Re: Voltage regulator

On Tue, 05 Apr 2005 03:46:56 -0700, Vladimir wrote:
> bigcat@xxxxxxxxxxx wrote in message news:<1112670756.951875.257730@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>...
>> Vladimir wrote:
>> > Hello!
>> > Suppose, i have a required voltage on the output of linear
>> > regulator(Vout=6.3),
>> > but output current is too high (Iout=7.5). I need Iout=5.5 A.

> First i am sorry for my bad English.

No problem. It is better than the English of some English-speakers. :-)

>
> I have this regulator:
> _________
> | Volt. | Output (Vout=+6.3; Iout=7.5A)
> ____|Regulator|_________
> | |
> |_________|
> (Adj)|
> |
> I can adjust output voltage using resistor divider (from 1.25v to 30v).
> But i need the other current value. Iout must be equal to 5.5 amperes.
> Can it be implemented, but without much complexity?

I think you misunderstand about voltage, current, and resistance. The
voltage is regulated at the set point, yes.

However: the load (what you're supplying power to) will only draw as
much current as it needs. The 7.5A spec on the existing regulator means
merely that it has the _ability_ to provide that much current. If the
load presents a resistance of, say, two ohms, then at 6.3 volts, it
will only draw 3.15 amps.

Think of a water faucet. When it's off, you still have full water pressure
(which will show on a gauge), but the amount of flow depends on how far
the valve is open. Even if the pipe is very big, and has the ability
to provide, say, one liter per second, if the valve is open only a
little bit, it might provide, say, half a liter per minute, but the
available pressure will be the same (except for losses in the system,
which are called "internal resistance").

So the supply itself can not control both voltage and current - the
load will only take what it needs.

Hope This Helps!
Rich

.

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