Calculating resistors required

[Danny T]

Why 2x?

Because I don't understand properly?

"The LM2940 1A positive regulator features the ability to source 1A
output current with a dropout voltage of typically 0.5V and a maximum of
1V over the entire temperature range."

I want to drop 1V - not 0.5V?

In any case, I'm gonna go with diodes for now, for both the PIC and the
motors, seems the easiest way

Don't use a regulator to power your motors. You could use a bunch of
diodes, you could tap power from the middle of the battery pack, you
could put the motors in series, or you could PWM the motors directly
with the +6V. There are other ways, but I'd probably go with center
tapping the PS (battery pack) or better yet using a different PS
altogether for the motors (helps prevent noise in the PICs PS).

I'll go diodes for now.

Thanks

 

[Miles Harris]

Didoes drop voltage not current. Silicon diodes (e.g. 1N4148 or 1N4001)
drop about 0.7V so one would probably be adequate; but two, in series, would
be safer.

Um, yeah, but they drop voltage _according_ to current! If they're
dropping 0,7V., they're not passing much current! Diodes are a crap
way to drop voltage unless the load is light and predictable!!!

miles

 

[Danny T]

Um, yeah, but they drop voltage _according_ to current! If they're
dropping 0,7V., they're not passing much current! Diodes are a crap
way to drop voltage unless the load is light and predictable!!!

http://www.kpsec.freeuk.com/components/diode.htm

says

"There is a small voltage across a conducting diode, it is called the
forward voltage drop and is about 0.7V for all normal diodes which are
made from silicon. The forward voltage drop of a diode is almost
constant whatever the current passing through the diode so they have a
very steep characteristic (current-voltage graph)."

 

[Peter Bennett]

Because I don't understand properly?

"The LM2940 1A positive regulator features the ability to source 1A
output current with a dropout voltage of typically 0.5V and a maximum of
1V over the entire temperature range."

The dropout voltage of a regulator is the point at which it can no
longer regulate. The LM2940-5.0 will maintain its output at 5.0
volts, as long as the input voltage is above 5.5 volts. (will probably
work with up to 30 volts or so on the input - but I haven't checked
the datasheet)


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[Danny T]

The dropout voltage of a regulator is the point at which it can no
longer regulate. The LM2940-5.0 will maintain its output at 5.0
volts, as long as the input voltage is above 5.5 volts. (will probably
work with up to 30 volts or so on the input - but I haven't checked
the datasheet)

Right, makes sense. I thought the output was based on the input, and 5V
was a min/max or something. I understand now. Thanks

 

[Byron A Jeff]

-
-Any chance you could look at my thread "Circuit & Component Check" to
-see if all looks well?

I'll go and take a look.

-
-> And you don't want 6V because then you'll have a problem regulating it. For
-> a 7805 you generally want at least 2.5V of headroom. So if you want regulated
-> 5V then you'll want an input voltage of at least 7.5V. That's why I said 8V
-> in the original post.
-
-Would something like this work? (if I cut the end off and attach wires I
-can plug into my breadboard!)
-
-http://www.argos.co.uk/webapp/wcs/stores/servlet/ProductDisplay?storeId=10001&langId=-1&catalogId=2501&productId=111269
-
-Or do I need something that says "DC" in it?

It would but it's extremely expensive. Almost everyone has old equipment
(phones, toys, small electronics) that are powered by wall warts. I have
a junk box full of them.

BAJ

 

[Miles Harris]

http://www.kpsec.freeuk.com/components/diode.htm

says

"There is a small voltage across a conducting diode, it is called the
forward voltage drop and is about 0.7V for all normal diodes which are
made from silicon. The forward voltage drop of a diode is almost
constant whatever the current passing through the diode so they have a
very steep characteristic (current-voltage graph)."

The forward voltage drop is entirely dependent on temperature (the
junction temp. of the p/n junction; which is in turn dependent upon
the current passed.) Higher currents equals higher temp. equals lower
voltage drop. It's a well known effect which can eventually destroy
the diode altogether. The physics of diodes is actually more complex
than a lot of texts would have you believe.

 

[Danny T]

It would but it's extremely expensive. Almost everyone has old equipment
(phones, toys, small electronics) that are powered by wall warts. I have
a junk box full of them.

I've ordered the item with order code 85-1820 from www.rapidelec.co.uk.
It's still expensive compared to the others I have lying around, but
it's 3A, and these are all x00mA! I doubt I'll need that, but it's
better to have it than not!

Thanks