Programmers like e.g. ICD2 offer the option to power the target system from the programmer. As I understand, this is always the case with this design.
I therefore consider adding a mechanical switch to the PCB, to allow for self-powered systems. Or is there another way? VDDU can obviously be turned on/off by software, but it looks like VDDU must be on to power the 3.3V regulator that is also used as reference to limit signal levels for the low voltage PICs.
I see no fuse in the design, but current to the target seems to be limited by Q2 so that none may be required.
What do you think?
Pascal
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Yes, VPPU is always used, although for some devices you could enter program mode even if they're always supplied.
If you are in this condition then the easiest thing to do is to leave VDDU not connected to the target board.
In case of 3.3V the voltage would be limited by the device protection diodes; anyways a 3.3V regulator does not limit the voltage, the load does it.
Max current is limited by Q2 and the USB supply.
Last edit: Alberto Maccioni 2015-02-19
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Thanks for the answer. What I meant by limiting voltage levels are the diodes D171 and D172 whose cathodes are connected to the output of the 3.3V regulator.
If you would like to refer to this comment somewhere else in this project, copy and paste the following link:
Yes, what limits the voltage is not the 3.3V regulator (which can only source current) but the load, i.e. the device, which is sinking current (1mA per pin is sufficient to lower the voltage from 5V to 3.3V+0.7V).
Those diodes are also in parallel with the intenal diodes (from every pin to the device VDD).
If you would like to refer to this comment somewhere else in this project, copy and paste the following link:
Programmers like e.g. ICD2 offer the option to power the target system from the programmer. As I understand, this is always the case with this design.
I therefore consider adding a mechanical switch to the PCB, to allow for self-powered systems. Or is there another way? VDDU can obviously be turned on/off by software, but it looks like VDDU must be on to power the 3.3V regulator that is also used as reference to limit signal levels for the low voltage PICs.
I see no fuse in the design, but current to the target seems to be limited by Q2 so that none may be required.
What do you think?
Pascal
Yes, VPPU is always used, although for some devices you could enter program mode even if they're always supplied.
If you are in this condition then the easiest thing to do is to leave VDDU not connected to the target board.
In case of 3.3V the voltage would be limited by the device protection diodes; anyways a 3.3V regulator does not limit the voltage, the load does it.
Max current is limited by Q2 and the USB supply.
Last edit: Alberto Maccioni 2015-02-19
Thanks for the answer. What I meant by limiting voltage levels are the diodes D171 and D172 whose cathodes are connected to the output of the 3.3V regulator.
Yes, what limits the voltage is not the 3.3V regulator (which can only source current) but the load, i.e. the device, which is sinking current (1mA per pin is sufficient to lower the voltage from 5V to 3.3V+0.7V).
Those diodes are also in parallel with the intenal diodes (from every pin to the device VDD).