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LM2679: Low effiency

Christian Mackin
Prodigy 40 points
Part Number: LM2679

We are only achieving 86% with an input voltage of 24V and an output of 12V. Worse yet, with the same input voltage we get even worse efficiency at 18V, approximately 83%. 

Is there a reference design that matches the efficientcy the data sheets list? 

  • 0
    TI__Expert 3400 points

    Hello Christian,

    For your criteria From Ray: 18-24Vin to 12Vout; 3.6A output current

    I would recommend first to use WEBENCH for your initial design support. I optimized with efficiency, but you can choose cost or size as well.

    https://webench.ti.com/power-designer/switching-regulator/customize/23?noparams=0

    In the meantime, can you please share your schematic and PCB layout? I will look into the reference design used to achieve the stated efficiency metrics.

    Best,

    Ryan

  • 0
    TI__Mastermind 34070 points

    Hello Christian,

    Assuming you are running higher currents at 12V out the efficiency seems quite low.  something must be getting quite hot with those losses.  Check your Diode you have selected, make sure its a schottky with Low forward voltage drop, also for a 12V output the inductance will be quite high and will require a high sat current, this means that the DCR could be high if you have opted for a smaller size, check the DCR and the losses incurred as a result.  

    If this does not resolve your issues please feel free to reopen thread providing more detail you design specification the current you were at when you measured the low efficiency and  components used, thanks.

    David.

  • 0
    Prodigy 40 points

    Here is a link to our files:  Google Drive Folder Link

    We are specifically converting from a maximum current of 1.8A at 24 VDC to 12VDC output for testing purposes. We are using a potentiometer to adjust the output voltage from 18VDC to 7.2. We are mainly concerned with achieving 40W output at 12VDC, considering that we can only pull 1.8A at 24VDC (43W). 40W/43W=93% efficiency wanted at 12VDC.

  • 0 in reply to Christian Mackin
    TI__Mastermind 34070 points

    Hello Christian, 

    I don't think you are going to make 93% total efficiency with the implementation you have.  Suggest a synch buck controller for highest efficiency. 

    Non sync Bucks are typically not as efficiency as sync Buck, you have a rectifying diode with a forward Voltage drop of ~0.5V in addition to that you have a diode in series with the load.  your POUT is 24W and you are losing 1W on the output diode which is already a 4% hit on your efficiency at 2A of output current.  you are then asking for only an additional 2% of loss for the power stage that includes the Diode, the internal MOSFT and the losses in the inductor (DCR losses and Core losses).  

    According to the DS for a 12V output from a 24V in you can expect 95% efficiency, assuming all things being equal, you can expect an efficiency drop to 91% with the series output diode. 

    As Ryan mentioned, use Webench for your initial design and then subtract 4% from the number it gives you.  Sugest you use the components Webench recommends.  Hope this helps.

    David.

  • 0 in reply to David Baba
    Prodigy 40 points

    93% was my ideal, but 91% is acceptable. Are you saying that if my electrical design team follows your recommendations that we should be able to achieve 91%?

  • 0 in reply to Christian Mackin
    TI__Mastermind 34070 points

    See link to WEBENCH design yielding higher efficiency (edited).  Hope this helps.

    https://webench.ti.com/appinfo/webench/scripts/SDP.cgi?ID=0BB97C6726083300

    David.