> Although the current cannot flow back to the battery, wouldn't it > still flow a large current back into the programmer's ground? For the > PIC to run and program, mustn't both it's ground and positive pins be > connected to the programmer? And shouldn't both of those pins be > hooked to the rest of the circuit for the PIC to run off the battery? Yes, BUT... 1. The battery won't be connected. Normally, the worry with a PMOS LDO is that the diode which inherently exists, reverse biased in parallel with the source and drain, will conduct if Vout > Vin. How can an LDO have a greater output than input? That's what happens in the case where the input bypass cap is "charged" to 0V, and the output is driven by another supply. Let's imagine a 3V output LDO driven by 3.5V worth of batteries (say, 3 marginally flat AA's). Now, hook your 5V PIC programmer up to it. Suddenly, that reverse biased diode is forward biased, and it's a direct connection between a 5V supply and a 3.5V supply. Say that diode drops ~.8V. That leaves .7V drop over the various resistances between the two sources. Disconnect those batteries. Now where does the current go? For a few micro- or milli- seconds, it goes into the input cap to charge it up, but then it droops off (asymptotically) to the leakage current of that cap. 2. Yes, they do need to be connected, but where's that huge current going to come from? And of course they will be connected, since the connection to the programmer is a header on the PCB. I think perhaps I was unclear in my original message. In the case of production grade ICSP, it is (usually) important that the PIC be run at 4.5V, to facilitate bulk erase. Then, it should be verified at the min and max voltages the system expects to see. This means that all components in the system which share a power bus with the PIC either A. must tolerate this excursion to 4.5V or B. must be isolated from this excursion to 4.5V. All the components in my system, save possibly the regulator, can tolerate the excursion to 4.5V. I was concerned about what would happen to the regulator if its output is raised to 4.5V with nothing on its input. An element in the design of this LDO is a reverse biased diode connecting the output to the input, which of course becomes forward biased if Vin