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LM5022: How calculate values for L1 and Rsns for a 20V-28V input to 180V 1mA Step-Up with minimal Iin

Bert Hennink24
Intellectual 335 points
Part Number: LM5022

I have to adapt a circuit previous designed by a former colleague . And there is no information how he calcuated the design.

I have a new requirement that the input current for this stepup converter should be as low as possible. It is driven from a current limitted source which can drive about 35- 40mA max.

I refer for the components to the designators in the datasheet.

I when calculating values using the equations, the result is that L1 should be about 8mH.

But the circuit does not reach the required 180V with such an inductor.

When I lower the inductor empirical to the intial values (in the range of 220uH to 680uH) and Rsns=2.2ohm from the design as it was implemented before the output voltage rises to the desired value and the current is below 35 mA so that would be good for my application.

But I want to understand why I need to use that smaller inductor value.

So I get it runnin, but I want to be able to calculate the optimum L1 and Rsns.

Greets

Bert

  • 0
    TI__Mastermind 40040 points

    Hello Bert,

     

    Without a schematic it is difficult to say exactly the cause?  But the inductor is a key component for stability and especially an inductor that is very large, this is due to the Right Half Plane Zero (RHPZ).  A right half plane zero gives you gain in the loop of 20dB per Decade and a phase drop of -45 degrees at the RHPZ frequency. This is very undesirable to loop stability because the phase shift will degrade your phase margin at the cross over frequency needed for stability (more details in the datasheet). It is possible that you are unstable with the original inductor value and when you reduce the inductor value the RHPZ frequency increases giving you stability.

    The procedure for calculating the inductor and Rsense value is also given in the Datasheet. 

    David.