integrated-circuitlevel-shiftingvoltage
I connected HC-SR04 ultrasonic sensor, to single-board computer BeagleBoard-xM. Trigger pin requires 5 V DC. However, control pins of BeagleBoard-xM can provide only 1.8 V DC.
What is the best way to convert voltage to 5 V? Is there anything else I should pay attention to?
For your wide range inputs you could consider a solution that I have used a number of times. Bring your inputs in through a circuit like this:
As shown the input threshold for high/low is set at 2.5V by use of the TLV431. The input signal coming through the 2.2K series resistors are clamped to the 5V supply to keep them in safe input range of the AM26LV32. Outputs swing from rail to rail of 0 to 3.3V for input to the FPGA.
A VL6180 board must have a stable 2.8V power supply (locally, at least the Sparkfun board has it), otherwise the sensor will have goofy unstable results. You need to use this rail to feed the low side of your interrupt level shifter instead of weak voltage divider.
Or you can try a diode in series to drop the VL_IN voltage from 3.3 into 2.6-3.0 area instead of resistive divider (replace your 100k with a 1N4148 diode).
And you will need a pull-up to 5V on the high-side of level shifter.
ADDITION: per suggestion of brhans, the min input level for ATMEGA32U4 is 0.2Vcc+0.9V = 1.9V, while even a fully loaded (8 mA) interrupt output from VL6180 should produce 2.4V, so no need in the level translator at all.
Best Answer
First of all, according to the datasheet for the HC-SR04 that you linked to, it appears to run on 5v, not 3.3v.
What you need is a level shifter, to convert the I/O signals on the BeagleBoard-xM expansion header from 1.8v to the 5v levels (and back again) as required by the sensor.
A device that will perform this function is the Logic Level Converter from SparkFun. It has 4 channels, which is more than you need. You could build your own with just two channels, but for the price ($3) its not worth it.