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TIDA-01494: Design and equations

Glidden Martin
Prodigy 30 points

Part Number: TIDA-01494
Other Parts Discussed in Thread: UCC28180

Hi,

    The design spec has Peak current of 30A for 3 seconds but I do not see the 3 sec used in any equations. Are you relying on the MOSFET SOA graphs to meet this requirement? I would like to change the peak current time, how would I do that? There are a number of equations where the variable name do not match its value used. Is this an error with variable name used or the value used?

Thanks

-Glidden

  • 0
    TI__Expert 4840 points

    Hi Glidden,

    The resonant tank is designed to handle 30A peak (720W). As shown in Figure 2, 720W gain curve peak is over 1 (nominal value), which means the converter is able to deliver 720W. While thermally, this design is designed to handle 480W continuously. Also please note that 720W at 390Vbus, primary MOSFET current is less than 2Arms, which is way lower than MOSFET SOA. If you can let me know which value in the design guide doesn't match with the schematic, I could look into it.

    Regards,

    Sheng-Yang Yu   

  • 0 in reply to Sheng-yang Yu1
    Prodigy 30 points

    Hi Sheng-Yang,

           In table 2, Pout(max) is 750W, this is the output power of the PFC to accommodate an output power of 720W (max) from the LLC.  

    In Equation (1) : Iout(max) =  Pout(max)/Vout(max)  but a value of 500W  is used for Pout(max)

    In Equation (2) : the same Pout(max) of 500W is used.

    In Equation (6): the value of 11.3 does not match Pout(max) of 500 or 750. Someone already mentioned this in one of the forums and the equation was corrected there.

    In Equation (15): the wrong value of Io_peak was used.

    In Equation 6 IL(max) is (sqrt(2) *(Pout(max)/Vinrms(in))) * (1 + (delta Ripple/2)). This is used to come up with the Boost Inductor value

    In TIDA -00779, which also uses the same chip (UCC28180) in the PFC, the equation used is Ipk = ((sqrt 2) * Po)/(efficiency * Vin_min)

    why is there a different set of equations used for a similar PFC?

    I understand that the converter can deliver 720W for 3 sec but I so not see how this is baked into the design. Unless there is some circuitry that is missing that would monitor the output, have some timer in place to trigger the OCP mode.

    -Glidden 

  • 0 in reply to Glidden Martin
    TI__Expert 4840 points

    Hi Glidden,

    Looks like the design calculation are based on 500W (nom) and 750W(peak) while the final specification is 480W (nom) and 720W (peak). Calculate based on a little bit higher power should still be good for the design.

    For equation 6, I also believe the calculation is wrong. Since the inductance has already fixed. We can estimate peak current by using Ldi/dt=dV.

    Here is my calculation, first determine the duty cycle at Vin,min peak: D = 1-Vin/Vout = 1- 85*1.414/390=0.692. Then iripple = Vin,min peak * D/(Fsw*L) = 85*1.414*0.691/(95k*420uH) = 2.085A.

    The maximum current at Vin,min peak can be calculated by Iin,rms,peak *1.414 +iripple/2 = Pout,peak*1.414/ (eff*Vin,min*PF) +iripple/2 = 750*1.414/(0.94*85**0.99) +2.085/2 =  14.449A. Please note that the actual inductance will be lower when current is higher. So the peak current will be a little bit higher.

    Thank you for the detailed review of the equations.

    Regards,

    Sheng-Yang Yu

  • 0 in reply to Sheng-yang Yu1
    Prodigy 30 points

    750W, this is the output power of the PFC to accommodate an output power of 720W (max) from the converter.

    In the equation that you provided, 750W is already the peak power so why are you multiplying it with sqrt(2)?

    The maximum current at Vin,min peak can be calculated by Iin,rms,peak *1.414 +iripple/2 = Pout,peak*1.414/ (eff*Vin,min*PF) +iripple/2 = 750*1.414/(0.94*85**0.99) +2.085/2 =  14.449A. Please note that the actual inductance will be lower when current is higher. So the peak current will be a little bit higher.

  • 0 in reply to Glidden Martin
    TI__Expert 4840 points

    I am assuming 750W power over full AC cycle. To calculate the peak inductor current. You first need to know the peak input current, which is Iin,rms * 1.414 (assuming sinusoidal current). That's why I multiply 1.414.

  • 0 in reply to Sheng-yang Yu1
    Prodigy 30 points

    look at this equation Pout,peak*1.414/ (eff*Vin,min*PF) +iripple/2

    so 750W is not really peak power, is that what you are saying?

  • 0 in reply to Glidden Martin
    TI__Expert 4840 points

    No, 750W is the peak power. The equation I'm using is Iin,rms,peak = Pout,peak /(eff*Vin,min*PF). Then Iin,rms,peak *1.414 to get the maximum value of the input current. I.e. Iin,rms,peak *1.414 = Pout,peak * 1.414/(eff*Vin,min*PF)