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'Energy in a capacitor'
1999\04\08@021141 by

I sent this yesterday, but the listserver says today that I am not
subscribed to the PICLIST and I couldn't find it in the archive so
here we go again.

Hello everybody,

I have a question about energy stored in a capacitor:

I have a capacitor, 220 nF, charged to 700V. This gives
a stored energy in the capacitor of 1/2 * C * V^2 = 54mJ.

I discharge this energy on the primary side of a ignition
transformer with a catch diode over it.

Now, my question is; on the secondary side of the
transformer I can not possible get a pulse with higher
energy than 54mJ, right? Even though the catch diode keeps
the current running through the primary of the transformer
when the voltage over the capacitor has reached 0 V (when
the capacitor is discharged)? With the catch diode removed
the resulting output pulse is much weaker. But still, I can
never create energy, only transform it, right. (And when
the catch diode is conducting no current is drawn from
the power supply, only circulating through the inductor
and the diode.)

And yes, the circuit is controlled by a PIC (or actually
a Scenix).

I have also thought of a way to actually measure the output
energy;
Rectify the output from the transformer and charge a
capacitor with a known number of pulses from the transformer,
then measure the voltage over the capacitor and calculate
the stored energy in the capacitor (its value is known) and
divide this with the number of pulses from the transformer.
Could this work, or is there a better way?

If You consider this off topic, please respond privately.
==============================
Ruben Jvnsson
AB Liros Elektronik
Box 9124, 200 39 Malmv, Sweden
TEL INT +4640142078
FAX INT +4640947388
ruben2.sbbs.se
==============================

Now, my question is; on the secondary side of the
transformer I can not possible get a pulse with higher
energy than 54mJ, right?

right, that's all there is, there ain't no more.

Even though the catch diode keeps
the current running through the primary of the transformer
when the voltage over the capacitor has reached 0 V (when
the capacitor is discharged)? With the catch diode removed
the resulting output pulse is much weaker. But still, I can
never create energy, only transform it, right. (And when
the catch diode is conducting no current is drawn from
the power supply, only circulating through the inductor
and the diode.)

Correct.

Ruben,

It would seem to me that the absolute most you could possibly
get would be everything applied to the primary.   But, that is
not possible because of losses.  You have loss resulting from
the conversion in the transformer, and you have losses in the
diode.  So I would suspect that you will get somewhat less out
than put in.

Regards,

Jim

{Quote hidden}

Just measuring the secondary voltage would not give you the right
information about transfered power... you really need to consume power
at the secondary to get the right information.  Measuring Voltage and
Current at the secondary is necessary, then using the PIC to do the
multiplication, or just use a quadrant multiplier...

When the secondary is rectifying to DC, you can play with the frequency,
dutty cycle and so on at the primary, to reach the best transformer
productivity.

circuits are consuming much power, the transformer would have its
magnetic flux at maximum and it would not be able to transfer more
power, so the "final productivity" drops down.

Some old TV transformer "stabilizers" used "saturated transformers", it
means that the transformer was plugged at 117Vac driving a TV, but it
already gets saturated early at 90Vac, and if the line voltage drops to
100Vac, its output would not change, since it still saturated...

Some IBM large mainframes use an external "motor-generator", bigger as a
fridge, an huge motor and generator. The motor runs in automatic
star-delta configuration attached to 60Hz-208Vac tri-phase, and it
mechanically drives a 400Hz 208Vac also tri-phase generator. At 400Hz,
the wires and cables can be smaller, also the transformers, and the
filtering capacitors. As the mainframe computer is almost 1/5 of its
size just power supplies, you can imagine the gain in space of just

Wagner Lipnharski wrote:

> Some IBM large mainframes use an external "motor-generator", bigger as
> a fridge, an huge motor and generator.  The motor ... mechanically
> drives a 400Hz 208Vac also tri-phase generator. At 400Hz, the wires
> and cables can be smaller,

OK, he *meant*, at 208V three phase, the wires and cables are smaller.

> also the transformers, and the filtering capacitors.  As the mainframe
> computer is almost 1/5 of its size just power supplies, you can
> imagine the gain in space of just using 400Hz instead 60Hz...

Indeed.  This was the 1960s version of the switchmode supply ;-)

Aircraft used 400Hz and/ or 1600Hz during and presumably, prior to
WWII.  Of course, in a plane, you don't notice the "singing" from the
mains, do you?

Sean Breheny wrote:

> I really don't think sound is the main issue.  After all,they could be
> sound-proofed.

Really?  It didn't matter too much in the early days, other than for
radios, but remember that communications bandwidth is generally
considered 300Hz to 3000Hz.  Using mains frequencies (400Hz or 1600Hz)
smack-bang in this range *would* be a *major* headache.  Can you imagine
soundproofing your fridge, toaster (Eh Lawrence?), everything with power
transformers?

> I think the biggest issue is radiation losses from the long run power
> lines.  I think that the radiation resistance of a 10 mile long wire
> is ALOT higher at 400Hz or 1kHz than at 60Hz.  After all, a half
> wavelength at 1kHz is only 468/.001, 468000 feet in copper.  That's
> only 88 miles, as opposed to 1477 miles for 60Hz.

And you are quite correct here.

> Other reasons might include typical rotational speed of large
> generators (could be wrong here),

Indeed you could be.  Multi-pole generators are quite easy to build,
that was the point initially made, they could *all* be one quarter the
size!  Wow!

Yep, that's involved too.
--
Cheers,
Paul B.

-----Original Message-----
From: Paul B. Webster VK2BZC <paulbMIDCOAST.COM.AU>
To: PICLISTMITVMA.MIT.EDU <PICLISTMITVMA.MIT.EDU>
Date: 09 April 1999 12:59
Subject: Re: Energy in a capacitor

{Quote hidden}

Aircraft today are still using 115 volt 400hz AC and 28 volt DC.

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