Calculating resistors required

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 :)

I'll go diodes for now.

Thanks :)

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Danny

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

-

-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!)

-

-

-

-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

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)."

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Danny

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

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Danny

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.

I've ordered the item with order code 85-1820 from

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 :-)

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Danny