> From:         William Chops Westfield[SMTP:westfw@MAC.COM]
> Sent:         Monday, March 15, 2004 12:04 PM
> To:   PICLIST@MITVMA.MIT.EDU
> Subject:      Re: [EE]: 5.5V 1F capacitor

> On Monday, Mar 15, 2004, at 05:43 US/Pacific, Alan B. Pearce wrote:

>>>  (a common disposable camera photoflash is about
>>> 8J at high voltages, and those are regularly discharged
>>> through pretty much a dead short.)
>>
>> A camera flash often uses the inductance of the wires from the
>> capacitor to the tube as the current limiting impedance when the tube
>> flashes. When my father was servicing camera equipment some of the
>> flash guns seemed to have inordinately long wires to the tube for this
>> reason.

> Huh.  The disposable flash units I was talking about typically have the
> flash tube soldered directly to the small PCB.  I've seen inductors
> mentioned as a way to get shorter flashes, but not really as a current
> limitting device.  An interesting concept (but inductors don't limit
> current, they limit di/dt, right?)

> BillW

The current from a capacitor discharging into a flash lamp is limited
only by the circuit resistance. This should be as low as possible so
that all of the energy in the capacitor is used to make light. To
preserve the energy and keep the current at a non-destructive level,
a series inductor can be placed between the capacitor and the lamp.
This forms a series resonant circuit with the capacitor. The current
waveform starts out with zero value (never mind how the lamp fires
with no current), and rises with a linear slope. As the capacitor
voltage falls, the derivative of the current decreases. With the
heavy damping of the lamp, the circuit typically makes one pulse,
rising linearly, rounding at the top like a sine wave, and falling
exponentially to zero. Ideally, all of the energy ends up in the
lamp, and the peak current is approximately (Vcap)/(Requiv). The
equivalent resistance Requiv is the square root of (L / C). This gives
Ipeak = V * sqrt( C / L). With L = 0, the peak current is infinite;
as L increases, the peak current decreases and the width
*increases* as sqrt(C * L).

John Power

--
http://www.piclist.com#nomail Going offline? Don't AutoReply us!
email listserv@mitvma.mit.edu with SET PICList DIGEST in the body