constant voltage, but it only differs from constant current because it feedbacks the voltage from a divider on the output rather than a shunt
Look at the datasheet of the XL6005 at the typical application circuit, notice the how there is a resistor below the LED and the voltage across it goes back to FB. That's a shunt, a resistor to measure a current by converting it to a voltage. The XL6005 compares that voltage to a threshold voltage (0.22V) and if it's too low it starts pulsing the internal MOSFET, which raises the output voltage, which raises the current through the diode and shunt, which causes the the XL6005 to use shorter or no pulses.
Now look at the datasheet of a constant voltage boost converter, lets say the XL6009. Instead of the shunt there is now a resistor divider across the output, which leads to FB.
The major difference between XL6005 and XL6009 is the threshold voltage, because the XL6005 is designed to develop that voltage across a shunt which sees a large current it requires a low threshold voltage, 0.22V. The XL6009 uses 1.25V because it gets feedback from the large resistor divider.
Regarding the video, i'm pretty sure that doesn't deliver constant current because the voltage changes, and if it does deliver constant current, how would I make it so it always delivers a max of 1a, no matter the voltage?
It was more to give you a general idea on how to construct a circuit around XL6005 if you want. These would be the changes you would need to make relative to the circuit he made :
- no need for the compensation network, the XL6005 has it build in.
- no need for the feedback divider, instead you need a shunt as explained above of 0.22 Ohm (0.22V/1A) ... making it adjustable would be slightly awkward, potentiometers at such low resistance get expensive.
- inductor is a little too small, needs to be around 22 uH (and needs to able to stand a DC current of around 1.5A)
PS. the supplies I linked earlier might actually have a problem ... the ebay page warns the constant current limiting stops working below Vin = 7V, I assume it's because of the opamp/comparator they used.