low voltage in, high voltage out? should defi physics?

I know its not true but id like to know how.

So an example would be the charging circuit in a disposable camera. A capacitor is rated at 300v and is charged by a 1.5 AA battery? I know they use a variety of methods, boost converts fly back transformers etc, but that sounds like a big increase. It just sounds a bit odd? could someone explain.

Also If I connect a 1.5 v battery to a 1ohm resistor I get a 1.5 amp surge? and its dead. if I connect a 300v capacitor over a 1 ohm resistor I get 300 amps right? so doesn't that mean I get more out from what I put in?

I know I'm missing something here please educate me

Thanks

Yep, You can't get anything nice for cheap.
1.5 amps for a few seconds or 300 amps for a few 1/1000s of a second.

OK so what defines the discharge time? i thought electricity travelled at the speed of light and 300Amps being a bigger quantity should take longer?

The capactitance of the cap, thus the energy stored in it.

WannabeaRobot,

I know I'm missing something here please educate me

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We'll try!

Also If I connect a 1.5 v battery to a 1ohm resistor I get a 1.5 amp surge? and its dead.

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Yes, but it will take several seconds for that to happen.

if I connect a 300v capacitor over a 1 ohm resistor I get 300 amps right?

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Yes, for a few milliseconds.

so doesn't that mean I get more out from what I put in?

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More what? Energy? No, you are de-energizing the capacitor at a high rate in a short time, and energizing it at slow rate over a longer time. The energy transferred remains the same.

OK so what defines the discharge time?

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The capacitance value and the resistance.

i thought electricity travelled at the speed of light and 300Amps being a bigger quantity should take longer?

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You are confusing how quickly something responds to how long it takes for the effect to occur.

Ratch

Power in = Power out. Power = Voltage * Current. Therefore, in order to have low voltage in and high voltage out, you must have high current in and low current out. When you boost voltage, you lose current. It doesn't defy physics at all.

It's similar to gear ratios: Low speed in and high speed out might sound like it defies physics. That is why you have to take into account torque. If you have two gears--one with 5 teeth and one with 20 teeth, and you drive the smaller gear with the larger gear, it will go 4 times as fast. But the output torque would be 1/4 of the input torque.

Same idea with transformers and other voltage boosters.

Add a flywheel to that and you have a way of storing and then supplying a short burst of high torque at that high RPM same as what the capacitor does with the high voltage.

No physics defied either way.

sounds related, what i am doing is charging from a 12v source @ 1.5amps, which i should mention takes about half a minute to get to my 300volts in my capacitors. which discharges in a fraction of a second, you are just not factoring in the time

12v*1.5a * 30seconds = 540 WATT * SECONDS


540 watt seconds = 300volts * 600amps * t(seconds)
t= 0.003s discharge

that is just average though, presuming that my resistor is 0.5ohms, and in reality capacitors are not linear

Hi
Wow some good answers there thank you but im still missing something. So I get you cant get out more than you put in, it all equals out when you factor time into the equation. If you charge the capacitor over time the power in is`equal to or greater than` what you get out but in a shorter amount of time, is that correct?

My questtion kinda relates to what your doing dr doggy, I want to build a capacitor bank. Im planning on getting alot of disposable cameras with there high voltage caps and puting them in parallel. Im just not sure how it charges them to 300v with a punie 1.5 v battery ?

If some one could give a quick explanation of a typical charging circuit like that or has a circuit diagram I could look at that would be great thanks

They typically use a flyback converter which uses a multi-winding inductor to convert the 1.5V to 300V or higher. Here are some typical circuits. An explanation of the circuit operation is here.

Thank you so much crutschow ill try and get my head round that, its exactly what im looking for. ive done about boost converts and buck converters but i never knew you could get such a big increase from 1.5 V.

Its actually quiet amazing and i still think there is some hidden potential there i.e. free energy haha (sorry to keep mentioning it). Come one 1.5v to 300v? even if it does discharge quicker and the overall power is the same i just cant get my head around it :O

what am i missing? here is what i know so far then.

1.5 volts x 1.5 amps =2.25 amps for a few seconds

300 volts x 300 amps = 90,000 watts for a few milliseconds

So the fly back converter using the number off turns on the transformer decides what current is induced in the secondary coil? can this be greater than the input current? this all to me sounds like getting more out than in.

when the transformer is in its off state the voltage across the inductor is reversed and you then see double the potential on the load? with more current?

im going to go read those notes now so hopefully it connects the dots for me i just think some people have a lot easier way of explaining it without going through the all the maths.

Thanks for your help.

DerStrom8 said:

When you boost voltage, you lose current.

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Hi isn't current decided by the amount of voltage you have i.e. high voltage means high current? i understand power in is ment to equal power out ( not taking into account the losses) but ive also been told V=i*R so no matter what the voltage if i put over a one ohm load i will get the equal quantity in current?

I think i am just not factoring in the time, and power.

Power in = power out holds only for transformers, because the primary and secondary are coupled by the core so they have to react at the same time. It does not hold for anything else, for example when you recharge a car battery the charger could be just a few watts, but when you crank the enging the power will be hundreds of watts.
The only rule that holds universally is energy in = energy out, that´s why it is called conservation of energy, not power or voltage or current. And because power=energy/time the power in power out holds in some cases, but not all.

RoboWanabe said:

Hi isn't current decided by the amount of voltage you have i.e. high voltage means high current? i understand power in is ment to equal power out ( not taking into account the losses) but ive also been told V=i*R so no matter what the voltage if i put over a one ohm load i will get the equal quantity in current?

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You are correct, V=IR. This means that if the resistance stays the same, when the voltage goes up so does the current. However, we're not talking about resistive circuits here, we're talking about a transformer. Therefore, you're using the formula for Watt's Law (not Ohm's Law) which is P=IV (power = current * voltage). If power in = power out in a transformer, that means P stays the same. If this is true, then I and V have an inverse relationship (basic maths here). If V goes up, I must go down, and vice versa.

kubeek said:

Power in = power out holds only for transformers, because the primary and secondary are coupled by the core so they have to react at the same time. It does not hold for anything else, for example when you recharge a car battery the charger could be just a few watts, but when you crank the enging the power will be hundreds of watts.
The only rule that holds universally is energy in = energy out, that´s why it is called conservation of energy, not power or voltage or current. And because power=energy/time the power in power out holds in some cases, but not all.

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Thank you for specifying that kubeek.

OK I'm beginning to under stand now. Thank you all for your help much appreciated ill ponder it abit more now and if i have any questions left ill probably be back lol

Thank you.

I should have mentioned this before, and it's very important to realize that the power in = power out only works for ideal transformers. For real-life transformers there will always be some power dissipated as heat, so power out does not exactly equal power in (for real transformers). This is where kubeek's comment about energy in = energy out comes in. Heat is a form of energy, so the "losses" are still covered.

Robo,

If you ////// energize the capacitor over time the ///// energy in is`almost equal to // /////// ////` what you get out but in a shorter amount of time, is that correct?

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Now it is.

what am i missing? here is what i know so far then.

1.5 volts x 1.5 amps =2.25 ///// watts for a few seconds

300 volts x 300 amps = 90,000 watts for a few milliseconds

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You are missing a basic knowledge of voltage, current, energy and power.

Hi isn't current decided by the amount of voltage you have i.e. high voltage means high current?

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It also depends on the resistance of the circuit. If the resistance is high enough, low current will ensue even at high voltage.

i understand power in is ment to equal power out ....

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No, in the case of the flash units, there is a lower rate of power input for a longer time, and a high rate of power output for a shorter time. The energy in and out is almost the same when losses are taken into account.

OK I'm beginning to under stand now.

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Do you? I wonder. Let's review. Energy is the ability to do work. Power is the rate of energy use. Power multiplied by time is energy. Current is the number of charge carriers per unit of time, electrons in your case. Voltage (joules/coulomb) is the energy density of the charge. If you have a higher voltage (energy density), then a smaller amount of charge will carry the same energy as a larger amount of charge at a lower voltage (energy density). Energy = charge x voltage.

Ratch

i'm going to be honest with everyone, this is a lot to take in. i'm going to have to sit and think about this and once i've got it ill try and explain it myself to see if i'm right.

cheers for the lesson Ratch if you think of anything else that might make it a bit clearer send it my way please.

Thank you