At 17:55 2018-02-05, Bob Blick wrote:
>Hi Mario,
>
>It depends on which PIC and what the supply voltage is. For the 16=20
>series PICs up to 5 volts you should be OK as long as you don't try to=20
>do it on very 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=20
>be, and that will bring the peak current up. You don't want the peak=20
>current to approach the rated limit of the entire port (is it 80 mA?=20
>100 mA? I forget).
>
>At 5 volts, the output resistance when sinking current is about 80=20
>ohms on a 16F88/887/etc. That puts your peak current at 63 mA, so you're O=
K
>
>At 5.25 volts or 5.5 volts supply that output resistance is lower and=20
>you will be shaving very close.
>
>Also, you want the output voltage to slew fast enough that the pin can=20
>reach the correct logic level within one internal clock cycle(4=20
>crystal cycles), so that limits the capacitance you can drive without=20
>the PIC getting confused and doing some weird read-modify-write stuff.=20
>So you want the pin voltage to change by a couple of volts in that time.
>
>BTW, this is all based on my experience, some of it may be technically=20
>or morally incorrect.

"morally incorrect" is fantastic. ;D

Cheers,
Mario


>
>Hope this helps.
>
>Friendly regards,
>
>Bob
>
>________________________________________
>From: piclist-bounces@mit.edu <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/si=
nked
>by a PIC pin.
>
>However, this in principle means that no capacitance can be directly drive=
n,
>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 s=
afe
>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 wo=
nder.
>While the peak current, without the external resistor, will exceed the max=
imum
>specified in the datasheet, this is meant as "continuous current",=20
>while MOSFETs
>do always exhibit a higher peak current capability.. so why not also PIC p=
ins?
>
>Thank you for your insights.
>
>Kind regards,
>Mario
>
>--=20
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>View/change your membership options at
>http://mailman.mit.edu/mailman/listinfo/piclist

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