Re: Exercise for a power supply design
- From: John Larkin <jjlarkin@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Wed, 24 Aug 2005 18:56:32 -0700
On Wed, 24 Aug 2005 22:04:35 +0200, "Frank Bemelman"
<f.bemelmanq@xxxxxxxxxxxxxxxxx> wrote:
>Assume this a very very long cable with a resistance of
>perhaps 10, 20 or 50 ohms. At the other end is a device
>that needs 12V. Current comsumption may vary between 0
>and 1A. Just some figures for the sake of the discussion.
>
>The powersupply, not yet designed of course, has an
>adjustment potentiometer. This potentiometer is set to
>the same resistance value as the long wire to the device.
>Ideally, with auto-setup, by shorting the wires at the other
>end. Hmm, this may require a small PIC. The PIC could
>disconnect the load, and check if a capacitor is at the
>other end. If the voltage drops immediately, there is
>a dead short on the other end, if not, the device is
>still connected.
>
>This miraculous power supply now regulates it's output
>voltage, so that the device get's the 12V that it wants,
>regardless the current it consumes.
>
>All we need is a brilliant designer who makes the schematic.
>
>(No, I don't need one, honestly)
Negative output impedance, easy. And as JW suggests, you could
compensate for ambient temperature effect on wire resistance. For
extra credit, simulate the dynamics of wire self-heating and
compensate that, too.
I've built several buck switchers (out of discrete parts, not
SimpleSwitcher IC's or such) that had mildly negative output
resistances; never bothered to find out why.
John
.
- References:
- Exercise for a power supply design
- From: Frank Bemelman
- Exercise for a power supply design
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