Understanding the most basic principles of current draw

Thread Starter

AlcoHelix

Joined May 15, 2017
20
Most LEDs will shine brightly with 1mA or lower. If you connect a 1kΩ resistor in series with the LED and connect it to three D-sized cells (4.5 total) the LED will be amply bright even though the current drawn is much less than 20mA.
Just to confirm, did I calculate the current correctly for the situation I had, with 2.8V and 10kΩ = 0.28mA? Just making sure I understood the variables.

And that Joule Thief circuit looks to be exactly what I need to learn for this particular project! Thanks!
 

WBahn

Joined Mar 31, 2012
29,976
I was trying some different voltage sources just to see what would happen, and noticed something that didn't make sense: when powered by the 2.8V AA batteries directly, no resistor in place, the LED lights very brightly, but I can add a 10kΩ resistor and the LED still faintly lights. How can it still be getting enough power to light at all with 10kΩ resistance in the way?
Connecting LEDs directly to a voltage source is a very good way to burn them out. You are relying on the internal resistance of battery to limit the current. You can get away with this on cheap designs using coin cells, but AA batteries tend to have very low resistance (relatively speaking and so they come on very bright -- but can burn out pretty quickly.

Is the math: I = 2.8V/10,000Ω, then current should be 0.00028A (0.28mA), which seems a long ways from the 20mA recommended (when I metered it without the resistor, it seemed to be pulling about 5mA?)
You don't have 2.8 V across the resistor. you have the difference between the 2.8 V of the battery and the forward voltage drop of the LED. Remember, Ohm's Law requires the voltage ACROSS the resistor.

With your ammeter in the circuit you are using the burden resistance of the meter to limit the current.

Or do I need to solve for volts here to see what's making it through the resistor?

For reference, these are the LEDs, and I'm using a green one: https://www.amazon.com/gp/product/B00UWBJM0Q/ref=oh_aui_detailpage_o06_s01?ie=UTF8&psc=1
There's not enough information there to really know what the forward voltage of your LEDs are. But it is trivially simple to measure it well enough for your purposes.

Take your AA batteries and connect them in series with a resistor and the LED. Measure the voltage across the LED and also the voltage across the resistor. Also make a notation whether the LED is dim, solid, bright -- however you want to describe it. Then calculated the current by dividing the voltage across the resistor by the value of the resistor. Use different resistances, starting it large resistances and working downward, until you get to a current of about 20 mA. You can either use this chart or you might want to plot the current as a function of forward voltage and break the chart into intensity regions based on your chosen descriptors. Now you have a good idea of how an LED behaves and a little reference guide to use, at least when using the green LEDs from your kit.

Speaking of metering, I've never done much with measuring current before, and I noticed that when I put my meter inline with the igniter to see how much current it might be drawing, it shows me 1.3A, but the coils on the igniter that normally glow hot orange when directly connected to power, do not glow. Is it a know situation that meters disrupt the current to this level? I knew they'd have to use a small amount of power to do their job, not a perfect observer, but I was surprised there seemed to be enough loss of current to keep the coils of the igniter from glowing.

For reference on the igniter: https://www.amazon.com/gp/product/B016I30XTU/ref=oh_aui_detailpage_o00_s00?ie=UTF8&psc=1
What would really be useful is the information on your meter -- the make and model and also what range you had it set on. Do you have the owner's manual?

What you want to find out is the burden resistance when it's on the range you wanted to use. It may well be higher than the resistance of the igniter.

A better way to make the current measurement is to use a low valued resistor, say 0.1 Ω, in series with the igniter and measure the voltage across it when the igniter is powered. If you are drawing 2 A, then you would have 0.2 V across that resistor and so it is upsetting your measurement by only about 10%.
 
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