawcross.ca
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[This followup was posted to alt.engineering.electrical and a copy was sent to the cited author.]

At 3V, his 3V LED may not light and with a 200-250 ohm resistor it's not likely to be visible in any case.

Right. Lighting a 3V LED (wonder what type) using a 3V battery isn't likely to work well. The only way its going to work is if the batteries' voltage is far enough above the LEDs threshold and the internal resistance, by some miracle, limits the current below where the LEDs go poof.

I didn't say you said anything wrong. Incomplete, perhaps, but not wrong. ;-) Though 3V for a 3V LED is a non-starter, at least in this simple circuit.
--
Keith

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