Difference between revisions of "Extension Explorer 16"

From Fixme.ch
Jump to: navigation, search
Line 33: Line 33:
 
=== Formula ===
 
=== Formula ===
 
* calculation of resistance R_LED_R :   
 
* calculation of resistance R_LED_R :   
−
 
−
 
 
<math>
 
<math>
−
R_{LEDR} = \sum_{n=0}^\infty a_n x^n
+
R_{LED_R}=\frac{U_r}{I}=\frac{VDD-U_{LR}}{I}=\frac{3.3V-1.8V}{10mA}
 
</math>
 
</math>
−
 
+
= ''' 150 ohm '''
−
 
+
* power dispersed by the resistance P_R_LED_R :
 
<math>
 
<math>
−
\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}.
+
\frac{R_{LED_R}^2}{I}=\frac{150^{2}}{10mA}
 
</math>  
 
</math>  
−
 
+
= ''' 15 mW '''  
−
= \frac{U_r}{I} = \frac{VDD - U_{LR}}{I} = \frac{3.3V - 1.8V}{10mA}
+
−
 
+
−
<math>0</math>
+
−
 
+
−
= ''' 150 ohm '''
+
−
* power dispersed by the resistance P_R_LED_R :
+
−
<code>$<math>\frac{R_LED_R^{2}}{I} = \frac{150 o ^{2}}{10mA}</math> = ''' 15 mW ''' $</code>
+
 
    
 
    
 
* calculation of resistance R_LED_O :
 
* calculation of resistance R_LED_O :
−
<code>$<math>\frac{U_r}{I} = \frac{VDD - U_{LO}}{I} = \frac{3.3V - 1.9V}{10mA}</math> = ''' 140 ohm ''' $</code>  
+
<math>\frac{U_r}{I}=\frac{VDD-U_{LO}}{I}=\frac{3.3V-1.9V}{10mA}
 +
</math> = ''' 140 ohm '''   
 
* power dispersed by the resistance P_R_LED_O :   
 
* power dispersed by the resistance P_R_LED_O :   
−
<code>$<math>\frac{R_LED_O^{2}}{I} = \frac{140 o ^{2}}{10mA}</math> = ''' 14 mW ''' $</code>  
+
<math>\frac{R_LED_O^{2}}{I}=\frac{140^{2}}{10mA}
 +
</math>  
 +
= ''' 14 mW '''   
  
 
* calculation of resistance R_LED_G :  
 
* calculation of resistance R_LED_G :  
−
<code>$<math>\frac{U_r}{I} = \frac{VDD - U_{LG}}{I} = \frac{3.3V - 2.2V}{10mA}</math> = ''' 110 ohm '''$</code>
+
<math>\frac{U_r}{I}=\frac{VDD - U_{LG}}{I}=\frac{3.3V-2.2V}{10mA}</math>  
 +
= ''' 110 ohm '''  
 
* power dispersed by the resistance P_R_LED_G :   
 
* power dispersed by the resistance P_R_LED_G :   
−
<code>$<math>\frac{R_LED_G^{2}}{I} = \frac{110 o ^{2}}{10mA}</math> = ''' 11 mW '''$</code>
+
<math>\frac{R_LED_G^{2}}{I}=\frac{110^{2}}{10mA}
 +
</math> = ''' 11 mW '''
  
 
= Layout =
 
= Layout =

Revision as of 00:57, 7 October 2014

Description

In French: Création d'une carte d'extension pour la carte Explorer (PROJET: Carte Explorer 16 & Exos sur les uC).

  • V1: un montage permettra de simuler un carrefour routier, composer de 4 feux (RGB)


in English: creation of one extansion board for the Explorer Board (see the links above)

  • V1: assembly will allow to simulate a crossroad (4 fires (RGB))

Digram - Schematic

Part Calculation of Electronics Digram

LED's for crossroad

  • the datasheet of the DSP implemented on the Explorer Board 16 (see pages 1 & 273) indicates that :
    • output current of I/O : I = 10mA
    • voltage I/O : VDD = 3.3V
  • the datasheet of the red LED (see pages 3) indicates that :
    • conduction voltage : ULR = VF = 1.8V
    • average current : I = 20mA
  • the datasheet of the orange LED (see pages 3 / 4 / 5) indicates that :
    • conduction voltage : ULO = VF = 1.9V
    • average current : I = 30mA
  • the datasheet of the green LED (see pages 3 / 4 / 5) indicates that :
    • conduction voltage : ULG = VF = 2.2V
    • average current : I = 30mA

Formula

  • calculation of resistance R_LED_R :

R_{LED_R}=\frac{U_r}{I}=\frac{VDD-U_{LR}}{I}=\frac{3.3V-1.8V}{10mA} = 150 ohm

  • power dispersed by the resistance P_R_LED_R :

\frac{R_{LED_R}^2}{I}=\frac{150^{2}}{10mA} = 15 mW

  • calculation of resistance R_LED_O :

\frac{U_r}{I}=\frac{VDD-U_{LO}}{I}=\frac{3.3V-1.9V}{10mA} = 140 ohm

  • power dispersed by the resistance P_R_LED_O :

\frac{R_LED_O^{2}}{I}=\frac{140^{2}}{10mA} = 14 mW

  • calculation of resistance R_LED_G :

\frac{U_r}{I}=\frac{VDD - U_{LG}}{I}=\frac{3.3V-2.2V}{10mA} = 110 ohm

  • power dispersed by the resistance P_R_LED_G :

\frac{R_LED_G^{2}}{I}=\frac{110^{2}}{10mA} = 11 mW

Layout

Components listing

Modification

Documentation

PDF FILE

required links

Contributor

Retrieved from "https://fixme.ch/w/index.php?title=Extension_Explorer_16&oldid=6029"