## Re: [Electronics_101] 101 you say..

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• ... Yup. It s a Law. ... As far as I know, voltage controls current, but it doesn t work the other way around. There are transistor circuits in which input
Message 1 of 3 , May 1, 2001
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>okay, v=ir, right.

Yup. It's a Law.

>did i miss this? is there any way to increase voltage, or are you stuck
>with the max. output of your ps?

As far as I know, voltage controls current, but it doesn't work the
other way around. There are transistor circuits in which input
current will control output voltage, but that's another story.

So, while it may work on paper using Ohm's law, it doesn't actually
work that way.

If you have your 1.5 volt battery, that voltage is a constant
(assuming it isn't wearing down and getting old...which will happen).
The resistance of the circuit and that voltage will determine the
current running through it. Voltage is a difference in electrical
potential...a pressure...a force. Current is the actual flow of
electrons which is 'pulled' by voltage and 'restricted' by resistance.

So if you have next to zero ohms (which is the resistance of
wire...nothing has zero ohms) you will not have exponentially
increasing voltage. It just doesn't work that way because the
current is not the constant you base your calculations (and real
situations) on.

>i'm trying to build some cute super mini synths and stuff for gifts, so
>a bit frustrated since i'm on a deadline. i'm trying to power circuits
>with single aaa batteries.. on it's own, not even enough to trigger most
>leds and transistors.

With capacitive discharge you can get an LED to fire off with even a
low voltage like that. Try looking for the ICs they use in camera
flash circuits. Some use small oscillators and flyback transformers
or the like to get the high voltages, but I think it is possible to
charge up a capacitor and let the low resistance of the LED allow a
huge surge current fire through the LED for a fraction of a second
allowing a huge discharge in light, but not for long enough to burn
it out. LEDs are very sensitive to being burnt out by too much
voltage or current.

If you use a germanium transistor, the voltage drop across the
base-emitter junction will be small enough that you'll still have
enough leftover voltage to work with. Try using a germanium
transistor to make a basic common-emitter amplifier with a 1.5 volt
battery as supply. If you don't know what that means, I'll look one
up for you and find you a shcematic online.

-Tavys
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