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9-volt batteries...

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9-volt batteries...

Unread postAuthor: paaiyan » Mon Mar 24, 2008 10:30 pm

Does anyone know how many amps the current from a 9-volt carries? I know amperage and voltage are inversely proportional.

Basically I need to know what kind of voltage I would need to reach to get the amperes down to, say, .003.
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Unread postAuthor: TurboSuper » Tue Mar 25, 2008 5:10 am

A quick search on google revealed this:

http://www.techlib.com/reference/batteries.html

Apparently 9V batteries are rated to 500mAh.

As for reducing the current...just use a resistor?
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Unread postAuthor: paaiyan » Tue Mar 25, 2008 8:12 am

Well that tells me how many mAh it has, but I'm wondering if your just use a copper wire to connect the poles, how many amps are flowing through that wire. And I don't want to use a resistor, I want to increase the voltage, while decreasing the amperage. Kind of like a taser.
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Unread postAuthor: dewey-1 » Tue Mar 25, 2008 10:07 am

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Unread postAuthor: paaiyan » Tue Mar 25, 2008 10:11 am

Right, as voltage increases, amps decrease right? I'm trying to use an inductor coil to create a high-voltage, low amperage spark kind of like a taser does. Heck, I may just end up buying a stun circuit to do it. It's kind of a security system, and may or may not find itself discharging into persons who go snooping in places they shouldn't be.
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Unread postAuthor: jimmy101 » Tue Mar 25, 2008 3:37 pm

A brand new 9V probably will source about 1 amp. But not for very long. (AA batteries will source 8-10 amps.)

Use ohm's law to calculate the resistance needed to drop the current to what you want.
V = IR
You have 9V and want 0.003A;
R = (9V)/(0.003A) = 3000 Ohms

The internal resistance of a brand new 9V is pretty low, probably about 9 ohms.
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Unread postAuthor: paaiyan » Tue Mar 25, 2008 3:39 pm

*sigh* I know how to do resistance. If I use a device to step up the voltage, will it lower to amperage to an acceptable amount?
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Unread postAuthor: starman » Tue Mar 25, 2008 4:42 pm

paaiyan wrote:*sigh* I know how to do resistance. If I use a device to step up the voltage, will it lower to amperage to an acceptable amount?

When you step up voltage through a transformer, the Voltage*current relationship stays the same (minus efficiency losses). So if voltage is higher on the output than input, current carrying capability at the output will be lower, if the voltage in the output is lower than the input, current capability at the output will be higher. Note that you don't carry any current at all unless there is a load on the ouput.

Unless you really want to do this as a learning exercise, I recommend you just order yourself a stun gun...you'll be much happier. Here's one I use, $14 before shipping. http://www.preventsecurity.com/productd ... b=45&p=540
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Unread postAuthor: Daegurth » Tue Mar 25, 2008 5:03 pm

V(coil 1) _ A(coil 2) _ No. turns (coil 1)
V(coil 2) -- A(coil 1) -- No. turns (coil 2)

if the maximum flow from the PP3 is 1A, and you're trying to step the voltage up to say 5000v. theoretically you could just use 9 turns on the primary and 5000 turns on the secondary, but this is unlikely to work due to the low power in the primary (9W)

however many turns you use, the current in this scenario would be about 1 x (9/5000) = 0.0018A, or 1.8mA.

you need to find out how much power (P=IV) the human body can take for different effects, and use the one you need. if you wanted half the current on the secondary, ie 0.9mA, you'd half the current on the primary with a resistor calculated with ohm's law; R=9/0.5= 18ohms.

is that what you were after?

ps. if this is a deterrent, stick with having a switched pp3. if you want a taser effect, use a 555 circuit or similar to create a waveform that will continue- since presumably you know that transformers can't transform DC continuously. :P
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Unread postAuthor: rcman50166 » Tue Mar 25, 2008 5:13 pm

My time to shine. I am an expert with electricity. (Not legally an expert mind you) First of all ohm's law is being interperetd wrong. If you put a resistor on a direct connection with a DC battery, the voltage decreases, never the current. V=IR is true but you must know that V is the element that changes along with R.

With transformers you're beginning to roam into Tesla territory. The quick answer to your transformer question, yes. However, if you want the current to .003A by means of only using a transformer, the voltage would be at 1500V ideally. This type of voltage is useless unless you want to use it for a gap or send electricity over long distances. The transformer, perfectly tuned, would need to have 167 turns on the secondary for every turn on the primary. (a rediculous ratio when it comes to transformers) Even before the transformer you would need to turn 9VDC into 9VAC at a frequency that matches the resonance of the transformer. This requires a signal generator which, unfortunately for you, isn't easy or cheap to make, especially if it's adjustable. After all of this you need a full wave rectifier rated at 1500V (also a rediculous value) to make it 1500VDC. A capacitor is needed to smooth out the pulsed DC that will come out of the rectifier. You then need a .5 megaohm resistor rated for the power you're using to bring it back down to 9 volts DC. All pretty complicated for what started off as a simple power source with a 9 volt battery.

Perhaps a current limiter is best? lol yes I went through all of that just to show that there is an easier way. A current limiter is a single coil transformer. The huge length of wire brings its voltage down but all the meanwhile the inductance keeps the voltage up. The end result: 9VDC at .003A. :D
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Unread postAuthor: Daegurth » Tue Mar 25, 2008 5:28 pm

1. the first part at least simply isn't true. go back over the applications of kirchoff's voltage law again. if the resistor is wired in series with the load, the current will drop- never the voltage- according to ohm's law. if the resistor is wired in parallel to the load, you create a potential divider and the voltage will drop, but not the current.

2. you've vastly overcomplicated it there. he doesn't need to touch tesla territory in order to make a simple shocking device. the basic high school equation at the top of my post will provide all the information he needs.

3. if you're going to call yourself an expert, be both correct and helpful.

4. real experts never call themselves experts- they let their work and imparted knowledge do the talking for them.

back to the original point, if you want 0.003A from a 9V, 1A source, you'll need a primary:secondary ratio of 1:333, and end up with a secondary voltage of 3000V. this is a power of 9W still (assuming you get 100% efficiency from your transformer, which is slightly unlikely ;)) check to see how dangerous/how this affects the human body.
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Unread postAuthor: rcman50166 » Tue Mar 25, 2008 5:39 pm

You have it backwards. Current drops off when multiple loops exist. Resistors will decrease voltage. And Ohm's basic laws only apply with DC current. You made it seem like these equations can be done with AC. Power doesn't calculate the same way. To keep it simple I used Volt-Amps in my equations. Wattage in AC is far too difficult to work with for the purpose I intended to prove. Which wasn't how to make a transformer circuit by the way, it was to show that a current limiter is the easiest option to lower current without affecting voltage.
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Unread postAuthor: Daegurth » Tue Mar 25, 2008 6:03 pm

yes, if multiple loops exist- you're talking about wiring the resistor the wrong way. if you place an 18ohm resistor in series with a 9v battery, you'll get a current of 0.5 amps. if you wire another 18 ohm resistor in parallel with that- acting as a load in your scenario- the total resistance would be the inverse of (1/18 )+(1/18 ), ie 9 ohms. this would mean that the current is now 9/9, or 1 amp- the maximum a pp3 battery can produce anyway, apparently.

if you'd read my first post, you'd have seen that i mentioned both a manually switched circuit and a 555-driven circuit. if you've ever tested an ignition coil with a 9v battery, you know what i'm talking about. this means that yes, i am talking about DC- in the switched circuit scenario. in fact, if you used the 555 circuit, it would be calcuated the same way, as it produces a square wave. obviously, it's not exactly square, so there'll be innacuracies. but it should do.

this isn't a degree project, just a simple deterrant, remember. there's absolutely no need to build a sine wave generator for no reason and a lot of money when a £2 555 job will work just as well. the best wave for tasing is an exponentially rising pulse, btw, but i'm not going to suggest that. KISS, remember?
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Unread postAuthor: TurboSuper » Tue Mar 25, 2008 7:44 pm

I'd go with a transformer from an old scanner personally. I was able to squeeze 600+V out of a 5V power supply, it's pretty cool 8)
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Unread postAuthor: jimmy101 » Wed Mar 26, 2008 2:15 am

rcman50166 wrote:My time to shine. I am an expert with electricity. (Not legally an expert mind you) First of all ohm's law is being interperetd wrong. If you put a resistor on a direct connection with a DC battery, the voltage decreases, never the current. V=IR is true but you must know that V is the element that changes along with R.

An expert eh?

9V battery with a 1K resistor from plus to minus. Voltage across the resistor is? Current through the resistor is?

Same scenario except a 10K resistor. Same two questions.

For scenario 1 the voltage across the resistor is 9V (assuming the battery is fresh enough). Current through the reistor is 9mA. For scenario 2 the voltage is still 9V but the current is 0.9mA.

How can you possibly say that "the voltage decreases, never the current". You've got it back-assward.

Like Daegurth said, but heck you barely need Kirchoff's law, common sense is sufficient. Voltage drops through an unbranched circuit such as several resistors in series with a DC source. Current is constant through all points in the circuit. That is not the same as saying the current is the same for all possible configurations of a circuit. The only value that is constant for all possible configurations of the circuit is the voltage drop.

paaiyan said:
Does anyone know how many amps the current from a 9-volt carries? I know amperage and voltage are inversely proportional.

Basically I need to know what kind of voltage I would need to reach to get the amperes down to, say, .003.

Common dude, get an f'in clue. What kind of responses do you expect for such a stupid question. If you meant using an inductor why the hell didn't you say so?

Common people, if you ask a question please try to supply enough f'in information so that people can respond. Lame ass questions like this just piss us old farts off. Get you head out of your ass and read your post before you hit the submit button. Is there enough info to get the answer you need. Have you described what you are doing in enough detail?

Damn, I'g getting cranky.

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