Limit current with fewer resistors

There are several 16-channel constant current LED driver ICs sold specifically for the kind of requirement described.

For instance:

• Texas Instruments TLC5940: 16 LED channels with PWM, up to 120 mA each. Perhaps the most well-known, with open source libraries available for many common microcontrollers.
• ST Microelectronics STP1612PW05: 16 LED channels with PWM, up to 60 mA each
• Linear Technology LT3754: 16 LED channels with PWM, up to 50 mA each
• Allegro A6282: 16 LED channels, up to 50 mA each
• OnSemi CAT4016: 16 LED channels with PWM, up to 100 mA each
• NXP PCU9955: 16 LED channels with PWM, up to 57 mA each

Each of these ICs is controlled using some form of serial interface, such as I2C or SPI, and at sufficiently high clock rates that a tachometer, even one with fading LED and peak highlight effects, can be implemented.

If one of these ICs is used, not only will all the LEDs be controllable with a single component, the need for individual current limiting resistors on all 15 LEDs will also be eliminated.

So, just one IC and a few supporting parts, plus serial control from your microcontroller.

Another alternative is to use some form of multiplexed LED driver, such as the Maxim MAX7219, which besides its well-known job as a 7-segment 8 digit display driver, can also be used to drive up to 64 individual LEDs, where any one LED will be lit and drawing current at a time, hence the concern about current draw will be taken care of.

1. 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

2. 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

3. 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.

4. 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.

5. And yes, you can safely remove the 1kΩ resistor from Fig. 2 without damaging your PIC.