LM5006 no-load current

Hi Dave,

Since you are using the minimum ripple configuration, changing inductor should not otherwise affect the circuit operation. What are you input/output currents/voltages? What frequency are you operating at? Are you using the synchronous FET or diode?

Regards,

Vijay

Hello Vijay,

  My output is 5V, 500mA (max).  I believe I am operating at 200KHz.  I am using the synchronous FET.

  My input voltage is anywhere between 6V and 60V.

  If changing the inductor will not have much effect, are there any other options for reducing the current draw?

Thanks,

Dave

How much input current are you seeing at no load? Does your application allow DCM (lower frequency at light load) operation? If it does then removing the FET and puting in a diode should reduce the no load current.

Increasing inductance can also help by reducing the peak currents etc.

Hi Vijay,

  I do not have the board in front of me at the moment but I believe I was seeing 4-5mA.  I will confirm this later today.

  Is DCM automatically enabled by using a diode instead of the FET?  My only fear for this is the loss of normal efficiency when there is a load.  I will have to measure the difference and see if it is acceptable.

  Do you have an approximate value of how much the current is reduced by using DCM?  I can, of course, measure this as well.

Thanks,

Dave

Great - I will try the diode and see if the DCM current draw is worth whatever I lose in normal efficiency.

Is there any way to get the best of both worlds?  The reduced light-load current and higher normal efficiency?  Would it help to reduce the frequency (and then increase the inductor to match)?

Thanks,

Dave

Vijay,

I changed the FET to a diode and the current did indeed drop - below 2mA, in fact. 

At this point, the dominant current draw is the 0.8mA passing through the feedback resistors.  I saw your recommendation of ~1mA through the resistors, but I could not find any mention of it in the datasheet.  Could you point me to the right place, or give me some more background on that number?  Every 100uA that I can save goes a long way!

Thanks,

Dave

Vijay,

  Thanks for the quick response.  The caveat in the LM5007 datasheet is very clear - maintain a 1mA load current, or the bootstrap cap will discharge and the regulator will shutdown, at least temporarily.  The LM5006 and LM5007 do seem like very similar devices, is there any way to confirm whether LM5006 needs the minimum load current as well, but it was accidentally left out of the datasheet?

  What would be the best way to confirm small-load operation of the LM5006?  Monitor the bootstrap capacitor to make sure it doesn't discharge?  Simply monitor the output voltage to make sure it stays at 5V?

  Interestingly - I'll never have an actual load below 1mA.  My load is much larger than 1mA, but it is removeable - so the load seen by the regulator can drop to just the FB resistors, but I don't particularly care what the regulator does in this situation, as long as it doesn't draw excessive current or get stuck in some bad state.  If there is a minimum load to maintain normal operation, and I do not meet the minimum load and the regulator shuts down or cycles back and forth - will the regulator recover when the load later rises above the minimum?  Or could it get stuck forever (well, until the input voltage is removed)?

Thanks,

Dave

Vijay,

  Great!  That document is very clear and I do think that the loss of regulation (which is recovered when the load becomes big enough) should be acceptable for my application.

  One minor question - the document references the "LM25000 series" of regulators.  The LM5006/7/8 are part of this family?

  Since the synchronous rectifier option available in the LM5006 does not use DCM, would you expect the minimum current issue to not apply if using the synchronous rectifier?  Since the small-load issue occurs due to the reduced DCM frequency, perhaps there is no particular minimum load requirement because the synchronous rectifier keeps you out of DCM?

Thanks,

Dave

Vijay,

  Great!  So, to summarize:

• With the FET, I can reduce the feedback current much lower (as long as I can provide the current needed by the FB pin).
• With the diode, I can reduce the feedback current, but the at some point the regulator may enter a "hiccup" mode and the output will cycle between ~2-5V.  However, this mode is not harmful (to the regulator) or permanent and normal operation is guaranteed to resume as soon as a larger load is connected.
• The reduced feedback current will not affect regulation during normal operation, when a larger external load is connected.

Sound correct?