NNTP-Posting-Date: Fri, 01 Nov 2002 19:11:03 -0600
From: email@example.com (John Fields)
Subject: Re: V regulator input cap size?
Date: Sat, 02 Nov 2002 00:22:30 GMT
Organization: Austin Instruments, Inc.
References: <firstname.lastname@example.org> <0001HW.B9D9CD54001A5D15163F7590@news.covad.net> <0001HW.B9DAE8B7003F6E14163F7590@news.covad.net> <3DC1A954.email@example.com> <1%hw9.136243$Q3S.firstname.lastname@example.org> <3DC1AF48.email@example.com> <3DC1C89E.firstname.lastname@example.org> <email@example.com> <3DC1DBEA.firstname.lastname@example.org>
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 Fri, 01 Nov 2002 01:41:24 GMT, Fred Bloggs
>It's easy to simplify things when you reduce the load by a factor 25, do
>not specify component values, and use totally new parts which may or may
>not be in the inventory. You need to get a life, and you can start by
>learning how to read or asking for a brain-transplant- you can only get
Well, Bloggsy, if you're talking about the OP changing his mind
about the relay current and stuff, then the original spec was
something like 20mA for two 555's, 30 mA for two LED's, 1mA for some
CMOS logic, and 200 mA for the relay coil, which gave us roughly
250mA. There is still a small question about the LED's, but since
he said they're running about 50%, let's say 15mA for them, 20mA for
the 555's, 1mA for the CMOS logic and, now, 25mA for the relay coil.
That adds up to 61mA. 250mA/61mA is only about a factor of 4, so
where did you come up with that factor of 25?
As far as specifying component values goes, this is
sci.electronics.design., not sci.electronics.basic., so it's really
only usually necessary to specify the values of components when
their value isn't intuitively obvious to the initiated. Perhaps
you'd be happier over in s.e.b.?
I think we all have to proceed assuming that we're using new parts
for which we can obtain spec's, if we need to, unless we're
specifically made aware of the difference between new and what's on
hand by the enquirer, don't you? I mean, how else would one go
about it? Any suggestions?
Finally, thank you for recognizing that I at least have the
capablity to get smarter, by whatever means necessary. After
reviewing _your_ recent posts, however, it seems to me that you've
gotten just about as smart as you'll ever be. Pity.
Now, if we can get back to the technical stuff, it seems to me that
we have four candidate circuits here (I've included your sow's ear
out of the goodness of my heart;^) and that we should be able to do
some analysis and pick one winner based upon some criterion which we
all agree upon. Or, perhaps, several winners based on different
criteria. (Cost, efficiency, etc...)
Let's take a look at Spehro's circuit, which I've taken the liberty
of redrawing and modifying slightly.
+----+ | | | |
28AC>--|~ +|------+--[CR1>]--+--[R1]--+-----+ [COIL] [CR3]
| | | | | | | |
28AC>--|~ -|--+ [C1] [C2] [CR2] [C3] +------+
+----+ | | | | | |
GND GND GND GND GND |
I've deleted the current limiting resistors for the LED's since
they'll probably be driven by the 555's, but it shouldn't matter
because they'll be on the 555 outputs and the current for them will
have to be supplied by VLOGIC in any case. Also, I've added R2 to
limit the current through the 12V relay coil. (A slight oversight
on Spehro's part, I'm sure!-)
Now, from the OP's description of the transformer output, we can be
fairly sure that it's a lightly loaded 24V control transformer.
These things are commonly available with ratings of from 4 to 40 VA,
so if we assume a 20VA unit it would need about an 830mA load to
bring the secondary down to 24V with a nominal 120V line. Our load
will vary from about 36 to about 61mA, so I think it's fairly safe
to say that the output voltage will stay pretty close to 28V with
any forseeable load variation. Looking at a 10% variation in the
mains (you've suggested 30%, which is clearly excessive) we can
expect the secondary voltage to vary from 25.2V to 30.8VRMS. Just
to provide a little extra window to take care of any change which
might happen for the load variation, let's say 25 to 31VRMS. We
haven't taken copper losses into consideration, but since our load
is so light I don't think we have to at this point.
OK, so with all that in hand what about the circuit?
For low mains we have 25VRMS, so VUNREG = (25*(sqrt2))-(2Vf) =
33.95V ~ 34V, and for high mains we have 31VRMS, so VUNREG ~ 42.43V.
Since C1 only has to supply current to the relay when the bridge
stops conducting, we have to size it so the low point of its ripple
is capable of supplying enough current into the relay to allow it to
remain energized continuously whenever KON is high, and we have to
make sure that we don't exceed the maximum allowable coil voltage
for an appreciable length of time.
Looking for 12V relays with 25mA coils, I found an Aromat SP2-DC12V
which seems to be a pretty typical "modern" small power relay with
contacts which would more than likely work for the OP. Sorry for
all the "maybes", but we don't have any contact requirements from
the OP. Maximum pickup voltage at 20°C is 8.4V, and minimum dropout
is 1.2V, which correspond to 17.5mA and 2.5mA, respectively, for a
480 ohm coil at 20°C. Since you seem to be concerned with what
happens when the coil temperature increases, let's take a look at
The resistivity (alpha) of soft-drawn copper is 0.00426, so if we
want to determine the resistance at 70°C we can say:
R70 = R20(1 + (alpha * deltaT))
= 480 * (1 + (0.00426) * (70°C - 20°C))
= 582.2 ~ 580 ohms
Now, since the maximum pickup current is 17.5ma, we will have to
place 0.017A * 580ohms = 9.86V across the relay coil to guarantee
pickup at 70°C. But wait... there's a resistance tolerance of +/-
10% associated with the coil, so its resistance could go as high as
638 ohms at 70°C, requiring a coil voltage of 11.16V to push that
17.5mA through it.
Since the minimum dropout current is 2.5 mA, the minimum voltage
across the coil would have to be 638ohms * 0.0025A ~ 1.6V, so it
seems we could use a very small capacitor for C1 and incur the 9.56V
ripple across it with no penalty.
Keeping up so far? Any comments?
More tomorrow if I can get into it.
Professional circuit designer