On Sat, 27 Nov 2004 01:05:03 -0600, the renowned Don Foreman
wrote:

I don't understand why a guy would post a question and then ignore
responses from others trying to help. Smacks of a "you watch my six
and I will too" attitude. Shrug, sigh and roger.

For others that may be interested, I've posted the outline of a
dirt-simple dirt-cheap "DC transformer" at
http://www.goldengate.net/~dforeman/...mple_24_to_12/

Looks good to me. The MOSFET worst-case power dissipation will be
higher because of Rds(on) positive tempco- maybe 50% higher. Still
fine, and the typical will be closer to your number.

Just for your interest, Don, your circuit is undoubtedly better for
this purpose, but just for fun, if it had to be made smaller and
dissipate less power.

http://www.st.com/stonline/books/pdf/docs/5651.pdf

A more expensive (almost double), but still very simple method could
use two MOSFETs (losing the Schottky and gaining some efficiency), an
ST L6384 bootstrapped half-bridge driver, 2 n-channel HRF3205's (0.008
ohms/$2.20 ea) and a 50% duty cycle 555 as previously. I count 11
parts total, just the same. Allied has the L6384 for $3.86 in 1's. So
total power dissipation would be in the 200mW range, if I figured it
right, not counting the dropping resistor.

(Actually I like my stone knives and bear skins transformer solution
too, but it's more like a 4" cube. ;-))


Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

On Sat, 27 Nov 2004 01:35:23 +0000, Ian Malcolm
wrote:


OK its a special case for 2:1 autotransformers but it *DOES* only need a
3.5A rating to supply 7A without overloading which would make a really
big difference to Gunner as it would be easier to source *and* to cram
in there.

Don Foreman wrote:
You're still talking 84VA.. One that comes close (80VA) from
Signal Transformer is 2.5" x 2.37" x 3". Don't know if that's
"small enough" or not. The $10 electronic solution could easily be
done in 1.5" x 1" x 0.5". A surfacemount version could be
considerably smaller.

There's been some good stuff in this thread. I hope Gunner can find you
that teflon liner and take up your generous offer. A long way back up
the thread I did mention solid state switching converters, but what I
thought he might be able to get is one of those modules truckers use to
drop 24 to 12 for their radios etc and I didnt really like the idea of
spending money. The transformer solution I hoped he could scab the bits
for out of his boneyard. As I said earlier, overrunning it slightly
shouldnt be a problem for intermittant duty.
--
Ian Malcolm. London, ENGLAND. (NEWSGROUP REPLY PREFERRED)
ianm[at]the[dash]malcolms[dot]freeserve[dot]co[dot]uk [at]=@, [dash]=- &
[dot]=.
*Warning* SPAM TRAP set in header, Use email address in sig. if you must.
'Stingo' Albacore #1554 - 15' Uffa Fox designed, All varnished hot
moulded wooden racing dinghy circa. 1961

Check out IR2153. It's a bootstrapped halfbridge driver with the
oscillator built in, so we can lose the 555. $1.68 at Digi-Key. It
also has an internal 15.6V zener so the PNP and zener in my circuit
would go away too. MOSFET IRFZ48V is $1.05 ea. Not as good as
the one you chose, but it'd work OK here at .012 ohms. I'll bet
we're still under $5 for the whole works. I do like the idea of
losing the shottkey.

On Sat, 27 Nov 2004 03:06:32 -0500, Spehro Pefhany
wrote:

A more expensive (almost double), but still very simple method could
use two MOSFETs (losing the Schottky and gaining some efficiency), an
ST L6384 bootstrapped half-bridge driver, 2 n-channel HRF3205's (0.008
ohms/$2.20 ea) and a 50% duty cycle 555 as previously.

Ian Malcolm wrote:

Ted Edwards wrote:
Ian Malcolm wrote:
Mike H wrote:
If I remember my tranformer theory correctly, the load current in the
lower half of the winding is opposed by the supply current so you only
need a 3.5A rated winding.

Sorry. No go. 7amps is 7amps. If you are getting 7A out of the
transformer, the wire of that half of the winding has to be carrying it.

So riddle me this. Visulize the autotransformer. 24V AC Live at the
top. 0V at the bottom, 12V at the centre tap. Supply connected top to
bottom. 3.5A flows from the supply into the top terminal of the winding
and has to flow back to the supply from the bottom terminal.

On further thought I, too, came to the conclusion that my above
statement was wrong. 3.5A in eac half in opposite directions.

Ted

Don Foreman wrote:

Nick's idea of half-wave rectifiying is simple, small, and will work
well if the transformer doesn't saturate. (It probably won't).

Another possible gotcha is the current. While the diode responds to
average current, the copper responds to rms current. It would need a
test to see if the transformer overheats. BTW, don't forget a catch
diode across the motor.

Another way that would be very small would be to chop the rectified 24
volts at a high frequency, using a MOSFET driven by a 555 timer.

That's certainly the most elegant solution, IMO.

Ted

On Sat, 27 Nov 2004 11:22:08 -0600, Don Foreman
wrote:

I've added a variant of Spehro's idea to the webpage at
http://www.goldengate.net/~dforeman/...mple_24_to_12/

24V RMS DC and the output would be about 12V RMS DC.

RMS DC????

Ted

Don Foreman wrote:

Check out IR2153.
...
On Sat, 27 Nov 2004 03:06:32 -0500, Spehro Pefhany
wrote:

ST L6384 bootstrapped half-bridge driver, 2 n-channel HRF3205's (0.008

I gonna hafta take a very serious look at those two parts. I have an
app where I want to drop 10 to 100V to about 7 at reasonable
efficiency. Thanks for the heads up.

Ted

Sure. Unfiltered full-wave rectified AC is DC, right? It doesn't
alternate polarity, so it must be DC! Any v(t) including unvarying
DC has an RMS value. With unvarying DC the RMS value, peak value
and nominal value are all the same.

Fulllwave rectified 24 VRMS AC is (about) 24VRMS DC, though it's
probably about 34 volts peak. Halfwave rectified 24VAC would be
12VRMS DC , still about 34 volts peak, but the resulting DC current in
the winding may cause saturation problems. Doesn't usually, but it
can happen.

On Sat, 27 Nov 2004 21:13:07 GMT, Ted Edwards
wrote:

24V RMS DC and the output would be about 12V RMS DC.

RMS DC????

Ted

Don Foreman wrote:

Sure. Unfiltered full-wave rectified AC is DC, right? It doesn't
alternate polarity, so it must be DC! Any v(t) including unvarying
DC has an RMS value. With unvarying DC the RMS value, peak value
and nominal value are all the same.

Agreed.

Fulllwave rectified 24 VRMS AC is (about) 24VRMS DC, though it's
probably about 34 volts peak.

Yup.

Halfwave rectified 24VAC would be
12VRMS DC , still about 34 volts peak,

Peak is still 34V but rms is 17V not 12V. Feed it to MathCAD and you'll
see:
Sqrt((1/2pi)Integral{(Sin t)^2}[0,2pi] ) = 0.707
repeat multiplying the integrand by Sin(t)0 for half wave and you'll
get
0.5 not 0.35. Not surprising since peakier waves have a higher rms/avg
value. The _average_ voltage of half wave is half the average value of
full wave but that doesn't work for RMS.

Ted

You're right. Don't need MathCAD to see it; square root of half is
..707. Half as many pulses deliver half as much power.

I think DC motors respond to average voltage rather than RMS voltage
to determine their no-load speed,

On Mon, 29 Nov 2004 20:56:04 GMT, Ted Edwards
wrote:


Peak is still 34V but rms is 17V not 12V. Feed it to MathCAD and you'll
see:
Sqrt((1/2pi)Integral{(Sin t)^2}[0,2pi] ) = 0.707
repeat multiplying the integrand by Sin(t)0 for half wave and you'll
get
0.5 not 0.35. Not surprising since peakier waves have a higher rms/avg
value. The _average_ voltage of half wave is half the average value of
full wave but that doesn't work for RMS.

Ted