digital-logicgpioraspberry pi
I'm sitting here with my Raspberry PI and some other equipment that requires digital signal (high/low, enabled/disabled).
But the Pi only has a GPIO so to get it to work I would need a microchip of some sort (guessing)?
I'm fairly new to all this, but like to play around on my breadboard.
Could anyone point me in the right direction here? I need the PI to control a device that operates with digital high/low signal but also "enabled" / "disabled"
I haven't seen the datasheet(*) of the Raspberry' BCM2835 SoC, but I would be surprised if it didn't have digital I/O. I guess the I/Os are configurable for either analog or digital, and input or output.
edit
This document describes the Raspberry/BCM2835 peripherals; on page 90 ff it describes how to configure I/Os for either input, output or one of up to 6(!) alternate functions.
(*) Apparently, Broadcom doesn't really want us to see the BCM2835 datasheet, since we have to a fill in an information request form to get at it. Not for me.
The best way to do this would be to use a transistor as a comparator to make the transition sharp.
Here is an example circuit:
It uses the LDR as the upper part of a voltage divider. When the LDR resistance drops the voltage at the transistor base rises and turns it on. The transistor can be any general purpose NPN.
We can calculate the resistor value based on whereabouts we want the turn on to happen.
Let's say the LDR resistances goes from 200Ω (dark) to 10kΩ (dark). We want the transistor to turn on when the LDR is at 5kΩ. The supply (V+) is at 3.3V. A typical NPN transistor turns on at around 0.7V, so if we do:
5,000 * (0.7 / 3.3) = 1060Ω needed for the base resistor. We can pick a 1kΩ resistor since it's near enough. Adjust your values to suit your turn on point.
Here is a simulation of the circuit:
The horizontal axis is the LDR resistance, and the blue line is the voltage at the Vout point (You connect this to the Rpi input pin - must be set to input. You can add a 1kΩ resistor between Vout and the Rpi pin to protect it in case of accidentally setting it to output) We can see the transistor turns on at around 5kΩ as predicted (won't be exact as the transistor base-emitter voltage will vary with temperature, etc but near enough for your purposes)
Note that the transistor output is low when it's light and high when it's dark, you can swap the LDR and resistor around and use 5,000 * (3.3 / 0.7) = 23.5kΩ for the resistor if you want it the other way round - this is actually a better configuration as it draws less current (due to higher resistances) so if that's important use this version.
Best Answer
GPIO means General Purpose I/O, and that's a digital input or output (though you often can assign an alternative function to it, like I2C signals, or ADC input). Digital outputs can control other devices by using high (logical "1") and low (logical "0") levels.
"enabled/disabled", both with the "d" at the end are terms which usually describe a status, and that will normally be an output of the device.
If it's indeed an input (which usually would be named "enable/disable", without the "d") then it's likely also a logic level input, which you can control with a second GPIO pin.