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LMR62014: SHDN pin does not reduce Vout to 0V

Part Number: LMR62014
Other Parts Discussed in Thread: TPS61178

Hello all,

I am using the LMR62014 as a boost converter to drive a fan in a circuit of mine, but I am unable to get the output Vout to go to 0V by use of the shutdown pin.

Here's my schematic: https://imgur.com/a/lBAeUK5

The two junctions J1 and J2 connect to the rest of my circuit.\

Vin is from a lithium ion battery, and is between 6.5V-8.4V.

EN comes from a microcontroller, and is either HIGH (3.3V) or LOW (pulled to 0V).

You'll notice I have 4 different options for the output voltage set by Ra1-4. Each of them is enabled-disabled with a shunt. For now, assume I am using the 116.5KΩ option for an output of 12V.

When EN is HIGH, the output  behaves as expected and goes to 12V, and powers my fan with no issues.

When EN is LOW, the output does NOT go to 0V as I expected. Instead, it equals Vin. This even happens if I manually short EN to GND with a jumper wire directly. And, this behavior is the same whether or not I have a load connected to Vout.

Am I misunderstanding the usage of the shutdown pin? Is it possible to completely set Vout to 0V in this way? The alternative would be to introduce a MOSFET before Vin in my circuit to enable and disable the voltage source itself, but I was hoping to use the shutdown pin. Let me know, thank you!

  • Hi Uzair,

    When LMR62014 SHDN pin voltage goes to logic low, the device stops switching. It doesn't mean the Vout = 0V. As LMR62014 is a non-synchronous boost converter, when device stops working, the Vin charges the output capacitor through inductor, rectifying diode to Vin. That's why you see the Vout voltage is same with Vin voltage.

    You may need a P-MOSFET before Vin to enable or disable the voltage source. Or use TPS61178, TPS61378 device which have true disconnection function.

    Hope that helps.