Has anyone tangled with the B6284 boost converter IC? It has an enable line that is supposed to shut down the device when connected to ground but I can only get it to drop the output from 12 V to 3 V. Seems bizarre for something as simple as a control pin not to work as advertised. Wondering if I'm missing something?
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Well - if you hooked it up in a circuit which resembles the Typical Application on page 1 of the datasheet, then yes you are missing something.
There's no way for the IC to prevent current from flowing from your supply thru L1 & D1, to the output.
The Enable pin is only intended to switch the Boost converter IC on & off, not act as an output switch.
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\$\begingroup\$ Oh yeah! So obvious now you point it out. All my experience to date has been with buck converters where tickling the enable kills the output too. So is there a boost design that does likewise? \$\endgroup\$ Commented Feb 16, 2019 at 3:50
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\$\begingroup\$ You could use a P-channel FET as a switch to cut power to the whole booster circuit. Put the FET between VIN and the booster circuit. Maybe use a pull-up resistor on the gate so it defaults to OFF and then use the enable signal you were going to use on the booster IC to pull the voltage on the gate low to turn the FET to ON and deliver power to the booster. Choose a FET that can handle the current draw and make sure the voltage drop across the FET isn’t a problem. \$\endgroup\$– Joe MacCommented Feb 16, 2019 at 6:24
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\$\begingroup\$ I'm gonna have to find a high side switch that is on when it's enable is hi to match the enable line which is active hi \$\endgroup\$ Commented Feb 16, 2019 at 23:27