PIC and Transistor

Without referencing a particular device or pin, generally speaking, you can drive the output of the PIC IO either high or low which means that you can use an NPN transistor. What you will need to consider, though, is how much base current you will need to flow in order to drive the transistor sufficiently and you need to be sure that you don't exceed the limits of the PIC's drive capability. If it turns out that the current drive is insufficient, you can use a FET instead and if necesary drive an NPN or PNP transistor with the FET.

Again, generally speaking, using an NPN or PNP device determines wether you use it as a "source" or "sink" device when used as a switch. In this case, by source or sink, I mean is it switching on the power side or the ground side of the circuit (load).

One thing to keep in mind if you do decide to use a PNP transistor, if you ground the base by turning on a PIC IO you could very well apply an extreme Vbe that could damage or destroy the device.

Transistors can be used as switches. You want a switch that connects your ECU to GND when it's active, and disconnects when it's inactive. You can use an NPN transistor to do this, as shown (view in fixed font or M$ Notepad):

| | ------. .----o | | | | | | | | | | | | | | ___ |/ | o-|___|-o-| | PIC | | |>

| | .-. | | | | | === | | | | GND | | '-' | | | | | === | | GND | | | | | ------' | (created by AACircuit v1.28.5 beta 02/06/05

I would assume your transistor will sink less than 20mA when on, so use

To turn the transistor ON, set the PIC output to "1".

Good luck Chris

No. You get to select which pins act as outputs and which act as inputs. If output is selected, you can order the pin to a logic high or logic low state. The output circuit has just a bit more current capability in the low state (sinking current) than it has in the high state (sourcing current), but the difference is not all that great. The easiest kind of transistor to drive is a logic level, N channel mosfet. You connect the gate to the active-high output pin and the source to ground. The drain connects to a load which is always connected to a positive supply. If the load is inductive, you may need a diode across it, to limit the turn off voltage.

In this case, you need a resistor between the output pin and the base, to limit the output current to what can safely pass through that pin. And the emitter of the PNP transistor must be tied to the same supply rail as the PIC.

Not to the PIC.

The transistor will pull up to almost the supply voltage for the PIC, probably 5 volts.

Simulate the ECU by connecting a 1k resistor between the collector and ground.

A little. Do you want the PIC to approximate the selector switch in the gear box (7 separate connections that each pull down to ground, in turn)? If so, I would use 7 mosfets. 6 of them could be small, but the one that represents the neutral contact has to also drive the neutral lamp, so it would be a bigger device.

actually I only need to approximate 1,2 and 3. Neutral and 4,5 and 6 will actually turn the PIC pin the opposite way and disable the transistor. Make sense?

Sorry

Nope. Please start back at the beginning and describe what circuit and signals you are dealing with. What part of the bike schematic is included and what additional stuff are you trying to add. I can't make any sense of your last several posts, put altogether.

Got it working, trial and error.