On Fri, 24 Dec 2010 11:58:12 -0600, email@example.com wrote:
Uh the problem is an led fires at a certain voltage. To dim the led you
need to apply that voltage via a PWM. Chop the firing voltage up into
pulses too fast for the eye to notice. The smaller the width of the
pulses the dimmer the led.
No, as has pointed out here by others, it doesn't "fire at a certain voltage".
It's light output is highly nonlinear WRT voltage, but there is no "on" or
"off", rather a continuum.
The output of an LED is a function of the average current through it. You can
either use PWM to change the average current or vary the current directly. The
light output is a pretty linear function of the current through the LED so
either method works.
No, the less the *ratio* of "on" to "off" the dimmer the LED. The important
variable is the *average* current.
On Fri, 24 Dec 2010 11:58:12 -0600, " firstname.lastname@example.org"
OK you have a string of LEDs dropping about 95% of the line voltage
and a resistor dropping the rest and limiting current now. How can
making that resistance more by adding a rheostat in series be more
On Sat, 25 Dec 2010 19:56:44 -0600, " email@example.com"
But the resistor will always be there. You are just making a bigger
resistor, the current will drop and the light will dim in a vary
The voltage you drop across your resistor will be the same no matter
how big it is. That is not like a rheostat on an incandescent where
you are changing the voltage applied to the filament.
On Sat, 25 Dec 2010 21:56:12 -0600, firstname.lastname@example.org wrote:
The voltage drop across a diode is relatively constant, so the
voltage across the resistor has to also be relatively constant, with
only the current being changed by the change of resistance. This is
not 100% accurate, but for this discussion I believe it is close
Not agueing with krw - just agreeing (to a point) with gfretwell. I
say "to a point" because it is not totally linear. Much more linear
than some would have you believe in a DC circuit.
On Sun, 26 Dec 2010 00:45:29 -0500, email@example.com wrote:
It is *NOT* "relatively constant" when you multiply the change times
the number of LEDs in a 120V string and compare that to the voltage
across the balast resistor. If you string a few together and have a
large balast resistor it matters less but you're simply wasting that
much more power, losing gains you made by using LEDs in the forst
Of course not. You could *NEVER* admit that you're wrong.
The circuit is not fixed. The conversation is that you change the resistor
value to change the brightness of the LEDs. There is very little change in
the voltage across the LEDs, but the big change is the voltage drop across
the series resistor as the value of it is changed to change the current that
changes the brightness of the LEDs.
On Sun, 26 Dec 2010 12:41:26 -0600, " firstname.lastname@example.org"
OK the science comes fast around here
I took a flashlight apart.
It appears they use a 4.5v LED, 9 in parallel with NO resistor at all.
The internal resistance of 3 AAA cells seems to be the limiter
With just the batteries in there the lights are pulling about 244ma
(they are fairly new batteries)
This is hurt your eyes bright.
I put my 1k pot in there and even all the way off I am dropping .05v,
current around 211ma. The slightest movement of the pot, only putting
a couple ohms in there rapidly starts dropping the current. Somewhere
around 900 ohms we are at 3ma, dropping 1.3v
and the light is "indicator bright".
OK so back to the junk drawer for some smaller resistors
With 10 ohms in there it is dropping 0.8v 77ma the light is noticeably
less but still pretty bright. When I double that with 20 ohms the
light dims quite a bit, current drops to 48ma and voltage 0.94v
I guess somebody has to crack open a 120v LED bulb or just do an
experiment like this to see how they work but I know what I need to
know about a low voltage setup like I want to make. I am thinking a 25
or 50 ohm pot will do the deed for me.
The problem that I have with it is counting on the internal resistance of the
battery. As long as the same *type* of battery is used there isn't a real
problem. If another type is used things can go very wrong.
they come with "heavy duty" AAA carbon-zincs,but I also tried alkalines
with them,and they were brighter with the carbon-zincs. they may be
I still have two more coupons for free ones,and the local HF store is only
a short walk away!
Plus,I see the ads are still coming out with the coupons.
they're nice little flashlights for the "price"! B-)
HomeOwnersHub.com is a website for homeowners and building and maintenance pros. It is not affiliated with any of the manufacturers or service providers discussed here.
All logos and trade names are the property of their respective owners.