LMZ10505 regulation under load

I am using the LMZ10505 adjustable regulator to generate +1.0V from a 5V input.  I've used the Table 2 in the datasheet to calculate the feedback resistors in order to set the output to 1.0V.  I chose 82.5K,1% and 330K,1% as the feedback resistor values so that Vout = 0.8V (1 + 82.5/330) = 1.00V.

I used all the other suggestions from the table as well for the other discrete components in the circuit and followed the layout design guide down to the letter. My concern is that when the regulator has no load it regulates the 1.00V but when we place the regulator under load (around 4A) the output drops  by approximately 30mV.

This 30mV drop under load seems too much of a drop.


Do you have any ideas as to what might be going on?


Is it possible that with 82.5K & 330K magnitude value, we do not have enough bias current for the feedback input pin? What values does National recommend for generating 1.0V (i.e. the lower output voltages)? 

In advance, thank you for helping.


  • The load regulation will depend on where the output voltage is sensed. A trace is then routed from the desired sense point to the ffedback resistor divider.

    If the feedback trace is routed to the point you want to regulate the load regulation will look closer to  the following which was measure on the eval board.  This has only a 10mV drop at 4Amps

    However if the trace to the top of the feedback divider is sensed far from the load you will have an IR drop based upon the resistance of the load traces multiplied by the load current. Since the load current is 4 amps. This means that you probably have around 5mOhms of trace resistance between your sense point and your load.





  • In reply to Marc Davis-Marsh:


       This is great feedback!  Thank you for sharing this information.



  • In reply to TIguru:


    Just so that I understand exactly what you are saying, let me elaborate some more on what we are doing. I read and understood your reply. Nonetheless, we followed the layout guidelines to the letter….

    1) Minimize area of switched current loops

    2) Have a single point ground

    3) Minimize trace length to the FB pin.

    4) Make input and output bus connections as wide as possible

    5) Provide adequate device heat sinking.

     The output of the regulator goes directly to the output cap and then (using multiple vias inside a copper pour) transitions directly to the power plane.

     And none of this would explain the 30mV drop we are seeing….the FB resistors / path is very short and placed / routed right next to the FB input pin……everything is extremely localized……and our current load is 2.8A at the moment.

     Are you saying that the top sense resistor needs to be connected to the feed point for the actual load?  If we are connected to a power plane is it feasible that the voltage drop we are experiencing is due to the power plane's resistance?


  • In reply to TIguru:

    It sounds like you are doing everything right.

    If your feedback trace is tied right at your output capacitor.  You would still have the following losses in your system.

    A typical 12mil via is around 1.5mOhms. 1oz copper is around 0.5mOhms / per square. So if we assume two squares for the plane (rectangular board) that is 1 mOhm.  Add this together and multiply for 2 for the return path and you get around 4mohms.

    For a place to start, I would look at your grounding.  This is the easiest place to introduce error. 


    I think for this level of detail it will probably be easier for me to take a look at your layout.  If you don't want to show it on a forum, you can send it to me directly.