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Original question:

PMP8740

PMP8740

anjana salgaonkar
Expert 1460 points
Other Parts Discussed in Thread: PMP8740

Hi Roberto,

1)Please explain how do you calculate values for L2 and Cin in PMP8740. L2 calculation is not given anywhere and if i calculate Cin as per application notes and datasheet of UCC28950 it gives different value. Please help.

2)can you share dc-dc main transformer details like no of turns of primary and secondary and wire gauge of both windings?, so that i can compare with my design.

3) In PMP8740

            a) why Iout signal used in voltage loop? Iout used in current loop as well as in voltage loop.

            b) after going through white paper i have one doubt why series and parallel combination is used between output connector and Vout at output capacitor?

Anjana 

  • 0
    TI__Expert 5305 points

    Hi Anjana,

    Answer 1:

    I believe you refer to L2 and C18 of the DC/DC stage, corect? In theory L2 is not necessary if the PFC and DC/DC boards are very close to each other (maybe few cm).

    I prefer to add L2 because in my prototype, I need ~ 20cm cable to connect the two main power boards. Without L2 a high-frequency current will flow through the cable, generating noise and worsening EMI. I calculated C18 by considering less than 0.5% peak-peak ripple voltage on it.

    The current waveform through this capacitor is close to a square wave. The worst case happens at maximum Vin, which is 420V: here the duty cycle is ~ 80% and the period of time the capacitor deliver full current to the bridge (which is 5A) is 1usec (please consider that on this capacitor we have double the frequency of the bridge, therefore 200 KHz). If you calculate Ic = C * dV/dT, where C = C18, Ic is 5A and dV is 0.5% of Vin (= 0.005 * 420 = 2.1V peak-peak) and dT = 1 usec, we can calculate C18 = 2.38 uF. I selected 10 uF to have more margin. This capacitor has also to withstand the RMS current of the full bridge, which is ~ 2.8A; therefore, best practice is to select it as polypropylene film capacitor.

    Now with 10uF capacitor we can calculate the peak-peak voltage ripple = "Ripple" = 0.5V pk-pk. The waveform of this ripple voltage is very close to a saw-tooth (also because the duty cycle is 80%). According to Fourier transform, the first harmonic (let's call it VR1, which is sinusoidal) has peak value = VR1 = (2 * (Ripple/2)) / π = 159mV peak.

    I consider only the first harmonic because all others are filtered even more by L2 thanks to the higher harmonic order.

    If the impedance of the output of PCF stage is low enough, we can consider that this ripple will be present on L2.

    Now I calculated L2 value by limiting the 200KHz ripple current below 1% of the input current of DC/DC stage (which is 5A, therefore Iripple < 50mA).

    The inductor L2 can now be calculated by the formula 2 * π * F * L2 > VR1 / Iripple, therefore L2 > VR1 / (2 * π * F * Iripple) = 2.53 uH.

    I selected the closest standard value I found, which is 2.3uH from Wuerth.

    Answer 2:

    Main DC/DC Transformer: I don't have the internals structure of the transformer, because the manufacturer holds it (they made the design and provided us with free samples). BTW I measured the diameter of the wires and we have Np = 19 turns, 1.6mm diameter (LITZ wire); Ns1 = Ns2 = 2 turns 6mm diameter (LITZ wire).

    Answer 3:

    a) Iout is used also in voltage loop to reduce the output impedance of the power supply, which helps when paralleling several unit.

    Nevertheless this resistor was not used in the final product, so you can remove it.

    b) Actually I don't understand the question here. Can you please explain it better?

    Best regards,

    Roberto

  • 0 in reply to Roberto S.
    Expert 1460 points

    Hi Roberto,

    Thank you so much clearing my doubts.

    regarding  to my question...

    1) when I am reffering 2KW parallelable power supply module document I got concept and clears about reverse polarity protection , but when I am looking to the PMP8740 circuit  in that there are Q12, Q17, Q20, Q25-Q31 MOSfets are connected between vout and Output connector J8, my doubt is why those MOSfets bank is used there. as per my understanding at normal operation those mosfets should be ON so that minimum resistance(neglisible resistance) will be come in the path of battery... and for that at the time of power supply turns ON, microcontroller first check for reverse connection of battery if battery connection is OK then only all MOSFETS get turn ON. Please correct if I am wrong.

    2) my another question is how do you calculate shim inductance (Ls) because there are two different formulas given in Application note SLUA560C and 2KW parallelable power supply module document, both are giving me negative results when I am calculating it for PMP8740 design. please explain how do you select 300uH of Shim inductance?

    Thanks again,

    Anjana

  • 0 in reply to anjana salgaonkar
    TI__Expert 5305 points
    Hi Anjana,
    Regarding question #1 you are right: these FETs should run always in ON condition, because they are placed in series with the battery.
    Their function is to avoid sparks on connector contacts, when the battery is connected. Without these FETs, the battery would try to charge the output capacitance of the power supply in "zero time", therefore generating sparks. Also these FETs are needed to provide reverse polarity connection: this is done by the microcontroller, watching Vout, as you already pointed out.
    Question #2: If my calculation (in the module document) and the SLUA560C result in a negative shim inductor, it means you don't need to add any shim inductor. If you add it anyway, you will get better efficiency at light and medium load, but slightly worse efficiency at full load.
    Regards,
    Roberto
  • 0 in reply to Roberto S.
    Expert 1460 points

    Dear Roberto,

    Thanks for your reply.

    can you please explain how do you calculate shim inductance and output capacitance of DC-DC converter? I am not getting any value closed to values used in PMP8740. I am considering system specifications as below. Please correct me if wrong.

  • 0 in reply to anjana salgaonkar
    TI__Expert 5305 points

    Hi Anjana,

    Regarding the shim inductor, I calculated it (in my white paper) by fixing a minimum load (in my case 15%) at which the converter has to meet zero voltage switching. This way the shim inductor is 11.3uH, and I selected 10uH. The excel sheet, based to the application note SLUA560c, calculates it in order to achieve ZVS down to 50% and not 15%. This way the shim inductor results into a negative value. The difference between my calculation (load min. = 15%) and excel sheet (load min. = 50%) is that with my value you get better efficiency in the load range 15%...80%, while with excel sheet the efficiency is optimized in the load range 50%...100%.

    Best regards,

    Roberto