gpio
Can someone point out the mistake I'm making in reading this graph (which is from the ADI Blackfin 70x datasheet)
If a driver was set high, to 1.8V, why would it be unable to source current? I'm reading up from 1.8V on the X-axis and going up to the middle Voh line, then over to the Y-axis.
So I'm not too sure I follow this first question here, but I'll take a stab at it: if Vin = 3.3V (i.e. your P-channel mosfet does not connect source and drain) then "TRIGGER" will essentially be disconnected, yes. This assumes that "TRIGGER" is connected to a high-impedance source, like an input pin on a standard microcontroller. As a quick note "essentially disconnected" means that there is a high impedance threshold to get across the source-drain junction, but it is not infinite. If Vin = 0V (i.e. source and drain are connected), then trigger will be close to (but slightly less than) 3.3V
So an input pin on a PIC24F is high impedance. This means that it will behave like a resistor with around 1MΩ of resistance to ground. If you check the "Electrical Characteristics" section in the "DC Characteristics: I/O Pin Input Specifications" subsection in the datasheet for the chip you're using, you should find a table which gives you the "Input Leakage Current." Using the dsPIC33FJ256GP710A as an example (it was a datasheet I happened to have on my machine), it lists 0 to 3.5 μA as the leakage current into or out of a pin within the limits of those pins.
So as a note, you'll also see that in the "Electrical Characteristics" section under the "Absolute Maximum Ratings" subsection, there will be a list which says something to the effect of "Voltage on any pin that is not 5V tolerant with respect to VSS(4) .... -0.3V to (VDD + 0.3V)". This is telling you that you can't put more than ~3.6V on any pin without damaging the pin. This doesn't mean that you should put a current-limiting resistor in front of the input pin, this means you should put a voltage divider in front of the input pin to limit the maximum voltage to less than VDD + 0.3V.
To tie this into question #1: this means that when a digital input pin is connected to a high-impedance output source like Fig. 1, you will see oscillating input values (1 and 0 randomly) based on when that very small drain removes or adds enough electrons to jump from one digital state to another. This is called a floating input
So I just answered this question in the previous one, but I should tell you: the "Maximum output current" is not internally limited. So although the PIC24F can "only output 18mA" this means that if you connect an output pin which is high directly to ground, it will output lots more than 18mA before it burns out! The 18mA is the maximum it can safely output without overheating/damaging the chip. This is true for both sourcing and sinking, there is no fuse which will stop you at 18mA, only the magic blue smoke.
Just to reiterate: you can safely connect any voltage to a digital input pin within its range (-0.3V to VDD+0.3V for the dsPIC33F). It would not blow the pin to connect it directly to +3.3V, nor would it blow the pin to connect directly to ground.
Just answered this in #3, but again, it's not the output impedance of the pin which limits this... Or, more accurately, it is the output impedance, but it will limit the currently by melting.
And yes, you can safely remove the 1kΩ resistor from Fig. 2 without damaging your PIC.
The more current you source or sink through a pin, the more the voltage deviates from the ideal voltage.
Here is what the datasheet of some random microprocessor says about the relationship between output current and output voltage (you should find such specifications in any datasheet):
The question is not so much how much current the pin can handle (if you go near the absolute limit, your circuit is likely to fail anyway), but how much of a voltage drop you can tolerate.
Furthermore, a higher drive strength can supply the needed current faster when switching, which means that the voltage also switches faster, i.e., you have shorter raise/fall times. This implies that higher drive strength can result in increased EMI.
Best Answer
Any amplifier - including this driver - has some output resistance. If you draw current from it, some voltage is dropped across this output resistance.
So the only way to see the open-circuit output voltage is with - well, an open circuit, i.e. to draw no current.
If you draw some current - say, 5 mA, the output voltage reduces by 0.4V. So the output resistance - or more properly, source impedance, is 0.4/0.005 = 80 ohms. (Approximately; as the load line isn't straight, the source impedance varies a little according to the load current).
Knowing this allows you to design the PCB to get good pulse shapes - you can design the PCB traces and any termination for a characteristic impedance that matches the driver.