Which Draws More Current, a Resistive Circuit or a PWM

If you make it so the LED is only on when you hold a push button, it will be cheap, easy, simple, relaible, and it won't draw too much current....apart from that the PWM is fine.

Two AA cells will probably be better than the 9v cell,(less power dissipated in the resistor than with the 9v where you will be throwing away about 75% of the power!),
but it depends on the LED you are using.

Del

_{( You could always design in one of those wind up torches)}

Hello Laughing Jaguar

At a guess you are using this for an optical sight, or sight glass tube, to illumine the graduations.

I made a simple unit just as Del the cat suggest above, with 2x AAA batteries fitted into a small waterproof battery holder tube, and a small switch.

The battery savings are not really worth the bother of the PWM system on a 9V battery - well the unit was excellent for my situation.

Kind Regards....

Your question in the subject line is a little off the mark. If you run an LED at 20 mA, regardless of the method you use, each circuit will draw 20 mA. However, a PWM (Pulse Width Modulated) circuit will use less power if the switching circuit is reasonably fast (efficient), because the PWM circuit is on only part of the time, whereas a resistor circuit is on all the time, and the power dissipated in the resistor is wasted as heat.

The current will not be going through the LED all the time, however, The PWM circuits will be drawing some power...allbeit minimal. The power saving is negligible. The only advantage a PWM gives is a faster, fine control of the power out.

In reply to Techno - "The power saving is negligible. The only advantage a PWM gives is a faster, fine control of the power out.

Actually the power savings and lamp life extension is quite significant if you use a dedicated LED driver module designed for such a task. You would be comparing a few milliamps current to drive the LED to a few microamps on standby for the circuitry.

There are a number of automotive LED driver chips that require minimal external components. These chips use current control to control the brightness levels of the lamps. Some have variable or preset intensity level controls as well a flash function (good for turn signals)

A quick search of some of the chip manufacturer sites or catalogues will give a wealth of choices for those who want to try out such systems. Some manufacturers do make evaluation kits available if you ask nicely.

I can name that tune in only ONE note!

http://www.joulethief.com/

Where is the intensity (brightness) control?

Hello dkwarner

It is actually invisible.

from me

For noting the invisibility of the brightness control.

Kind Regards....

For a single LED I don't imagine that you'd need one. If you wanted to you could put two LEDs in parallel thru a switch, and have a low/high option. Or use a low output LED and a high output LED, thru a switch.

I'm a mechanical designer and not well versed in electronics so I'm limited to kits or extremely simple circuits.

Spark Station got it right: this device is being built so as to illuminate the graduated reticule in a polar alignment scope.

The head on my Atlas EQ mount is equipped with a small scope like that so it can be properly aimed but I can't read the graduations in the dark to facilitate polar alignment.

I'll take the simple route.

Thanks all for responding.

If you use a low-current (1 - 2 mA) LED, serial resistor is the simplest way. At higher current I would use a step-down switching regulator (like this one:http://www.dimensionengineering.com/DE-SWADJ.htm)and a small (some 10 Ωs) serial resistor. The minimal output voltage would be set a bit lower than the ignition voltage of the LED, a maximal one is to ensure the maximal LED current with the serial resistor.

I know the regulator mentioned above is a high-power one but you can build similar low power switchers with 555 or discrete components or you can use other low power LED brightness circuits like this one from Maxim: http://www.maxim-ic.com/appnotes.cfm/an_pk/3866.

There is about 100:1 variation in the efficiency of available LEDs. If you buy a slightly more expensive high brightness LED you can run it at a very low current.

The load draw for a typical LED is so low that the power consumption of the PWM circuit "MAY" be greater than the power savings of the switching action. Ideally a PWM circuit simply turns a switch on and off to supply pulses of power that are run through a filter to control either an average output current or average output voltage. The ideal averaged pulses provide 100% efficiency. In reality these circuits consume a fair amount of power and are often but not always inefficient at very low output power. My gut feeling is that the low power consumtion of a single LED is not going to warrent a PWM circuit.

Great answer dude! Often the extra complexity designed into systems to "save" power end up doing exactly the oposite due to the power consumption of the control circuits.

The right tool for the job.

In this case the lighting system is easy access and intermittent use. I suggest the easiest and cheapest solution is as suggested, a led+switch+battery. may I suggest maybe one of these? http://www.dealextreme.com/details.dx/sku.1100 You could always extent the leads for the LED to have it where you need it and add a dropping resistor if you need to limit the brightness.

The PWM solution, while elegant, is not practical for this application. Such a solution is warranted when access to the lighting system and replacing batteries cost more in man-hours-labour plus special equipment or access being required, than the parts being replaced are worth.

While for current going into the LED, the PWM will draw less current (since 20ma @ 10% is normally brighter than 2ma continous), the losses and expense of the PWM curcuit is probably not worth it. As other mention, the use of 2 AA/AAA batteries would probably give better results as you will have much less power lost in the resistor (both the 9v or 3v pack will need to supply the same current, but the 3v pack will have 1/3 the power draw so should last longer).