Hi Mario, It depends on which PIC and what the supply voltage is. For the 16 series P= ICs up to 5 volts you should be OK as long as you don't try to do it on ver= y many pins and don't do it on multiple pins at the same time. The higher the supply voltage, the lower the output resistance will be, and= that will bring the peak current up. You don't want the peak current to ap= proach the rated limit of the entire port (is it 80 mA? 100 mA? I forget). At 5 volts, the output resistance when sinking current is about 80 ohms on = a 16F88/887/etc. That puts your peak current at 63 mA, so you're OK At 5.25 volts or 5.5 volts supply that output resistance is lower and you w= ill be shaving very close. Also, you want the output voltage to slew fast enough that the pin can reac= h the correct logic level within one internal clock cycle(4 crystal cycles)= , so that limits the capacitance you can drive without the PIC getting conf= used and doing some weird read-modify-write stuff. So you want the pin volt= age to change by a couple of volts in that time. BTW, this is all based on my experience, some of it may be technically or m= orally incorrect. Hope this helps. Friendly regards, Bob ________________________________________ From: piclist-bounces@mit.edu on behalf of Mario Sent: Sunday, February 4, 2018 11:53 PM To: Microcontroller discussion list - Public. Subject: [EE] Driving a 10nF capacitance Hello, this is for an extremely space-constrained application, so every component counts. The PIC datasheet specifies what is the max current that can be sourced/sin= ked by a PIC pin. However, this in principle means that no capacitance can be directly driven= , as any capacitor with low ESR (eg. ceramic) will look almost like a short circuit, although only for a very brief time. Regardless of what the datasheet strictly says, in your experience is it sa= fe to drive a 10nF capacitance directly from a PIC pin (for few hundreds Hz's, i.e. without thermal or average current excesses, only peak current), or a resistor is always needed? The internal resistance of the pin is in the tens of ohms already, so I won= der. While the peak current, without the external resistor, will exceed the maxi= mum specified in the datasheet, this is meant as "continuous current", while MO= SFETs do always exhibit a higher peak current capability.. so why not also PIC pi= ns? Thank you for your insights. Kind regards, Mario --=20 http://www.piclist.com/techref/piclist PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .