0
\$\begingroup\$

I'm trying to use a XL6009e1 based module with an arduino to control whether the input voltage, or the boosted voltage is sent on output.

As per the XL6009 Datasheet or XL6019 Datasheet, the ENable pin 2 is tied/floating high with VIN. For the chip to be disabled, it needs to be 0.8v or below, or 1.4v or higher to be enabled (default).

Currently I have one of these DC-DC modules, and have disconnected the second pin from the board.

Here's my question - With a 12v input on VIN, what exactly do I need to do so I can control the output by the enable pin by setting an arduino pin high or low? Obviously I can't connect Enable pin 2 directly to a GPIO since it's outputting 12v, and will fry something on the arduino. But where I'm quite confused is what exactly I can use, and secondly how I'd connect it up.

Even if I didn't have the Arduino, how exactly would I pull this pin "low"? I thought perhaps connecting directly to GND but I feel like I'm confusing something else here. I imagine using a pull down resistor, but how exactly would this be connected?

To be clear, I want to bypass the regulator and have the 12V input flow to the output when the regulator is off.

\$\endgroup\$

1 Answer 1

0
\$\begingroup\$

Here's my question - With a 12v input on VIN, what exactly do I need to do so I can control the output by the enable pin by setting an arduino pin high or low?

Use an NPN BJT like this: -

enter image description here

A 10 kΩ resistor will be fine.

\$\endgroup\$
6
  • \$\begingroup\$ Hey thanks. I just realized I failed to mention when the enable pin is LOW, I want 12v to flow through to output. I got the impression from this thread that this is what should occur when voltage is low. So i.e. desired behavior going between 12V and boosted voltage (14V) controlled via Arduino. I've tried a 2N2222 NPN BJT with a 10k resistor to the Arduino, but I'm finding when Enable is low (and being tied to GND) I've got no voltage on output. \$\endgroup\$
    – Matt
    Nov 16, 2020 at 21:15
  • \$\begingroup\$ 12 volts will flow through to the output when the chip is disabled. Look at the circuit and tell me what can prevent this happening. \$\endgroup\$
    – Andy aka
    Nov 16, 2020 at 22:28
  • \$\begingroup\$ I'm certainly pretty stumped. I just realized I linked to the module I'm waiting for in the post. The actual module I have right now is DSN6000AUD. While the layout is slightly different, I can't understand why VIN doesn't parse through with EN low. \$\endgroup\$
    – Matt
    Nov 16, 2020 at 23:41
  • \$\begingroup\$ Well, your question is pretty clear and if in fact you meant to ask about something else that doesn't appear to have a data sheet or schematic then that's an impossibility for me to foresee. You appear to have failed to mention one thing and then linked to the wrong module. Not a very effective start on this site. I suggest you ask a new question about the module you have just mentioned but only if you can find a data sheet for it or a schematic. If you can't find any data sheet or schematic then nobody's going to be able to answer. \$\endgroup\$
    – Andy aka
    Nov 17, 2020 at 8:18
  • \$\begingroup\$ @andy it seems this is the proper schematic image.easyeda.com/histories/… from easyeda.com/johngr/xl6009-buck-boost-converter or similar circuitdigest.com/electronic-circuits/…. There is an inline capacitor between two of the three inductors, wouldn't that block the flow? \$\endgroup\$
    – Passerby
    Feb 9, 2022 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.