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I am using an LM2577 as a step-up boost converter (to boost 7V to 12V).

I would like to be able to turn it on/off (there are certain times when it needs to be turned off).

I have tried a transistor (BC516, BC517, 547B):

  1. on Vin pin: only get 1V instead of 7V so the converter only outputs 2V
  2. on Switch pin: converter only outputs 6V instead of the 12V I want.
  3. on GND pin: transistor overheats
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    \$\begingroup\$ Matt, it's always a good idea to post a link to the datasheets of the parts in question. It saves time and clicks. Folks out here appreciate it. \$\endgroup\$ Feb 27, 2014 at 21:36

1 Answer 1

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The problem here is that it is a boost. So, if you were to just turn off the LM2577 somehow, there would still be Vin on the output. You will need to use a P-channel FET to turn off the input voltage to the whole boost circuit.

Something like this could work.

enter image description here

As long as Vin is more than 5V and less than 10V the FET should switch correctly. Of course, now you have to explicitly turn the boost circuit on by applying ~5V to "On" .

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  • \$\begingroup\$ Excellent point. \$\endgroup\$ Feb 27, 2014 at 21:58
  • \$\begingroup\$ Yup good answer. I was thinking/musing to raise the feedback pin a volt (or so) by injecting a dc level but, ya still got Vin coming thru the inductor and diode. +1 \$\endgroup\$
    – Andy aka
    Feb 28, 2014 at 0:21
  • \$\begingroup\$ Why P-channel FET? is it important? \$\endgroup\$
    – Roh
    Feb 28, 2014 at 4:15
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    \$\begingroup\$ @Roh P-channel FET is the easiest, but a pnp BJT would work too. \$\endgroup\$
    – gsills
    Feb 28, 2014 at 20:06
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    \$\begingroup\$ @Roh: Since this is high-side switching, a P-type device is easy mode. \$\endgroup\$ Feb 28, 2014 at 21:51

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