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LM3481: Changing parameter of PMP20183

na na78
Guru 16770 points
Part Number: LM3481
Other Parts Discussed in Thread: PMP20183,

Hi

Our customer refers PMP20183 to design the circuit for following requirement.

V_INPUT : DC 9V

V_OUTPUT : 260V (Max)

I_OUTPUT : 10mA(Max)

According to the requirement, PMP20183 looks to be good reference design while a little change might be needed.

So, what should we modify PMP20183 to meet the requirement?  Could you please tell us your view?

BestRegards

  • 0
    TI__Guru* 78320 points
    Hi na na78,

    I think the ref design can support the requirement. Since you do have lower input and higher output, you may consider to increase the switching frequency by reducing R10 if needed after test.

    Thanks,
    Youhao Xi, System and Applications Engineering, BMC-BCS Product Line
  • 0 in reply to Youhao Xi
    Guru 16770 points

    Hi

    To achieve Vout=~260V,  I think the charge pump circuit should be added.

    I attached sample schematics using charge pump with N=5. (Not tested)

    260/5 = 52V, so we will adjust output of LM5481 for 52V.

    Is this understanding correct?

    Is it is correct,  what should be considered for adding charge pump?

    And switching frequency will be adjusted as you mentioned.

    BestRegards

  • 0 in reply to na na78
    Guru 16770 points

    Hi Youhao

    I want to confirm the minimum off time of LM3481.

    According to datasheet, the maximum duty 81% (min) is characterized with RFA=40Kohm.
    RFA=40K means Fsw=~480KHz.
    In other words, ~400nsec can be calculated as the off time.

    Is it possible to consider the minimum off time as 400 nsec?

    If it is correct, I can understand that you said PMP20183 can support 260Vout without any design change.
    And we don't need have 5 charge pumps. Is it right?

    My understanding is following.

    If 260V output is assumed, duty can be calculated using (260/3) - 9/ (260/3) = 90% duty.
    However, PMP20183 offers 100KHz switching frequency.

    The off time can be calculated as 1/100K * 0.1 = 1usec > 400nsec.

    So, PMP20183 can support Vout = 260V.

    BestRegards

  • 0 in reply to na na78
    TI__Guru* 78320 points

    Hi na na78,

    The max duty cycle is determined by a time ratio circuit and it is basically independent of the switching frequency. So, at 100kHz, the max duty would also be 81%.

    We do have some circuits using the fixed forced off time to limit the max duty, but it is not the case of the LM3481.  Hope this clarifies.

    Thanks,

    Youhao

  • 0 in reply to Youhao Xi
    TI__Guru* 78320 points
    There are some typos in my previous post, and I have corrected it. Please check the reply online. Sorry for the confusion this may cause.
  • 0 in reply to Youhao Xi
    Guru 16770 points

    Hi Youhao

    Thank you for your reply.

    Sorry, I'm still confused.

    1.
    Finally, can PMP20183 as is boost from 9V to 260V?
    (Duty = ((260/3) - 9 )/(260/3) = ~89%)

    If yes, please tell us the reason why it can support requirement regardless of 89% duty.

    2.
    If not supported, could it be the solution by increasing charge pump to N= 5?
    (Duty = ((260/5) - 9)/(260/5) = ~82%)

    BestRegards

  • 0 in reply to na na78
    TI__Guru* 78320 points

    Hi na na 78,

    The duty cycle calculation is correct, but it is only valid when the circuit operates in CCM.  If the circuit is designed in DCM, the required duty cycle is determined by:

    D_DCM = sqrt ( 2x Fs x Lm x Po ) / Vin

    Where Fs is the switching frequency, Lm is the boost inductor, and Po is load power, and Vin is the input voltage.  It shows that smaller the Lm the smaller the duty required.  However,  smaller Lm also leads to higher peak switch current.  This is why in the PMP20183 the designer uses additional charge pump.  I believe the number of charge pump stages in the PMP20183 should be enough. 

    Thanks,

    Youhao

  • 0 in reply to Youhao Xi
    Guru 16770 points
    Hi Youhao

    Thank you for your reply.

    I calculated duty using your equation.
    Could you please check if the following understanding is correct?

    Vin(min)=9V
    Vout=260V
    Iout=10mA
    Fs=100KHz (from PMP20183)
    Lm=68uH (from PMP20183)
    Po = 10mA x 260/3 = ~0.866

    Duty_DCM=sqrt(2 x Fs x Lm x Po) /Vin = 0.38, so PMP20183 should be enough.

    BestRegards
  • 0 in reply to na na78
    Guru 16770 points
    Hi

    I greatly appreciate if you could my last question.

    BestRegards
  • 0 in reply to na na78
    Guru 16770 points
    Hi

    Should I calculate with Po=260V x 10mA = 2.6 ?
    The customer is asking how to calculate Po in PMP20183.

    I believe Po could be calculated by 260V x 10mA, but could you please tell us your calculation to avoid misleading of customer?

    BestRegards
  • 0 in reply to na na78
    TI__Guru* 78320 points
    Hi,

    Yes, you are right: the output power is the output voltage x the output current. However, in this ref design, the Po should be the boost stage output power, which also includes the power losses of the charge pump. As a quick estimate, use 50% as the efficiency, then your Po in that formula would be 5.2W.

    Best Regards,
    Youhao