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Step 4Selecting the LEDs and calculating the resistors
Selecting the LEDs is the creative part of this instructable. So the following text is just a suggestions from me to you. Feel free to vary and change them, I will tell you how to do this.
Colours:
It is difficult to smoothly switch a strip on or off with LEDs of a complete new colour. So my recommendation is that each strip contains LEDs of all colours but in changing quantities.
If we imagine the sunset reversed the first strip would contain a lot of red LEDs and maybe one white, a blue and a UV one. So let's say 5 red ones, 2 yellow, 1 warm white and 1 UV. If you like you might replace one of the red or yellow LEDs by an orange one (Strip 2 in the schematic)
The next brighter strip would then have a few red ones substituted by yellow ones. Let's say 2 red, 5 yellow and 2 warm white (strip 3 in schematic)
In the next strips a few more red ones will be substituted by yellow ones or even white ones. Let's say 1 red, 1 yellow, 4 warm white and 1 blue. (strip 4 in schematic)
The next strip might consist of 3 cold white, 2 warm white and 1 blue LED. (strip 5)
This would be four strips for sunset so far. For Sunrise we could use the leftover three strips with mainly cold white and blue LEDs. If you connect the 7th and the 8th input together you could also use 4 strips for sunrise, or give sunset a fifth strip, just as you like.
You might have noticed that the strips containing red LEDs have more LEDs per strip than the pure white ones. This is caused by the difference in minimum voltage for red and white LEDs.
As the LEDs are really bright and even dimming them down to 1% is quite a lot, I calculated strip 1 with 3 reds, 2 yellow and a warmwhite LED to have only 5mA of current. This makes this strip not as bright as the other ones and therefore suitable for the last hint of sunset. But I should have given this strip an UV-LED too, for the last glance.
How to calculate the LEDs and the resistors:
The LEDs need a certain voltage to operate and even the darlington-array uses 0.7V per channel for its own purpose, so to calculate the resistor is very simple. The FET practically doesn't cause any voltage loss for our purposes. Let’s say we operate at 24V from the power supply. From this voltage we subtract all the nominal voltages for the LEDs and 0.7V for the array. What is left must be used by the resistor at the given current.
Lets look at an example:
first strip: 5 red, 2 yellow, 1 warm white and 1 uv LED.
One red LED takes 2.1V, so five of them take 10.5V.
One yellow LED also takes 2.1V, so two of them take 4.2V.
The white LED takes 3.6V, the UV LED takes 3.3V and the array 0.7V.
This makes 24V -10.5V - 4.2V - 3.6V - 3.3V - 0.7V = 1.7V which must be used by some resistor.
You surely know Ohm's law: R = U/I. So a resistor that uses 1.7V at 25mA has a value of 1.7V/0.025A = 68 Ohm which is available at electronic stores.
To calculate the power used by the resistor just calculate P = U*I, this means P = 1.7V * 0.025A = 0.0425 W. So a small 0.25W resistor is enough for this purpose. If you use higher currents or want to burn more volt in the resistor you might have to use a bigger one!
That’s the reason why you could only operate 6 high voltage consuming white LEDs on 24V.
But not all LEDs are really the same, there might be big differences in the voltage loss from LED to LED. So we use the second potentiometer (300 ?) and a current-meter to adjust the current of each strip to the desired level (25mA) in the final circuit. Then we measure the value of the resistor and this should give us something around the calculated value.
If the result is something in between two types then choose the next higher value if you want the strip to be a little darker or the next lower value for the strip to be a bit brighter.
I installed the LEDs in an acrylic glass board which I fixed to the power-source-housing. Acrylic glass can easily be drilled and bend if heated to around 100°C in the oven. As you can see on the pictures I also added the sunrise – sunset selection switch to this display. The potentiometer and the reset-button are on the circuit-board.
Colours:
It is difficult to smoothly switch a strip on or off with LEDs of a complete new colour. So my recommendation is that each strip contains LEDs of all colours but in changing quantities.
If we imagine the sunset reversed the first strip would contain a lot of red LEDs and maybe one white, a blue and a UV one. So let's say 5 red ones, 2 yellow, 1 warm white and 1 UV. If you like you might replace one of the red or yellow LEDs by an orange one (Strip 2 in the schematic)
The next brighter strip would then have a few red ones substituted by yellow ones. Let's say 2 red, 5 yellow and 2 warm white (strip 3 in schematic)
In the next strips a few more red ones will be substituted by yellow ones or even white ones. Let's say 1 red, 1 yellow, 4 warm white and 1 blue. (strip 4 in schematic)
The next strip might consist of 3 cold white, 2 warm white and 1 blue LED. (strip 5)
This would be four strips for sunset so far. For Sunrise we could use the leftover three strips with mainly cold white and blue LEDs. If you connect the 7th and the 8th input together you could also use 4 strips for sunrise, or give sunset a fifth strip, just as you like.
You might have noticed that the strips containing red LEDs have more LEDs per strip than the pure white ones. This is caused by the difference in minimum voltage for red and white LEDs.
As the LEDs are really bright and even dimming them down to 1% is quite a lot, I calculated strip 1 with 3 reds, 2 yellow and a warmwhite LED to have only 5mA of current. This makes this strip not as bright as the other ones and therefore suitable for the last hint of sunset. But I should have given this strip an UV-LED too, for the last glance.
How to calculate the LEDs and the resistors:
The LEDs need a certain voltage to operate and even the darlington-array uses 0.7V per channel for its own purpose, so to calculate the resistor is very simple. The FET practically doesn't cause any voltage loss for our purposes. Let’s say we operate at 24V from the power supply. From this voltage we subtract all the nominal voltages for the LEDs and 0.7V for the array. What is left must be used by the resistor at the given current.
Lets look at an example:
first strip: 5 red, 2 yellow, 1 warm white and 1 uv LED.
One red LED takes 2.1V, so five of them take 10.5V.
One yellow LED also takes 2.1V, so two of them take 4.2V.
The white LED takes 3.6V, the UV LED takes 3.3V and the array 0.7V.
This makes 24V -10.5V - 4.2V - 3.6V - 3.3V - 0.7V = 1.7V which must be used by some resistor.
You surely know Ohm's law: R = U/I. So a resistor that uses 1.7V at 25mA has a value of 1.7V/0.025A = 68 Ohm which is available at electronic stores.
To calculate the power used by the resistor just calculate P = U*I, this means P = 1.7V * 0.025A = 0.0425 W. So a small 0.25W resistor is enough for this purpose. If you use higher currents or want to burn more volt in the resistor you might have to use a bigger one!
That’s the reason why you could only operate 6 high voltage consuming white LEDs on 24V.
But not all LEDs are really the same, there might be big differences in the voltage loss from LED to LED. So we use the second potentiometer (300 ?) and a current-meter to adjust the current of each strip to the desired level (25mA) in the final circuit. Then we measure the value of the resistor and this should give us something around the calculated value.
If the result is something in between two types then choose the next higher value if you want the strip to be a little darker or the next lower value for the strip to be a bit brighter.
I installed the LEDs in an acrylic glass board which I fixed to the power-source-housing. Acrylic glass can easily be drilled and bend if heated to around 100°C in the oven. As you can see on the pictures I also added the sunrise – sunset selection switch to this display. The potentiometer and the reset-button are on the circuit-board.
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