Re: CFL effciency?

On Wed, 26 Nov 2003 14:38:06 +0000 (UTC), [EMAIL PROTECTED] (Don
Klipstein) wrote:

>In <[EMAIL PROTECTED]>, Victor Roberts wrote:
>>On Wed, 26 Nov 2003 13:53:07 +0000 (UTC), [EMAIL PROTECTED] (Don
>>Klipstein) wrote:
>>
>>>  The way I envision it, if you put an inductor in series with a diode and 
>>>apply AC, you don't get runaway current.  The current builds up during one 
>>>half cycle and decreases to zero the next half cycle.  If there is also 
>>>resistance, then the current decreases to zero before the second half 
>>>cycle is over and stays at zero for a little while.
>>>  One thing I just realized:  The peak current is (theoretically) twice 
>>>what it would be without the diode, and that would normally saturate a 
>>>ballast.
>>>
>>If you place a diode in series with the line you have created a DC
>>component to the supply voltage. The magnetic ballast cannot "impede"
>>a DC current so there is no current control for the lamp.
>
>  During the half-cycle that applies voltage in the direction opposite 
>that of current flowing through the diode and inductor, the line voltage 
>will be across the inductor in the direction opposing current flow.
>  If you want halfwave rectified voltage to appear across the inductor and 
>no opposing voltage during the other half cycle, you will have to put a 
>second diode in parallel with the inductor.  That would result in DC 
>current limited only by resistance.

I disagree with your statement about the voltage across the inductor.
During the half-cycle that applies voltage in the direction opposite 
that of current flowing through the diode and inductor [to reuse your
own words], the line voltage will be across the diode (assumed here to
be perfect), not the inductor, since the diode has infinite impedance
and the impedance of the inductor is finite. So the inductor does
indeed see a DC voltage. 

However, I agree that you (and also I) can be right about the
operation of the lamp depending upon the ratio of line voltage to lamp
voltage. We know that fluorescent lamps can be operated on pulsed DC
power sources if the voltage and pulse width are carefully controlled.
I can provide a number of US Patents that describe such ballasts. The
reason for this is that it takes the electron density a finite amount
of time to build up to high enough values to carry the current
necessary to damage a lamp, even when operated from a non-current
limited voltage source. If the power is turned off soon enough and
held off long enough, the electron density in the lamp will decay and
the lamp current can be held under control. All the DC pulse ballast
circuits I know of operate with pulse widths significantly smaller
than 8.33 msec, usually less than 100 usec. However, in this case we
have both inductor, which serves to limit the rise of the current, and
the lamp, which needs a finite amount of time to build up the higher
electron density needed to carry higher currents so on times of 8.33
msec should be possible.

So, bottom line, for some combinations of lamp voltage and line
voltage I agree that the a magnetically ballasted system fed with
half-wave rectified 60hz power will operate without destroying the
lamp. However, for systems that have a very low lamp voltage relative
to the line voltage, I still believe, but have not yet proven through
tests, that the lamp current will rise to destructive values during
the half-cycles that the power is applied.

-- 
Vic Roberts
http://www.RobertsResearchInc.com