LM3671 hysteresis enable/disable

[jimbostlawrence]

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!

Cheers

Jimbo

[jimbostlawrence]

Circuit layout for above issue...

[pebe]

I assume you tried out your circuit but it didn't work.

But you didn't say whether it wouldn't switch on, or whether it wouldn't switch off.

[jimbostlawrence]

It would switch on when a magnet was brought near to the reed switch. Potential at the EN pin correctly jumps to about 3.5V and the LED switches on, however, when the magnet is removed, the resistor divider has a potential of about 0.5V and the LED switches off. I was hoping that as the potential doesn't drop below 0.4V the regulator would remain in an on-state. Naive.... Any ideas on how I can correctly plumb in some components to make it do what I'm after?

[pebe]

Your Idea should work OK. The reason for the 'deadband' between 0.4V and 1V is because of production spreads in manufacturer. The way it is quoted means that the manufacturer guarantees that all devices will switch off below 0.4V and will switch on above 1V. But the actual changeover voltage of a single item can be anything between those two figures.

The best way would be to measure that voltage of your device with a variable pot instead of the fixed resistors. Adjust it so the device is on, then lower the voltage till it just turns off, and measure the voltage on the Enable pin. Then you can replace the pot with suitable value fixed resistors.

[jimbostlawrence]

Ah ok, cheers, it makes sense what you're saying about spread parameters due to manufacturers etc. I posed the same issue ib abither forum http://forum.allaboutcircuits.com/showthread.php?t=73927&page=2 and someone rightly pointed out that this chip is a down converter and so the output will track down with Vin once it falls below 3.3V for Vout3.3V. I've found a buck boost chip (AMS AS1331 http://www.ams.com/eng/Products/Power-Management/DC-DC-Buck-Boost-Converters/AS1331 that might do the trick with regards to producing 3.3V out, for anything as low as Vin2V or so, but still left with the issue of the enable and getting it to shut off and stay off once a minimum Vin is reached by the depleting battery, as the battery is the only supply to the circuit and so using it to shut itself off is a weird one for me.

[pebe]

It appears from Fig 1 in the datasheet for the AS1331 that you can use the LBI pin to control the device. LBI threshold is accurately defined (1.25V±3%) so it is easy to set R4 & R5 to give that voltage for the lowest battery voltage.

But looking through the data sheet I cannot see any mention of switching off the output if LBI goes below threshold. From Note 3 that follows Table 3, only LBO is affected.

I found the description on page 11 very vague when it described the ‘Power OK’, especially when it talks of LBO but does not show it in the block diagram of Figure 23.

So if you are getting the chip you may need to experiment. But if I were doing it I would try this:
1. Use the potential divider from Vin so LBI operates to measure the battery voltage, as in the datasheet. LBI going low would make LBO low.
2. Connect the LBO and Enable pins together, with a common pullup. That would take Enable low and switch off the chip when LBI went low.

That way, the chip would revert to normal operation when a new battery was fitted.