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Hi all,

I thought I had this sorted but I was wrong...

I have an LM3671 3.3V regulator which has an EN pin which...

... enables the regulator when presented with >1.0 V ... shutsdown the regulator when presented with <0.4 V

I thought that if I pulsed 3.5V onto the EN pin to switch on the regulator, and also had a potential divider of 100k (lower) and 620k (upper) with 3.5V at the top, with center-tap also going to the EN pin that with the regulator pulsed on, the potential divider would hold the regulator on as it was above the 'off' voltage of 0.4V (at anything above 2.9 V, the potential divider will have a center value of above 0.4 V). What is this 'no man's land' between 0.4 V and 1.0 V ?

To pulse the regulator on initially I have a REED switch which connects the 3.5 V from the battery-in (regulator Vin) to the regulator EN. What I would like is for the regulator to continue to run producing 3.3 V at Vout until the battery voltage drops to about 2.8-2.9 V, and then the EN pin will have 'something' to tell it to shutdown...

Any help would be gratefully received. Was up until 530 am trying to solve this!

I've just drawn the circuit up in CircuitLab (superb app'!!!) and will attach it this thread as soon as I work out how...

Cheers

Jimbo

@jimbostlawrence,

I think I get the hang of what you would like to do but you cannot do it the way you would like.

The enable input has a range of 0.4V to 1V for which the operation of the regulator will not be defined and therefore may be unstable. It may be on or off or oscillate between the two or go into some other state that could even damage it.

To turn it off safely as the input voltage reduces, you will need to use a comparator with some hysteresis to ensure that the EN input snaps cleanly from >1V to <0.4V.

However, there's a second problem with the voltage levels you describe. The LM3671 is a step down regulator which can also operate as a linear regulator for input voltages only very slightly greater than the output voltage. The key here is that the input voltage must be slightly greater than the output for the chip to operate as a regulator.

As soon as the input-output voltage difference drops below Vdropout (referred to in several places in the datasheet but not specified) then the output will fall out of regulation and start to reduce. By the time the input has fallen to 2.9V the output will already be down to just below that voltage.

It may be that you do not need to build your enable with hysteresis because the device already has a built in undervoltage lockout (UVLO) function.

The LM3671 is not well specified in the datasheet. It appears to have an UVLO lockout function (see block diagram) but the voltages at which it operates is not given in the specifications tables. It is referred to in the section on SHUTDOWN MODE at the bottom of page 16 where it implies that the lower turn-off voltage is 2.7V.

You might be able to infer the UVLO behaviour and voltage levels from the TI Webench online simulation tool (but I doubt it ... I've found it to be a pretty useless tool).