Excuse my noobishness....
I'm buying these 10mm white LEDs on Amazon, and I want to connect 3 of them in series to a 9v battery. Is the 9v battery enough to power all 3 of these LEDs?
Is this a circuit that I could add a resistor to? If so, how many ohms should the resistor be? Thanks!!
With a series connection, the same amount of current flows through all devices.
You'll want to look closely at this circuit and ask yourself "what will keep these 3 LEDs from flowing more than their rating of 20ma?
The number on the data-sheet won't do it. That only admonishes you not to design a circuit which would allow them to draw more than 20ma.
If you're thinking "well it says 20ma draw therefore it must surely self-regulate to 20ma", nope. Components don't work that way. Consumer appliances do; a light fixture that claims 120ma draw on AC power will, because it's engineered as a system (and approved and listed by UL) to do so. The constant-voltage AC power system forces appliances to work in a linear way (self regulate their draw).
Components do whatever is their nature to do. LEDs are highly non-linear and a small change in voltage will result in a high change in current.
So if you engineer for worst case, a 9v battery outputting 9.7V and the LEDs dropping you say 3V each, that's 0.7V to drop in the resistor, and you wnt 20ma flow.
E=IR is Ohm's Law.
0.7 = 0.02 x R
50 x 0.07 = R
35 = R
35 ohms is a place to start. You'll want to test it real-world to see what it actually does.
These are very cold white (bluish) LEDs that I expect are 3.0 to 3.1 at 20 mA but have an ESR of about 16 Ohms each. {ESR~1/60 mW)
Meanwhile 9V alk. batteries range around the same or less ESR, so you dont need any series R. but it will end up being a battery voltage meter than a flashlight.
But you will likely get <20mA except with fresh Panasonic Alkaline batteries.
If the LEDs measure 3.1V each then with 9V across 3 in series, you get <20mA
.e.g. the Vf best case is {3.0V- (20-I(mA)]* 16R} * 3 LEDs = Vf for diode string.
If the 9V cell has pulsed short circuit current of 1A or an ESR of 9 Ohms then 10 mA then voltage as much as 90 mV .
You will get far better results with CR2 battery cells x3 (cheap online)
Battery powered LEDs should not use a current limiting resistor.
The LEDs you chose have a forward voltage specified at 3.0V-3.2V, so a 9V battery would be borderline. You'd have to get them and measure the actual forward voltage. Many white LEDs are less than 3 volts, more like 2.8v typical. If the actual Vf is below 3V then the current needs to be regulated.
Battery powered LEDs are best driven with a CCR rather than a current limiting resistor.
Example CCR: On-Semi NSI45020
3 white LEDs and a 9V battery is not recommended. The LEDs will be constantly getting dimmer.
Better would be to power them with a 3.6V battery and use a 3 output CCR such as the ON-Semi CAT4003B or Microchip CL320
A 9V battery, or any battery really, is a non-perfect voltage source which is better represented as a voltage source and a resistor in series. For an alkaline 9V battery these have a high Equivilant Series Resistance compared to say AA batteries.
Leds will draw a forward current at a given forward voltage.
If the voltage is lower than your target current, then it van be connected without any other type of current restriction. In this case, the 9V battery also acts like a resistor, not that it is needed.
You can put 3x 3V @ 20mA leds across an alkaline 9v battery without any issues.
I regularly put 2x AA batteries without a resistor with 3.2V blue leds.
