Sir, i want to ask. I have a project to control 11 LED RGB using microcontroller.
The circuit is as you can see below :
The RGB Led i'm using with is SMD 5050 RGB. Or you can see the datasheet in the attachment file.
LED 1 = Red LED
LED 2 = Blue LED
LED 3 = Green LED
So basically, the total of the RGB LED SMD 5050 is 11 LED i'm using with because in 1 package of IC, contains red, green, and blue.
3 of them are in 1 IC led package.
And the specification is :
Red LED needs Vf = 2,1V, If = 20mA ;
Green LED needs Vf = 3,2V, If = 20mA ;
Blue LED needs Vf = 3,2V, If = 20mA ;
So at the green LED and blue LED, i use 25 ohm resistor.
This is because :
36 Volt - 11*(3,2V) - 0.3V (assume when the 2N222A transistor or in reality, i'm using MMBT2222A, is in saturation region ) / 20 mA = 25 ohm
And at RED LED, i use 630 ohm resistor, because :
36 Volt - 11*(2,1V) - 0.3V (assume when the 2N222A transistor or in reality, i'm using MMBT2222A, is in saturation region) / 20 mA = 630 ohm.
What i want to ask is that, AT THE RED LED circuit, when i applied KVL equation, 36 Volt - 11*(2,1V) - 0,3V = 12.6V ( the transistor is in the saturation region ). Where the 12.6 V will be dissipated ? Will it be dissipated at the resistor of 630 ohm ? Or will it be dissipated at the transistor MMBT2222A ? My guess is the rest of the voltage ( 12.6 V ) will be dissipated on the resistor. So, i need at least P = 12.6V * 20mA = 0.252 Watt resistor ( i think i'm going to use 1/2 Watt resistor )
Am i right sir ?
Thanks before.
The circuit is as you can see below :
The RGB Led i'm using with is SMD 5050 RGB. Or you can see the datasheet in the attachment file.
LED 1 = Red LED
LED 2 = Blue LED
LED 3 = Green LED
So basically, the total of the RGB LED SMD 5050 is 11 LED i'm using with because in 1 package of IC, contains red, green, and blue.
3 of them are in 1 IC led package.
And the specification is :
Red LED needs Vf = 2,1V, If = 20mA ;
Green LED needs Vf = 3,2V, If = 20mA ;
Blue LED needs Vf = 3,2V, If = 20mA ;
So at the green LED and blue LED, i use 25 ohm resistor.
This is because :
36 Volt - 11*(3,2V) - 0.3V (assume when the 2N222A transistor or in reality, i'm using MMBT2222A, is in saturation region ) / 20 mA = 25 ohm
And at RED LED, i use 630 ohm resistor, because :
36 Volt - 11*(2,1V) - 0.3V (assume when the 2N222A transistor or in reality, i'm using MMBT2222A, is in saturation region) / 20 mA = 630 ohm.
What i want to ask is that, AT THE RED LED circuit, when i applied KVL equation, 36 Volt - 11*(2,1V) - 0,3V = 12.6V ( the transistor is in the saturation region ). Where the 12.6 V will be dissipated ? Will it be dissipated at the resistor of 630 ohm ? Or will it be dissipated at the transistor MMBT2222A ? My guess is the rest of the voltage ( 12.6 V ) will be dissipated on the resistor. So, i need at least P = 12.6V * 20mA = 0.252 Watt resistor ( i think i'm going to use 1/2 Watt resistor )
Am i right sir ?
Thanks before.
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