I got a few 2SD1048's, as I wanted to add a switch to an LCD project I have, so that I could turn the LCD screen on and off via a microcontroller pin.
The LCD screen is this and pulls ~100mA (rounded up). I'm operating it at 3.3v and the 2SD1048's were selected for their very low Vce(sat), since I can't afford to drop much voltage across the switch into the screen. I'm still pretty new to reading transistor datasheets, but it looks like the Vce(sat) for the 2SD1048 is 80mV at max, so probably 3.22V should reach the LCD screen. Marginal, but it should power on and work.
However, what I'm seeing is instead 3.3V at the collector, but 2.60V at the emitter, which isn't enough to power the LCD enough to function.
What am I doing wrong here, or what am I misinterpreting in the datasheet? Here is my circuit, voltages measured at Vin and Vout.
Best Answer
You should use a PNP eg. 2SB815 and invert the drive. Emitter to +3.3, and collector to the load.
simulate this circuit – Schematic created using CircuitLab
The way you are doing it will not saturate the transistor, because you would need more than 3.3V at the base resistor- it’s an emitter-follower which will always have at least one diode drop.
Something like this would give you the datasheet performance for the 2SD1048, however it's less convenient because you probably don't have a 6.6V supply or a way to level shift the control signal. There's also a potential issue if the load is removed the 3.3V supply could rise in value and damage something.
simulate this circuit
P.S. If by "screen" you mean only the backlight, you can simply use the D1048 as a low-side switch, but that's not a great way to switch the entire display power.