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I'm making a portable device which is powered by a lithium battery and boosted with a FP6293 boost converter IC. I want to turn on and off the device from a switch that activates and deactivates the FP6293, I know I can do this from the EN pin, but the documentation is not really clear in that aspect:

enter image description here

Following the application circuit, the EN pin is directly connected to the power input (VIN), so I guess the boost IC activates when Vin enters the circuit. Can I put a switch between the Vcc and EN pin to activate and the activate the IC to I can control the shutdown and power up of the device? I don't know if I should put a resistor, connect it to the ground, the documentation is really poor in that aspect...

Here is the datasheet I'm using: https://datasheet.lcsc.com/szlcsc/Feeling-Tech-FP6293XR-G1_C83576.pdf

Thanks for your attention.

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  • 2
    \$\begingroup\$ My golden rule is that chips with poor documentation don't get used in a design unless there are no feasible alternatives..... \$\endgroup\$ – Andy aka Sep 11 at 11:15
  • \$\begingroup\$ Can I put a switch between the Vcc and EN pin to activate and the activate the IC to I can control the shutdown and power up of the device? Probably yes, a pull down resistor (like 100 kohm) and a switch to Vcc might also work. But since you need a toggle switch (EN does not work like push-on, push-off) you could just put a toggle switch in series with Vcc. \$\endgroup\$ – Bimpelrekkie Sep 11 at 11:30
  • \$\begingroup\$ Following your answer, I put a switch just in front of the EN pin like this: i.snipboard.io/4a6rJM.jpg So I can leave the EN pin open and when I want to activate the circuit I push the switch. \$\endgroup\$ – Electrical Voyager Sep 11 at 11:53
  • \$\begingroup\$ What controls the voltage level on the EN input with the switch open? \$\endgroup\$ – Andy aka Sep 11 at 12:01
  • \$\begingroup\$ When the switch is open EN is connected to ground, so there's no input until the switch is down and feeds EN with the input voltage, which I guess activates the circuit. \$\endgroup\$ – Electrical Voyager Sep 11 at 12:42

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