10V during auto-reset ?

I set up my breadboard (i.e. atmega168, ftdi232r chip) based on the diecimila schematics and everything’s working as it should. Previously while troubleshooting though I observed some odd behavior on the DTR / auto-reset line. With a 100nF cap connected to the reset line and a 15k pullup the line goes to 0v immediately and then slowly charges back up to 5v - so far so good. Right after that, however, the line immediately goes up to 10v and then slowly goes back down to 5v. The same behavior was observed here (note that I am using USB only, not serial though) : http://www.arduino.cc/playground/Learning/AutoResetRetrofit

My questions are:
a) how can there be 10v if the supply is only 5v from the USB bus ?
b) is this pulse (which lasts more than 100 uSecs) harmful to either the FT232R or the atmega168 ?

Normal signal response of passing a square wave pulse through a capacitor. I don't think it can cause any harm as it's been a standard design for quite a while now. By the way the AVR reset pin is rated to handle higher voltages as it's used with the high voltage programming mode not used by the Arduino.

Lefty

Care to point me to some further information ? I only expect to see voltage spikes like this from an inductor, not a capacitor. My (basic) textbooks claim that the voltage should be within the envelope of the square wave, not exceeding it.

Well the signal your seeing is normal and I can see it on my scope also. I'm not sure I can explain it in text book quality but it has to do with there being active 5vdc voltage sources on both sides of the cap, the pull-up resistor on the Arduino side and the FTDI DTR driver output pin. The changing pulse voltage charges and discharges the cap and these voltage can add or subtract from there steady state value. So 0vdc and +10VDC are the transient values to be seen from the reset pin input which is normally pulled-up to 5vdc.

This is not unlike capacitor/diode voltage doubler and tripler circuits operate that allow voltage charges to add to make a higher voltage then the source voltage that is driving the input.

Lefty

When you suddenly changes the voltage on one side of a capacitor the other side instantly experiences the same voltage change.

It has to to that because at the instant of the change there is no extra charge in the capacitor and so if the far side didn't follow the same rise there would be a change in charge in zero time and that would be impossible.

So suppose you had a capacitor with side A at zero volts and side B at +5V. Then you make side A go to +5V, then side B must experience the same rise and so it goes to +10V. Then it starts to discharge until both sides are at 5V. So with a capacitor you can generate a voltage doubling effect, two of these and you will multiply the voltage by 4. This is used in chips like the MAX232 to generate 12V from 5V, put some diodes in and you can generate -ve voltages as well, again like the MAX232.