NNTP-Posting-Date: Tue, 22 Oct 2002 19:21:18 -0500
From: email@example.com (John Fields)
Subject: Re: V regulator input cap size?
Date: Wed, 23 Oct 2002 00:18:21 GMT
Organization: Austin Instruments, Inc.
References: <0001HW.B9D991AD000C5917163F7590@news.covad.net> <firstname.lastname@example.org> <0001HW.B9D9CD54001A5D15163F7590@news.covad.net> <0001HW.B9DAE8B7003F6E14163F7590@news.covad.net>
X-Newsreader: Forte Agent 1.5/32.451
X-Abuse-and-DMCA-Info: Please be sure to forward a copy of ALL headers
X-Abuse-and-DMCA-Info: Otherwise we will be unable to process your complaint properly
On Tue, 22 Oct 2002 11:36:24 -0700, DaveC wrote:
>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
Ah, well, that's fish of a different color!
How about my favorite, the reactive voltage dropper?
This is what you started with:
| | | 7812
| +-----------+---IN OUT--+--->>--+
| | | GND | |
+-[CR]-+-[CR]-+ [470µF] | [22µF] [RL]
| | | | |
If you do this:
| | | | GND | |
+--+ [CR5] [C2] | [C3] [RL]
| | | | | |
Then you can use the reactance of C1 to drop the voltage into the
Working backwards, and assuming that your load is going to be 76mA
all the time means that we can use C1 to pass the current we need
for the load while dropping the voltage which would otherwise go
higher than the 7812's limit. Unfortunately, however, if the load
current goes below 76mA the input voltage to the 7812 will rise, and
if the load current goes to zero, the input voltage will go to the
peak voltage minus 2 diode drops in the bridge, which will be
(28VRMS*sqrt2) - 1.4V ~ 38V. Fortunately, we can fix this problem
easily by choosing CR5, a Zener diode, correctly.
OK, the first thing we need to do is determine the dropout voltage
of the 7812, and from the data sheet it's 2.5V max, so we have to
make sure its input never drops below 14.5V and never rises above
For the 14.5V part we have to make sure C2 is large enough to supply
current to the load when the mains voltage supplied through C1 falls
below 14.5V, and for the 35V part we have to choose CR5 so that it
will clamp the output voltage from C1 should it rise to 35V.
We have quite a large margin to work with (35V-14.5V = 20.5V) so
where to start? Well, the data sheet for the 7812 spec's
everything with a 19V input, so let's start there for a first cut.
with a 19V input and a 12V output pulling 76mA through the 7812 it
will disspate (19V-12V)*76mA = 532mW, or a little over a half a
watt. Not bad.
Now for C2. If we charge C2 up to 19V and it discharges to 14.5V
before we can start to charge it up again, we can say
C = I*dt/dv = (76mA)*(1/120Hz)/(19V-14.5V) ~ 140µF. Again, not bad,
but we have to be careful here. If we use a +/- 20% electrolytic
that means that the capacitance could be 150µF - 20% = 120µF, which
means the voltage to the input of the 7812 could fall to 11.6V,
which clearly won't do. We need at least a 187.5µF cap to keep the
low end voltage above 14.5V, and more would be better. 200µF would
be fine. Its voltage rating should be at least equal to the Zener
voltage of CR5, and what we need to do there is to choose a voltage
which will never allow the 7812 input to go to 35V no matter what
happens to the load. Since we're already allowing the input to rise
to 19V we'll want something higher than that, so let's say 25V just
to have something to work with. So, C1 needs to be 200µF, 25V (at
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!
Now we can go back to the Zener.
Tomorrow. Right now, She Who Must Be Obeyed has supper on the
Professional circuit designer