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I am wanting to know if there is a simpler way to enable/disable a switching voltage regulator I'm using. The voltage regulator's datasheet says to pull the EN pin to GND to turn it off, and let it float to turn it on.

The DC jack is being used as the on/off switch in this product. The middle pin is the shunt, and is normally contacting the sleeve/GND when the charger isn't plugged in. When the charger is plugged into the jack, the shunt disconnects from the sleeve, and the NPN transistor base is pulled high through a resistor and turns on, turning off the voltage regulator.

When the charger is unplugged, the shunt pin connects back to GND, pulling the NPN's base to GND and turning the voltage regulator on.

I feel like I've overlooked something and there is a more simple way? Thank you for your help.

The product, charger, and jack are center pin positive.

VCC is a 12V battery.

Schematic:

enter image description here

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  • \$\begingroup\$ Have you tried looking for a regulator that has the behavior you desire? \$\endgroup\$ Commented Mar 23, 2017 at 5:27
  • \$\begingroup\$ I know there are regulators that you pull up high to turn them off, but the regulator I have chosen has been working perfectly in my design and I don't really want to change it. It is inexpensive and has a low count BOM. \$\endgroup\$
    – klcjr89
    Commented Mar 23, 2017 at 5:29
  • \$\begingroup\$ Not an answer to the question, but have you confirmed with the datasheet that EN is in a well-defined state when Q1 is off? \$\endgroup\$
    – user133493
    Commented Mar 23, 2017 at 5:31
  • \$\begingroup\$ @replete yes, and in the real world confirmed on my test PCB. \$\endgroup\$
    – klcjr89
    Commented Mar 23, 2017 at 5:31

2 Answers 2

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Simpler than what you have? Without changing the regulator? No. A NPN plus resistor is as simple as you can get. The other option would be a voltage comparator, and a diode as a reverse protection and isolating node when the charger is connected. Which your battery may not like.

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  • \$\begingroup\$ I was looking for a pre-biased transistor to reduce the component count, but that wouldn't work as they don't give you a pin after the base resistor. \$\endgroup\$
    – klcjr89
    Commented Mar 23, 2017 at 5:44
  • \$\begingroup\$ Have you seen any pre-biased transistors what I'm looking for? \$\endgroup\$
    – klcjr89
    Commented Mar 23, 2017 at 5:50
  • \$\begingroup\$ There are transistors with built in base resistors, but you want a pull-up and a pull-down. I doubt it's cheaper than an individual resistor. \$\endgroup\$
    – Passerby
    Commented Mar 23, 2017 at 6:03
  • \$\begingroup\$ Would this work? mouser.com/ds/2/308/DTC143E-D-278514.pdf I have some on hand. There isn't a pin at the two resistors junction though; would that matter? \$\endgroup\$
    – klcjr89
    Commented Mar 23, 2017 at 6:08
  • 1
    \$\begingroup\$ If you can find an optocoupler with a built in led resistor that would be a four pin single part solution. Make sure it's sized for 12V operation. \$\endgroup\$
    – Passerby
    Commented Mar 23, 2017 at 6:28
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It's not 103% obvious whether the break jack acts as an on/off switch to other components as well but, if not.

You could reverse the jack polarity if that's acceptable.
The break contact jack is now in the +ve lead.
See diagram below. enter image description here

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