Subject: Re: V regulator input cap size?
Date: Tue, 22 Oct 2002 11:36:24 -0700
References: <0001HW.B9D991AD000C5917163F7590@news.covad.net> <email@example.com> <0001HW.B9D9CD54001A5D15163F7590@news.covad.net>
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On Tue, 22 Oct 2002 8:10:08 -0700, Lizard Blizzard wrote
(in message ):
Circuit current requirements:
555s: 10mA each x2
4020: 1mA max including output drive current
LEDs: 15mA each x2
Relay coil: 25mA (I was a bit off on this figure!!)
Supply voltage measures at 28vac rms. (for some reason it's running lower
than normal for HVAC) which will give ~ 37vdc(?). The max input for the 7812
> The schematic at the top of page 12 of this URL
> shows a current amplified zener on the input of the regulator to drop
> the voltage. Use a power transistor on a heatsink and a 1W zener, say
> 12V for a V drop of about 12.6V. Better yet, use two 9V zeners for a
> total V drop of about 19V. 100 ohms seems like a rasonable value for
> the resistor. This should also cut down the power dissipation of the
Like a TIP-31A (60V, 1A, $0.49)? When trying the 100 ohm resistor, what
criteria (ie, voltage, current measurements) should I look for to decide that
it is or is not appropriate? Is 1/4W O.K. for this resistor?
I'm guessing that the best design is to drop half of the voltage across the
TIP-31 and half in the 7812? This would mean a 12v/1W zener setting the TIP's
voltage resulting in ~ 12.6v drop here. Input to 7812 would be ~ 24v.
This looks like the most straightforward solution. I'll have a bit of
heatsinking to do (26v x 0.076A ~ 2W, max, to dissipate between the two
devices), I'm sure.
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