LED Cube/Column Resistors in Detail - Versionsgeschichte

Marvin am 15. April 2012 um 03:06 Uhr

2012-04-15T03:06:43Z

← Nächstältere Version		Version vom 15. April 2012, 03:06 Uhr
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	===Resistor===		===Resistor===
	Now, we are finally able to calculate our resistor value. This is simply done by subtracting all ~~dropuout~~ voltages from the supply voltage and then apply U = R * I. So:		Now, we are finally able to calculate our resistor value. This is simply done by subtracting all dropout voltages from the supply voltage and then apply U = R * I. So:
	R = (Vcc - Vio - Vta - Vled)/I		R = (Vcc - Vio - Vta - Vled)/I
	= (5.0 V - 0.5 V - 0.7 V - 1.9 V)/(20 mA)		= (5.0 V - 0.5 V - 0.7 V - 1.9 V)/(20 mA)

Marvin am 15. April 2012 um 03:05 Uhr

2012-04-15T03:05:43Z

← Nächstältere Version		Version vom 15. April 2012, 03:05 Uhr
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	__NOTOC__		__NOTOC__
			==Introduction==
	Choosing the right column resistors for your LED cube requires you to consider several factors:		Choosing the right column resistors for your LED cube requires you to consider several factors:
	* Suitability of the cube base for different colors		* Suitability of the cube base for different colors
Zeile 6:		Zeile 7:
	* Voltage drop on the transistor array		* Voltage drop on the transistor array
	* Voltage drop on the LED		* Voltage drop on the LED

			This page will only cover some basics on how to do these calculations for your own requirements. If you are only building an LED Cube you should refer to the values given in the construction manual.

	For now, we are only interested in the resistor value for one specific LED color (and also only one type as a color may be created using different techniques which results in different forward voltages), thus only one LED forward voltage has to be taken into consideration. Also we assume that the USB voltage is 5.0 Volts which is okay, as we will see. Now we are at a situation where we have five volts which will be divided onto our components. Also, we are going to make this calculation only for 20 mA current per LED, however, you can make your own calculations quite easily. Let's have a look at the different I/V characteristics:		For now, we are only interested in the resistor value for one specific LED color (and also only one type as a color may be created using different techniques which results in different forward voltages), thus only one LED forward voltage has to be taken into consideration. Also we assume that the USB voltage is 5.0 Volts which is okay, as we will see. Now we are at a situation where we have five volts which will be divided onto our components. Also, we are going to make this calculation only for 20 mA current per LED, however, you can make your own calculations quite easily. Let's have a look at the different I/V characteristics:
Zeile 68:		Zeile 71:
	= '''40 Ohm'''		= '''40 Ohm'''
	But now let's assume that the USB voltage increased to 5.5 V for whatever reason, which is well within specs. If we assume that this will result in a voltage increase of 0.3 V at the resistor, our current is now 28 mA. For 9 enabled LEDs and the supply current of 20 mA for the uC, this results in a current through the Vcc pins of 268 mA, which is relatively close to the absolute maximum ratings of 300 mA.		But now let's assume that the USB voltage increased to 5.5 V for whatever reason, which is well within specs. If we assume that this will result in a voltage increase of 0.3 V at the resistor, our current is now 28 mA. For 9 enabled LEDs and the supply current of 20 mA for the uC, this results in a current through the Vcc pins of 268 mA, which is relatively close to the absolute maximum ratings of 300 mA.

			==Conclusion==
			We can see that it is quite easy to calculate a useful resistor value for one situation if all parameters are available. However, finding a good balance between brightness (which is perceived proportionally to the logarithm of the current) and a good operating range has to be considered for each situation.

Marvin: /* Resistor */

2012-04-15T03:00:50Z

Resistor

← Nächstältere Version		Version vom 15. April 2012, 03:00 Uhr
Zeile 41:		Zeile 41:
	===Resistor===		===Resistor===
	Now, we are finally able to calculate our resistor value. This is simply done by subtracting all dropuout voltages from the supply voltage and then apply U = R * I. So:		Now, we are finally able to calculate our resistor value. This is simply done by subtracting all dropuout voltages from the supply voltage and then apply U = R * I. So:

	R = (Vcc - Vio - Vta - Vled)/I		R = (Vcc - Vio - Vta - Vled)/I
	= (5.0 V - 0.5 V - 0.7 V - 1.9 V)/(20 mA)		= (5.0 V - 0.5 V - 0.7 V - 1.9 V)/(20 mA)
Zeile 48:		Zeile 47:

	Of course, we do not need to use the full 20 mA, which are quite on the limit for the used components. A more reasonable value would be 15 mA:		Of course, we do not need to use the full 20 mA, which are quite on the limit for the used components. A more reasonable value would be 15 mA:

	R = (Vcc - Vio - Vta - Vled)/I		R = (Vcc - Vio - Vta - Vled)/I
	= 1.9 V / 15 mA		= 1.9 V / 15 mA
Zeile 54:		Zeile 52:

	or 10 mA:		or 10 mA:

	R = (Vcc - Vio - Vta - Vled)/I		R = (Vcc - Vio - Vta - Vled)/I
	= 1.9 V / 10 mA		= 1.9 V / 10 mA
Zeile 60:		Zeile 57:

	For a blue LED (3.0 V drop) at 10 mA we get:		For a blue LED (3.0 V drop) at 10 mA we get:
			R = (Vcc - Vio - Vta - Vled)/I
			= (5.0 V - 0.5 V - 0.7 V - 3.0 V)/(10 mA)
			= 0.8 V / 10 mA
			= '''80 Ohm'''

			Here is something interesting. We could also calculate with 20 mA:
	R = (Vcc - Vio - Vta - Vled)/I		R = (Vcc - Vio - Vta - Vled)/I
	= (5.0 V - 0.5 V - 0.7 V - 3.0 V)/(20 mA)		= (5.0 V - 0.5 V - 0.7 V - 3.0 V)/(20 mA)
	= 0.8 V / 10 mA		= 0.8 V / 20 mA
	= '''80 Ohm'''		= '''40 Ohm'''
			But now let's assume that the USB voltage increased to 5.5 V for whatever reason, which is well within specs. If we assume that this will result in a voltage increase of 0.3 V at the resistor, our current is now 28 mA. For 9 enabled LEDs and the supply current of 20 mA for the uC, this results in a current through the Vcc pins of 268 mA, which is relatively close to the absolute maximum ratings of 300 mA.

Marvin: /* Resistor */

2012-04-15T02:54:45Z

Resistor

← Nächstältere Version		Version vom 15. April 2012, 02:54 Uhr
Zeile 58:		Zeile 58:
	= 1.9 V / 10 mA		= 1.9 V / 10 mA
	= '''190 Ohm'''		= '''190 Ohm'''

			For a blue LED (3.0 V drop) at 10 mA we get:

			R = (Vcc - Vio - Vta - Vled)/I
			= (5.0 V - 0.5 V - 0.7 V - 3.0 V)/(20 mA)
			= 0.8 V / 10 mA
			= '''80 Ohm'''

Marvin: /* Resistor */

2012-04-15T02:51:35Z

Resistor

← Nächstältere Version		Version vom 15. April 2012, 02:51 Uhr
Zeile 47:		Zeile 47:
	= '''95 Ohm'''		= '''95 Ohm'''

	Of course, we do not need to use the full 20 mA, which are quite on the limit for the used components. A more reasonable value would be 15 ~~or 10~~ mA~~. Here~~:		Of course, we do not need to use the full 20 mA, which are quite on the limit for the used components. A more reasonable value would be 15 mA:

	R = (Vcc - Vio - Vta - Vled)/I		R = (Vcc - Vio - Vta - Vled)/I
Zeile 53:		Zeile 53:
	= '''127 Ohm'''		= '''127 Ohm'''

			or 10 mA:

	R = (Vcc - Vio - Vta - Vled)/I		R = (Vcc - Vio - Vta - Vled)/I
	= 1.9 V / 10 mA		= 1.9 V / 10 mA
	= '''190 Ohm'''		= '''190 Ohm'''

Marvin: /* Resistor */

2012-04-15T02:51:16Z

Resistor

← Nächstältere Version		Version vom 15. April 2012, 02:51 Uhr
Zeile 46:		Zeile 46:
	= 1.9 V / 20 mA		= 1.9 V / 20 mA
	= '''95 Ohm'''		= '''95 Ohm'''

			Of course, we do not need to use the full 20 mA, which are quite on the limit for the used components. A more reasonable value would be 15 or 10 mA. Here:

			R = (Vcc - Vio - Vta - Vled)/I
			= 1.9 V / 15 mA
			= '''127 Ohm'''


			R = (Vcc - Vio - Vta - Vled)/I
			= 1.9 V / 10 mA
			= '''190 Ohm'''

Marvin: /* Resistor */

2012-04-15T02:49:17Z

Resistor

← Nächstältere Version		Version vom 15. April 2012, 02:49 Uhr
Zeile 40:		Zeile 40:

	===Resistor===		===Resistor===
	Now, we are finally able to calculate our resistor value. This is simply done by subtracting all dropuout voltages from the supply voltage and then apply U = R * I.		Now, we are finally able to calculate our resistor value. This is simply done by subtracting all dropuout voltages from the supply voltage and then apply U = R * I. So:

			R = (Vcc - Vio - Vta - Vled)/I
			= (5.0 V - 0.5 V - 0.7 V - 1.9 V)/(20 mA)
			= 1.9 V / 20 mA
			= '''95 Ohm'''

Marvin am 15. April 2012 um 02:46 Uhr

2012-04-15T02:46:12Z

← Nächstältere Version		Version vom 15. April 2012, 02:46 Uhr
Zeile 40:		Zeile 40:

	===Resistor===		===Resistor===
			Now, we are finally able to calculate our resistor value. This is simply done by subtracting all dropuout voltages from the supply voltage and then apply U = R * I.

Marvin: /* Transistor Array */

2012-04-15T02:41:50Z

Transistor Array

← Nächstältere Version		Version vom 15. April 2012, 02:41 Uhr
Zeile 37:		Zeile 37:
	The forward voltage drop in the transistor array is a little bit more tricky and shown in the following curve:		The forward voltage drop in the transistor array is a little bit more tricky and shown in the following curve:
	[[File:Uln2003-uce.svg]]		[[File:Uln2003-uce.svg]]
	This is a little bit annoying. Depending on how many LEDs we switch on, the voltage drop changes. For only one LED we have a voltage drop of about 0.7 V. For 9 LEDs we are already at 1 V. We will continue to calculate with the '''0.7 V''' figure as we are interested in bright LEDs for now and we do not want to risk the increased current that will result ~~out~~ of a reduced voltage drop.		This is a little bit annoying. Depending on how many LEDs we switch on, the voltage drop changes. For only one LED we have a voltage drop of about 0.7 V. For 9 LEDs we are already at 1 V. We will continue to calculate with the '''0.7 V''' figure as we are interested in bright LEDs for now and we do not want to risk the increased current that will be the result of a reduced voltage drop.

	===Resistor===		===Resistor===

Marvin: /* Transistor Array */

2012-04-15T02:41:38Z

Transistor Array

← Nächstältere Version		Version vom 15. April 2012, 02:41 Uhr
Zeile 37:		Zeile 37:
	The forward voltage drop in the transistor array is a little bit more tricky and shown in the following curve:		The forward voltage drop in the transistor array is a little bit more tricky and shown in the following curve:
	[[File:Uln2003-uce.svg]]		[[File:Uln2003-uce.svg]]
			This is a little bit annoying. Depending on how many LEDs we switch on, the voltage drop changes. For only one LED we have a voltage drop of about 0.7 V. For 9 LEDs we are already at 1 V. We will continue to calculate with the '''0.7 V''' figure as we are interested in bright LEDs for now and we do not want to risk the increased current that will result out of a reduced voltage drop.

	===Resistor===		===Resistor===