Re: Voltage shedding question

On Tue, 09 May 2006 07:23:42 GMT, Lars Torben Wilson
<torbenwilson@xxxxxxx> wrote:

Hi all,

I have a simple question on a power supply I'm working on
in order mostly to learn how to design simple power supplies.
Working my way through Art of Electronics.

Project: simple linear power supply.

Basics:

120V => XFMR (36VCT) => NTE166 => 4400uF 50V => LM7812CV

There are a couple of other things: a lighted switch and a
fuse between 120V and XFMR, and a resistor/LED between the
caps and the regulator. The transformer is dual-primary
dual-secondary, and the primaries are wired in parallel and
in phase. I'm only using one leg of the secondaries for 18VAC,
which leaves me with 25VDC after rectification and filtering.

I originally designed this to power a device which
usually runs off a 12VDC/800mA wall wart. When powering the
device from my bench supply, I observe it needing no more
than 0.5A (and that's when I run it as hard as I can).

For this purpose, the power supply runs fine. The regulator
gets hot (I bolted on a bent piece of aluminum with some
heat sink compound) but not over 100C.

However, I would like to build it up to a full 1A output.
For the regulator and transformer I'm using, I don't even
think that's possible (the heat sink on the regulator would
seem to need to be HUGE). If I'm reading right, the 7812
would need to dissipate 13W at 1A when I'm pushing 25V into
it.

So my question is: without swapping the transformer, is there
an easy way to shed a few volts before the regulator so it
doesn't have to shed so much itself? I know the energy has
to go somewhere, but what about just moving some of the
dissipation off the regulator?

I don't mind getting a new transformer for this thing--in the
end it will be built as well as I can with as few bandaids as
possible. But I would like to know what the bandaid is when
you are faced with a too-high input voltage to a regulator.

---
You could do it by putting a suitably sized capacitor in series with
the primary. That would bring the transformer's output voltage down
to something the regulator could handle, and the capacitor wouldn't
dissipate any power except that due to its ESR.

Here's the circuit: (view in Courier)

7812
+----+ +-----+
120AC>--[C1]--+ +--|~ +|-----+-----| |--->12V>--+
P||S | | |+ +--+--+ |
R||E | | [4700µF] | [12R]
I||C | | | | |
120AC>--------+ +--|~ -|-----+--------+------>GND>--+
+----+

To find the value of C1, we'll work backwards from the output.

Since you want 12volts out at 1 amp, the worst-case load will look
like a 12 ohm resistor, and it'll dissipate 12 watts.

The 7812 needs at least 2 volts of headdroom, and if the 4700µF cap
is at the low end of its tolerance it'll be generating:


I dt 1A * 0.0083s
dV = ------ = ------------- ~ 2.2V
C 3.76E-3F

of ripple and that, added to the 2V headroom spec for the 7812 means
that the cap needs to charge up to at least 16.2V on peaks.

On the other side of the cap we've got a couple of diode drops that
need to be added in, so that brings the peak output voltage needed
from the transformer to 16.2V + 1.4V, which is 17.6V. Divide that
by 1.414 and you get 12.44V, (say 12.5) which is the RMS voltage you
need to get out of your transformer.


Right now your transformer puts out 18VRMS for 120V in, so to get
the output voltage down to 12.5V you'll need to get the input
voltage down to:


120V * 12.5V
Vin = -------------- = 83.3V
18V

Now, you'll need an amp from the transformer to feed the 12 ohm
load, plus, since you have a capacitive filter, you'll need another
about 0.8 amp to charge the cap on peaks, for a total out of the
secondary of 1.8A

Since the transformer's turns ratio (for the windings you're using)
is:


Vp 120
N = ---- = ----- = 6.67
Vs 18


the current in the primary will be:


Is 1.8A
Ip = ---- = ------ ~ 0.27A
N 6.67


And the impedance looking into the transformer will be:


Vp 83.3V
Z = ---- = ------- = 308.5 ohms
Ip 0.27A


Since the voltage into the transformer needs to be 83.3V and we have
120V mains, the reactance of the capacitor needs to drop about 37V.


Since current in a series circuit is everywhere constant, if we
simplify the circuit to:


120AC>--+
|
[C1]
|
+
|
[R1]
|
120AC>--+


we'll have 120VRMS supplying 0.27A into the circuit, so the
impedance of the whole circuit must be:


E 120V
Z = --- = ------- ~ 444 ohms
I 0.27A


In order to determine what part of that impedance is due to the
capacitor's reactace we can write:


Z = sqrt (R² + Xc²)


and then rearranging to solve for Xc we have:


Xc = sqrt (Z² - R²) = sqrt(444² - 309²) ~ 319 ohms


Now, since the reactance of a capacitor can be found from:

1
Xc = ---------
2pi f C

and since we know the reactance and frequency we can solve for the
capacitance by rearranging:


1 1
C = ---------- = ------------------- = 8.32E-6F
2pi f Xc 6.28 * 60Hz * 319


So, unless I made a mistake somewhere, it looks like if you put
about 8µF in series with the transformer's primary it'll get you
where you need to be without dissipating any power (other than about
3 watts in the regulator, which can't be helped with the headroom
spec and the ripple being what they are.

If you decide to try it, use good quality mains-rated caps and make
sure the transformer and the cap(s) aren't resonant at 60Hz!


--
John Fields
Professional Circuit Designer
.

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