From: Fred Bloggs
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Subject: Re: V regulator input cap size?
References: <0001HW.B9D991AD000C5917163F7590@news.covad.net> <firstname.lastname@example.org> <0001HW.B9D9CD54001A5D15163F7590@news.covad.net> <0001HW.B9DAE8B7003F6E14163F7590@news.covad.net> <email@example.com>
Date: Wed, 30 Oct 2002 13:55:03 GMT
NNTP-Posting-Date: Wed, 30 Oct 2002 05:55:03 PST
Organization: EarthLink Inc. -- http://www.EarthLink.net
John Fields wrote:
>>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!!)
> If you do this:
> 28RMS>--+-[CR1>]-+-[ | | | 7812
> | +-------------+-----+---IN OUT--+--->>--+
> | | | | GND | |
> +--+ [CR5] [C2] | [C3] [RL]
> | | | | | |
> Then you can use the reactance of C1 to drop the voltage into the
> 7812 losslessly!
> Now, how about C1?
> Well, looking at the input of the regulator we can see that we have
> an input voltage which varies from about 19V to, say, 15V, so if we
> split the difference it'll be at 17V, and when the load pulls 76 mA
> it'll look like 17V/76mA ~ 224 ohms.
> Since we have a 38V source from which we'll be pulling 76mA, the
> impedance of the entire circuit needs to look like 38V/76mA = 500
> ohms, of which the load looks like 224 ohms. The series capacitor
> will have to supply the reactance to provide the proper impedance,
> and since Z = sqrt (R²+Xc²), we can rearrange and solve for Xc =
> sqrt(Z²-R²) = sqrt(500²-224²) ~ 447 ohms. Now, since
> Xc = 1/2pi*f*C, C= 1/2pi*f*Xc = 1/6.28*60*447 ~ 5.9µF
> So, C1 needs to be about 5.9µF, and it needs to be rated for about
> 40V. Not bad at all!
Oh yes, it is VERY bad. Your reasoning is a gross error.
> Now we can go back to the Zener.
> Tomorrow. Right now, She Who Must Be Obeyed has supper on the
The simple reactance calculations are inapplicable to this highly
non-linear circuit, and the reason for this is the limited conduction
angle of C1 due to the voltage on C2 which must be exceeded to forward
bias the bridge. To make a long story short, it can be EASILY
demonstrated that for the case of predominately capacitor-limited
current, the equivalent DC source resistance of the series capacitor is
1/(4*C1*F). For C1=5.9uF this computes to ~700 ohms at 60Hz FW
rectification, which means that for 76mA loading, the DC voltage across
C2 will be Vi,pk-2xVdiode-700x0.076=-15V. This ridiculous number means
you don't come close to providing enough current at nominal line. Also,
only one LED should be on at a time, and the relay load is intermittent.
So the real load is 36mA +25mA intermittent. At 36mA, the DC voltage
with your 5.9UF is 13V so you are getting close, but no cigar as they
say. In any case, the 25mA of relay unknown puts your DC voltage
excursion at 700x0.025=17.5V which is way too much to justify the added
expense of the capacitor for the so-called energy savings of your
so-called lossless technique with uncomputed advantage.
The most reasonable thing to do in a case like this is to use a flea
power regulator for the CMOS and run the relay off a current source to