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dazed & confused...
Posted: Wed Jul 16, 2008 7:36 pm
Guest
I am building a charger for large (20-40 AH) lead acid batteries to
power kid-size electric cars. I'm using an L200CV voltage regulator
that's a step up from the 78XX, or LM317 types. It has 5 pins, and
offers more control. Unfortunately, when I hooked it up to a battery
in reverse polarity, I fried my L200CV. I decided to add a diode
between the negative output and the battery, to avoid doing that
dumbness twice. It happened again. How can that be? Shouldn' the
diode on ONE output prevent the massive current flowing backwards that
I caused? I've now put a 2nd diode on the positive output. NOW I'm
safe--right?

Frying in Fresno,
Bruce
Mike V...
Posted: Thu Jul 17, 2008 7:29 am
Guest
On Wed, 16 Jul 2008 22:36:29 -0700 (PDT), "dazed & confused"
<ehsratcliffe at (no spam) gmail.com> wrote:

Quote:
I am building a charger for large (20-40 AH) lead acid batteries to
power kid-size electric cars. I'm using an L200CV voltage regulator
that's a step up from the 78XX, or LM317 types. It has 5 pins, and
offers more control. Unfortunately, when I hooked it up to a battery
in reverse polarity, I fried my L200CV. I decided to add a diode
between the negative output and the battery, to avoid doing that
dumbness twice. It happened again. How can that be? Shouldn' the
diode on ONE output prevent the massive current flowing backwards that
I caused? I've now put a 2nd diode on the positive output. NOW I'm
safe--right?

Frying in Fresno,
Bruce

It's not clear to me where you put the diode.

If you added it to the L200CV circuit then it should have worked OK.
If you added it to the battery then it would not have added any
protection.

What diode? What battery voltage?

Mike
 
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