On Wed, Mar 24, 2004 at 12:40:03PM +0200, Lucian wrote:
> First, thank you all who replied to me.
>
> The comparators are feeded at the voltage provided as 12V, so if it is
> 10V, they are feeded at 10V, so I think it could be a voltage drop
> bellow 12V.

As Jinx has asked in another post, take some time and explain the purpose
of the comparators. Why are they in your circuit? What are their inputs?
What are their outputs?

The reason is that there's almost always another way to solve the problem.
But right now we don't have enough information to solve the problem.

> You suggested that I put the battery first, after the transformer and
> the rectifier bridge, if I understood well?

Yes.

> Shouldn't it be some charging supervisory circuit there ?

Maybe. If the output voltage of the transformer/bridge circuit is really 13.8V
you can simply float charge the battery without any charging supervisor.

However if the voltage is significantly higher than 14V under load, then at
the very minimum you need a regulator to 13.8V followed by a current limiting
resistor.

But it is probably better to match the transformer/bridge so that the battery
can be directly connected. Anything near 13.8V should be fine.

I do apologize for not being clear. I was addressing the point that in your
circuit that the battery comes before the regulators, not after

> It's not very clear to me how the schematic looks if I admit that I
> don't need exacty 12V, but it can drop to 10V.

I still don't know why it is that you need the 12V or the 10V. If you can
describe the purpose of the comparators, maybe it is possible to get this
down to a single regulated voltage source.

> 13.8V unrectified may power some sensors ? Isn't this dangerous for them
> ?

I don't know. What are the specifications on the sensors?

Also the battery will serve as a form of regulator, as batteries tend to pull
the input voltage towards its terminal voltage.

I am still looking for more information.

BAJ

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