NNTP-Posting-Date: Wed, 23 Oct 2002 10:03:16 -0500
From: email@example.com (John Fields)
Subject: Re: V regulator input cap size?
Date: Wed, 23 Oct 2002 14:57:21 GMT
Organization: Austin Instruments, Inc.
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On Wed, 23 Oct 2002 00:18:21 GMT, firstname.lastname@example.org (John Fields)
>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:
>28RMS>--+-[CR1>]-+-[ | | | 7812
> | +-------------+-----+---IN OUT--+--->>--+
> | | | | 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.
Since the reactance of the capacitor is 447 ohms, we would expect
that if the output of the bridge were shorted the capacitor would
allow I = E/Xc = 38V/447ohms ~ 85mA to flow through it, which it
does. (I just hooked up a 5.8µF cap and a full wave bridge, and
shorted the output of the bridge to make sure.-) With a 15V and a
10V Zener in series replacing the short and the load removed,
however, the current flowing through the cap, the bridge, and the
Zener is about 63mA. That's about 1.6W, so a couple of 1N4743A's in
series would do the trick there and keep the 7805 safe no matter
Professional circuit designer