ledresistors
I am trying to power a number of different color LEDs with a 5V wall-wart power supply able to deliver 2 Amps.
I tried to make them light with equal intensity and I like them to be bright. (They get quite irritating to the eyes)
I get the following resistances:
blue: 5K, yellow: 350Ω, red: 150Ω, orange: 1K, green: 50Ω
Do the values make sense? The 150Ω and the 50Ω resistances get quite hot.
Am I doing anything wrong?
The simplest analog circuit I can come up with is this:
V1 represents the temperature sensor output value.
The values of R1 and R3 may need to be adjusted specially if you use other transistors (you can use variable resistors to find out the correct values then replace them with fixed value resistors).
You may also need a voltage divider on the base terminal of Q1.
This is the output signal analysis.
This assumes you are using a common anode RGB LED.
If the LEDs are to be directly powered by the batteries, the blue, white, and possibly the green LEDs (depending on which type of green LED it is) will not work at all on 3 Volts: They require a forward voltage of 3+ Volts, 3.4 Volts as specified in the question.
Also, once the voltage starts dropping due to depletion, even the marginally lit ones (e.g. the green, may be the orange) will no longer work.
Therefore ideally only red LEDs should be used: Red LEDs typically have the lowest operating voltage amongst visible colors. Alternatively, as noted, 3 cells would be needed in series.
Now about power:
LEDs can be operated typically at half the nominal current or less. So 20 LEDs operated in parallel at around 10 mA each will require 200 mA current, well within the capabilities of typical AA Alkaline cells.
For instance, the Energizer E91 Alkaline AA battery is rated for around 3 to 4 hours at 200 mA current. Put three of those in series, add in an appropriate current limiting resistor for each LED, and it will all work.
Best Answer
This diagram may help.
Figure 1. Figuring out the required voltage drop across the current-limiting resistor for a green LED at 20 mA. Source: LEDnique.
The graph shows the VF (forward voltage) of various LEDs at currents between 0 and 50 mA. We can see that at 20 mA the green LED will drop about 2.25 V. You are feeding from a 5 V supply so that means that the voltage drop across R is 5 - 2.25 = 2.75 V.
From Ohm's law we get \$ R = \frac {V}{I} = \frac {2.75}{0.02} = 137 \ \Omega \$. Pick the nearest standard value.
You can easily work out the resistor values for each of the other colours too.
For other currents slide the diode and resistor vertically to the desired value.
An alternate approach using the same graph is to draw the load-lines for a range of resistors.
Figure 2. Various 5 V resistor load-lines overlaid on IV curves.
Let's plot these points on the load lines.
Figure 3. The OP's resistor values found to give reasonably even brightness on a range of colours.
The plots indicate to me that there is a very large discrepancy in the efficiency (or possibly the optical focus) of the LEDs. If they were all the same efficiency the points should be close to the same height off the horizontal axis. From the results it appears that the blue is super-high efficiency but the green (which is in the most sensitive region of human vision) is terrible.