I want to get some clarification about output current and output voltage from a PIC12F675. As I read in datasheet, max. current for an output pin is 25 mA.
Is it always 25 mA, or will it depend in some settings? (Vdd = 5 V)?
And also, if Vdd is 5 V, the outout voltage will be Vdd - 0.7 V = 5 V - 0.7 V = 4.3 V. Or did I miss something?
Another question is: what does Ioh mean? I saw the sign "-" means current is sourced by a pin.
25mA is for absolute maximum rating.
The table you're showing is from "DC Characteristics", which means a normal operating condition. It means when the load is drawing 3.0mA, the pin voltage will be VDD-0.7V. Thus, the voltage drop of P-channel MOSFET of output buffer will be 0.7V, between the VDD line and output pin.
m new in this things, and cant get the point with current.
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If you look at the datasheet, you'll see the "maximum output currents" are in the "Absolute Maximum Ratings" section, and there is a note beneath the table that explain if you exceed these values the part may well suffer permanent damage.
Values in this section aren't therefore what you should be planning to use in normal operation, rather they are what you should avoid exceeding under all operational states.
For current limits, this normally relates to thermal damage due to internal resistances. Operating voltage does not normally significantly change this limit.
In terms of the Voh rating, each output will have drive structures for high and low, and these will have voltage/current relationships that vary according to supply voltage. The datasheets will commonly have graphs showing these relationship at various supply voltages. There will also generally be tables giving minimum, typical and maximum under certain conditions. These ranges allow you to produce designs with the required tolerance for part-to-part variations.
Ioh is the pin current (I) in output mode (o) at in the high (h) state. The equivalent for low output current would be Iol.
The entry you quote says that the high output voltage (Voh) will be a minimum of Vdd-0.7V when the output (Ioh) is supplying 3mA and the supply (Vdd) is 4.5V.
The negative sign is simply convention, with currents being quoted in terms of current flowing into the pin (hence Iol are generally positive, and Ioh negative).
