• State Resolved
  • Locked Locked
  • Replies 12 replies
  • Answers 1 answer
  • Subscribers 67 subscribers
  • Views 3067 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

SIMPLE SWITCHER negative output?

Tom Waschura
Prodigy 220 points
Other Parts Discussed in Thread: LMZ21701, LMZ14201H, LMZ14203H, LMZ14203

Can I use the TI LMZ14201H type SIMPLE SWITCHER MODULE (or, more generally, all of the SIMPLE SWITCHER MODULES including the high-density Nano Modules --e.g. LMZ21701) in an inverting configuration to create a negative voltage output? 

Thanks,

Tom

  • 0
    TI__Genius 15810 points
    Yes. You can use any of the power modules in an inverting topology.
  • 0 in reply to Akshay Mehta
    Intellectual 670 points
    I am having issues simulating the negative output configuration in TINA as well as LTSpice and Webench. Anyone have a working simulation model? The positive output configuration works just fine in TINA and Webench. LTSpice simulation struggles with that too (using the non-encrypted LIB file in LT and TINA).
  • 0 in reply to Akshay Mehta
    TI__Genius 11085 points
    You can find application notes in the product folder --> Technical documents
    www.ti.com/.../technicaldocuments
  • 0 in reply to Saurabh Tewari
    TI__Genius 15810 points
    The simulations tools forum could better help you out here. Sorry!
  • 0 in reply to Akshay Mehta
    Intellectual 670 points

    Not a problem, thanks!


    I think I have figured it out -- looks like TINA model is sensitive to the pin the ground node is connected to. I got the attached to run in inverting mode simply by changing GND to COM and connecting GND (Node 0) to GND pin on the switcher.

  • 0 in reply to Akshay Mehta
    Prodigy 90 points
    Hi, I have the following question in addition to this;


    When using LMZ14203H in negative output mode,
    If Vin = 12VDC can vout be -15V?
    and what is the current rating in this case?

    if LMZ 14203H is used in possitive mode i read:
    Vin should be at least Vout + 3 VDC, so minimum negative voltage outpus is -3VDC?
  • 0 in reply to Daan Stelma
    Prodigy 220 points

    There are many who know this more than me, but the LMZ14203 is a step-down DC-DC converter.  If you apply 12V you can only create something less than 12V (in fact, they say 9V by saying the Vin should be at least Vout + 3VDC).  It is possible to level-shift the 9V so that it runs from 0 to -9V (instead of the more typical 9V to 0), but you will still only create an output voltage difference of up to 9V.

    If you need to get a larger voltage than what you start with, you'll need a "boost" DC-DC converter.

    Hope this helps...

    Tom

  • 0 in reply to Daan Stelma
    TI__Genius 10095 points
    The LMZ14203H can be use to create 12V to -15V. The design procedure is the similar to designing a 27V to 15V design using the part. Set the output voltage resistors etc for 15V and connect the ground and vout pins as you have shown. This is documented in www.ti.com/lit/an/snva425a/snva425a.pdf
    The current will be derated to a little under 2A because in this configuration the input current and output current run through the switch.

    Regards,
    Marc
  • 0 in reply to Marc Davis-Marsh
    Prodigy 220 points
    Marc,

    Are you sure about this? Independent of inverting the converter's output about ground to get a negative voltage (e.g. the application note you reference), 12V into a Buck Converter and you can get 15V out? Perhaps I'm missing something.

    Tom
  • 0 in reply to Tom Waschura
    TI__Genius 10095 points
    Its not 15V out its negative 15V out. Therefore the device sees 27V (12Vin-(-15Vout) across the device from Vin to the Ground Pin. So you can think of the device as having 27Vin to 15Vout with the whole thing level shifted by 15V. So now ground is -15V, Vout is 0 and Vin is 12V. There is a 12V difference between Vout (0V) and Vin (12V) which is enough to meet the duty cycle limitation you mentioned before of 3V.

    Regards,
    Marc
  • 0 in reply to Marc Davis-Marsh
    Prodigy 90 points

    Thank you for the clear explination.

    So in negative configuration +12V in gets -15V out, @ 2Amp in stead of 3Amp.

    in negative mode maximum operating input voltage will drop from 42V to 42-15V = 27VDC 

    Learn from mistakes:

    In a design i made a mistake, there are used 2 LMZ14203H to create +15V and -15V that supply some high current opamps.

    The circuit is supplied with 12VDC.

    Now the problem is, what was supposed to be +15V = +12V because LMZ14203 output cannot be higher than the input.

    Solution increase the supply voltage.

  • 0 in reply to Marc Davis-Marsh
    Prodigy 220 points
    Marc,

    Thank you. That's very clear and, of course, now makes perfect sense. Sorry for the question, but you taught me something. Tom.