I'm trying to saturate a 3904 to drive 6 LEDs in parallel at about 65mA. So I gave a buffer of 35mA and called it .1A. The 3904 has an hFE of 250 and the circuit is 5V. So here is my equation for the base transistor:
5v / (0.1A / 250hFE) = 12.5Kohms
I went down to 10K because I don't have anything quite that close and I'm just prototyping. So it seems to me that the transistor should be fully saturated for this circuit. But I find that the LEDs are still brighter if I don't use a resistor at the base at all. Am I doing something wrong? Does it matter that I'm using the 3904 to switch the anode of the LEDs?(Unfortunately I cannot switch the cathode because the LED's are already grounded on the PCB they are mounted on. I only have access to the anode.) I realize this means there's a voltage drop before the LEDs but even taking that in to account, the LEDs are still brighter with no base transistor rather than 10K.
Thanks for any help.
5v / (0.1A / 250hFE) = 12.5Kohms
I went down to 10K because I don't have anything quite that close and I'm just prototyping. So it seems to me that the transistor should be fully saturated for this circuit. But I find that the LEDs are still brighter if I don't use a resistor at the base at all. Am I doing something wrong? Does it matter that I'm using the 3904 to switch the anode of the LEDs?(Unfortunately I cannot switch the cathode because the LED's are already grounded on the PCB they are mounted on. I only have access to the anode.) I realize this means there's a voltage drop before the LEDs but even taking that in to account, the LEDs are still brighter with no base transistor rather than 10K.
Thanks for any help.