The datasheet says "Css is charged via a PNP transistor that is connected to the error amplifier." Does this mean that current flows from the emitter through the base?
I have drawn on this diagram to show what I think must be happening:
Electronic – Does current flow from the emitter through the base in this circuit
transistors
Related Solutions
Okay, looking at the picture I think you may have the transistor the wrong way round.
Try turning it round.
See this picture for reference:
As you can see the collector is on the right with the flat part facing you, so you have the collector connected to the LED in your circuit (if the 2N2222A part you are using has the same pinout)
I got the picture from here.
EDIT - It's actually a 2N222A, but the above advice still goes as the pinout appears to be the same from the picture posted.
As Russell mentions the more standard way is to connect the LED to the collector, but your circuit should work if set up correctly.
"Emitter Base Voltage" is the maximum voltage that may be applied when the base-emitter diode is in reverse; not conducting. This is generally much lower than a small signal diode in reverse can handle.
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Best Answer
It does, but it is negligible compared to the current supplied by Vref. Not only is the base current typically at least 100 times smaller than the collector current, the error amplifier can only source 1mA anyway. So the base current will be something less than 10µA.
At least based on the snippet you've provided, the datasheet is not claiming Css is charged via the PNP, only that the PNP is connected to Css. That connection actually affects the PNP more than it does the capacitor.
Specifically, while the capacitor is not charged, the PNP is hard on and sinks all the current from the error amplifier (dumping it out the collector to ground), effectively starving the feedback network and inhibiting the IC. As the capacitor charges, the PNP is pinched off and no longer sinks the current from the error amplifier. The error amplifier is then free to supply the feedback network and the IC begins normal operation.