LM3671 hysteresis enable/disable

Discussion in 'General Electronics Chat' started by jimbostlawrence, Sep 1, 2012.

  1. jimbostlawrence

    Thread Starter New Member

    Sep 1, 2012
    8
    1
    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
     
  2. #12

    Expert

    Nov 30, 2010
    16,298
    6,811
    You kind of answered your own question. The specifications actually mean, "no mans land" or, "undefined". The enable is guaranteed to go one way if the input is below .4 volts and it is guaranteed to go the other way if the input is above 1 volt. What's in the middle is not dependable, not guaranteed, and not a good place to design a circuit. Betting you can use the undefined input range for a voltage regulator is just wrong.
     
  3. jimbostlawrence

    Thread Starter New Member

    Sep 1, 2012
    8
    1
    Yup, fair enough. I was just told in the past that I should be able to sort this enable issue with a potential divider and that was what I came up with.

    So I'm still stuck as to how to make use of this 0.4/1.0V facility. Surely there is a simple method to make use of this divide in potential to allow a clean cut off of the regulator? I don't want the regulator to oscillate on/off as the battery input voltage comes down...

    Essentially what I'm after is for the battery voltage to come down towards the end of its life, and for the regulator EN pin to be given a shut off, and 'stay off'....

    Any ideas?

    Cheers

    Jimbo
     
  4. #12

    Expert

    Nov 30, 2010
    16,298
    6,811
    You could use a CMOS opamp as a very low power comparator to slap the enable pin high or low but, there are chips dedicated to this job. I think they are called battery monitors.

    This looks like the wrong one but, you get the idea.
     
  5. jimbostlawrence

    Thread Starter New Member

    Sep 1, 2012
    8
    1
    Yeah, figured it'd have to involve a comparator in the end. I know it's not complicated but was just hoping for simple-simple :p Have played with battery monitors before, will try to apply that/op-amps to this situation... cheers.
     
  6. bwack

    Active Member

    Nov 15, 2011
    109
    10
    Hello Jimbostl.

    You are asking for Vout < Vin on a step-down DC-DC converter..I think...
    Looking at the datasheet, you'll find in the electrical characteristics:
    "input voltage (Note 10)":
    "Note 10: V_in = (V_out + V_dropout) to 5.5V for 1.8 V <= V_out <= 3.3V
    where Vdropout = I_load*(R_DSON_PFET + R_INDUCTOR )"
    Say your load and R_Inductor i fairly low, you'll get Vout =~ Vin, So if the battery goes below 3.3V the output does too.
     
  7. jimbostlawrence

    Thread Starter New Member

    Sep 1, 2012
    8
    1
    So this chip won't produce 3.3v out for any input down to 2.7V?
     
  8. #12

    Expert

    Nov 30, 2010
    16,298
    6,811
    Page 17, minimum supply voltage, higher than the output voltage, per current needed.

    Oops. Too big to upload.
     
  9. jimbostlawrence

    Thread Starter New Member

    Sep 1, 2012
    8
    1
    Damn, ok cheers.

    Well, I guess, even at 3.3V minimum, I need something to switch the regulator off as and when the battery output starts to drop...

    Edit:

    I'm using a 3.6V non-rechargeable battery to power the regulator, and the current drain by the main circuit after the regulator is only going to be 10 mA or less.
     
    Last edited: Sep 1, 2012
    bwack likes this.
  10. bwack

    Active Member

    Nov 15, 2011
    109
    10
    You could do that, but also take into account the V_dropout noted above and the the voltage drop in the battery (internal resistance). For all of this you need to specify the load. Is it only the one led you will drive ?

    Or you could go for a buck-boost converter. They allow you to have the fixed output voltage inside the voltage input range.

    If your Li-ion battery doesn't have any protection circuit module on it, it is important that you select a low quiescent current circuit for your low-battery lock out function. Stick to cmos and pay attention to quiescent currents). There might be more ideal all-in-one converters that have this protection circuitry allready, I don't know.
     
  11. jimbostlawrence

    Thread Starter New Member

    Sep 1, 2012
    8
    1
    It's not actually an LED, I was just giving a low load example. The main circuit consists of a PIC, the odd pulsed LED (rare) and a microSD card which is written to sporadically.

    So a buck-boost converter would help to maintain a fixed 3.3V for anything from ~3.6V down to say 2.5V or so?

    Have been looking at http://www.ams.com/eng/Products/Power-Management/DC-DC-Buck-Boost-Converters/AS1331 which I can configure for 3.3V output. Can I assume that from page 1, this will operate at 3.3V output with 1.8V<Vin<5.5V ?

    If that chip (AS1331) is good, I'm still left with the issue of the shutdown pin which will shutdown the device if EN <0.4V or enable it if >1.4V.......

    Beginning to lose the will here but desperately trying to find a solution to all this. Want to get the most out of a battery as possible, but mainly, shut the regulator off when the battery cannot achieve a 3.3V output thus stopping the PIC from corrupting anything on the SD card etc.
     
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