Voltage spike on capacitor charge

I'm trying to understand something in a circuit where there is a capacitor in series between two pins on a couple of ICs, A and B:

A o-------------||----------------oB

Pin A is the active low reset pin of a PIC 16F876 and pin B is an output from an FTDI USB interface. A and B are, in normal use, held at +5V.

For programming, a signal is sent to the FTDI chip that causes B to be taken to 0V, and this causes a very rapid voltage rise to around +10V at A, which then declines asymptotically back to +5V.

Unfortunately I'm better with digital electronics than analogue and I'm a bit out of my depth so here are a couple of questions:
a) what causes the sudden voltage rise at A, and
b) (more important) is there any easy way, given the 5V starting values, to get the voltage to hit something closer to 13V rather than 10V?

The rise in voltage is used as a signal for the PIC to go into programming mode, but the PIC spec says this should be around 13V. The 10V or so has worked on the PICs that we have used up to now, but some newer PICs are not responding to the 10V and I reckon it's because they are less tolerant of low programming voltages so I need to get A to rise to something closer to 13V.

All suggestions welcome!

What you have described doesn't make sense to me. If A and B are both at 5V, and one is pulled low, the other will go low.

To work like you say, one would have to be at 0V, the other at 5V. Then when you take the 0V one to 5V, the 5V, one will go to 10V.

To make it go to 13V would either require more than one pulse and some diodes, or maybe some inductance.

Keith

You didn't tell, in which programming adapter you have found the said strange circuit. I can only say, that common PIC programming
adapters, e.g. from Microchip, are using a filtered and stabilized VPP voltage source. You can refer to the PICkit 2 schematic, that
has been published from Microchip, or serious third party designs.

I can add a bit more clarity to this, as well as a (fuzzy) snap of the scope trace during the reset / programming pulse on the PIC.

To recap the circuit, I have this:

A o---------||-----------oB

where A is the reset / programming pin of a PIC16F876 and is pulled up to 5V by a 10K resistor. B is an output pin of a FTDI FT2232L, with no (external) pull-up or -down.

In the attached pic, the top line is at A and the bottom line is at B.

The normal situation is that both sides of the capacitor are at 5V. For programming, and this is my interpretation of events, pin B is brought down to 0V, which causes the sudden drop at A as the capacitor discharges.

The capacitor then regains charge until B is brought back to 5V, causing the jump to 10V at A as the excess charge in the capacitor is dumped.

The 10V peak is used as the programming trigger for the PIC, except that the PIC is expecting a 13V trigger. Nevertheless, this has worked up to now but I suspect that the batch of PICs we are using now are less tolerant of a low programming voltage. The circuit is one that I have inherited.

My plan at the moment is to bias the supply voltage to the board, currently at 5V, to something closer to 6V. This shouldn't affect any of the components, but it should hopefully bring the peak on the programming pin to something like 12V and that might be enough to do the job.

Any other suggestions welcome!