Hi guys, I need some help.
I'm trying to make a led driver with a transistor(BC548C) but something is wrong in my opinion, and I dont understand why.
So, as I said i'm trying a transistor as a switch (BJT) where I want to control a 20mA led when voltage is applied Base,
My calculations to get the Saturation:
1º - In my Collector I have 12.53V so the first thing I did is to calculate the collector resistor:
(12.53V - 1.8V (led forward voltage drop) / 0.02A) = 536 Ohm
2º - Once calculated the collector resistor I need to know how much current base need to saturate my transistor so:
Assuming Beta = 420

Base Current = Ic/Beta (0.02A/420 = 47uA)
That means that I need 47uA in my transistor base to saturate right?
3º Base resistor Vin with 3.3V
(3.3v - 0.7) / 47uA = 55K
The Problem is even with this calculation I cant get the saturation and I dont kow why, still have voltage in Vce as I show bellow.
View attachment 236980
I hope I explained it well to make you understand my doubts. Thanks everyone! (And sorry for my bad english)
I'm trying to make a led driver with a transistor(BC548C) but something is wrong in my opinion, and I dont understand why.
So, as I said i'm trying a transistor as a switch (BJT) where I want to control a 20mA led when voltage is applied Base,
My calculations to get the Saturation:
1º - In my Collector I have 12.53V so the first thing I did is to calculate the collector resistor:
(12.53V - 1.8V (led forward voltage drop) / 0.02A) = 536 Ohm
2º - Once calculated the collector resistor I need to know how much current base need to saturate my transistor so:
Assuming Beta = 420

Base Current = Ic/Beta (0.02A/420 = 47uA)
That means that I need 47uA in my transistor base to saturate right?
3º Base resistor Vin with 3.3V
(3.3v - 0.7) / 47uA = 55K
The Problem is even with this calculation I cant get the saturation and I dont kow why, still have voltage in Vce as I show bellow.
View attachment 236980
I hope I explained it well to make you understand my doubts. Thanks everyone! (And sorry for my bad english)