Blinking Nightlight (by request)

Step 2 Calculating the desired resistor value for our LEDs

LEDs are current-driven. They require a current in order to operate. The average red LED has a normal operating current of about 20 mA, so that's a good place to start. Because they are current-driven, the brightness of the LED depends on the amount of current flow, and not the voltage drop across the LED (which is about 1.5-1.7 volts for your average red LED. Others vary).

This sounds great, right? Let's just pump a ton of current through, and we'll have super-bright LEDs!

Well... in reality, an LED is only able to handle a certain amount of current. Add much more than that rated amount, and the magic smoke starts leaking out :(

So what we do is add a current-limiting resistor in series with the LED, which fixes the problem.

For our circuit, we'll have 4 LEDs in parallel. We have two options for our series resistor(s):

Option 1 - Place a resistor in series with each LED

With this option, we treat each LED separately. To determine the series resistor value, we can simply use the formula:

(V_s - V_d) / I = R

V_s = Source voltage (In this case we're using two AA batteries in series, which is 3 volts)
V_d = The voltage drop across our LED (We're figuring about 1.7 volts)
I = The current we want running through our LED in Amps
R = Resistance (the value we want to find)

So, we get:

(3 - 1.7) / 0.02 = 65Ω

65 ohms is not a very standard value, so we'll use the next size up, which is 68 ohms.

PROS: Each resistor has less power to dissipate
CONS: We have to use a resistor for EACH LED

I checked this value in the following way:

I measured each LED for resistance, and determined each was about 85 ohms. Adding that to the resitor value gets us about 150 ohms on each of the 4 parallel nodes. The total parallel resistance is 37.5 ohms (remember that resistance in parallel is lower than resistance of any single node).

Because I = E/R we can determine that 3V / 37.5Ω = 80mA

Divide that value by our 4 nodes, and we see that we're getting about 20 mA through each, which is what we want.

Option 2 - Place a resistor in series with the entire group of 4 parallel LEDs

With this option, we'll treat all of the LEDs together. To determine the series resistor value, we have to do a bit more work.

This time, using the same value of 85Ω per LED, we take the total parallel resistance of our LEDs (without and additional resistors), and we get 22.75Ω.

At this point, we know the current we want (2mA), the source voltage (3V), and the resistance of our LEDs in paralles (22.75Ω). We want to know how much more resistance is needed to get the value of current we need. To do this, we use a bit of algebra:

V_s / (R_l + R_r) = I

V_s = Source voltage (3 Volts)
R_l = LED resistance (22.75Ω)
R_r = Series resitor value, which is unknown
I = Desired current (0.02A or 20mA)

So, plugging in our values, we get:

3 / (22.75 + R_r) = 0.02

Or, using algebra:

(3 / 0.02) - 22.75 = R_r = 127.25Ω

So, we can put a single resistor of about 127Ω in series with our LEDs, and we'll be set.

PROS: We only need one resistor
CONS: That one resisor is dissipating more power than the previous option

For this project, I went with option 2, simply because I wanted to keep things simple, and 4 resistors wehre one will work seems silly.