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TIDM-1000: Efficiency estimate - loss calculation

arun jeyaprakash
Prodigy 80 points

Part Number: TIDM-1000

In the Excel sheet provided to estimate efficiency, the calculated total loss is directly subtracted from total input power, which means it is the total loss. How to calculate the loss per switching device (diode and mosfet) from that data ?

  • 0
    TI__Guru* 84660 points

    Arun,

    I've assigned your post to a C2000 expert, but due to the US holiday we will get back to no later than July 8th, by end of day US Central Time.

    Best regards,
    Matthew

  • 0 in reply to MatthewPate
    TI__Mastermind 43730 points

    If you look at the excel sheet the losses are estimated for a single phase , the losses are per switch already. 

  • 0 in reply to Manish Bhardwaj
    Prodigy 80 points

    I understand that the losses are estimated for a single phase. If the losses are per switch, then shouldn't the losses be multiplied with number of switches per phase before calculating the efficiency. Or could you send me the single phase Vienna circuit for the correct understanding.

  • 0 in reply to arun jeyaprakash
    TI__Mastermind 43730 points

    Because efficiency is calculated per phase there is no multiply. A phase for vienna will be the T network , i.e. 2 SIC Diodes and 2 FETs that connect to the neutral point. 

    Also note only one diode and two fets are in the power flow at any given point of time. The sheet uses for D1 for the SiC Diode, T1 and Da2 for the switch and the body didoe of the switch.

    Now looking back, because we used MOSFETS the loss on the FETs should simply just be the conduction loss as well for the values to be more accurate. 

  • 0 in reply to Manish Bhardwaj
    Prodigy 80 points

    Since there are 2 diodes and 2 mosfets in single phase circuit, should't the loss used to calculate the efficiency be divided among these 4 devices. If we use per switch losses wouldn't we be missing out the losses on other devices.

  • 0 in reply to Manish Bhardwaj
    Prodigy 80 points

    Let me rephrase my doubt. I understand that the total loss calculation is correct. In reality both diodes are going to be used, but not simultaneously. Does this mean that the diode loss be distributed among these 2 diodes. This would imply that per switch loss is different.

  • 0 in reply to arun jeyaprakash
    TI__Mastermind 43730 points

    Arun,

    For losses we have to take a instant of time, 

    at any instant of time only one SiC diode is in the circuit, hence estimating the loss in power which is watt/time is the same as estimating the power loss in the complete converter for that leg. Losses during the other half will be similar , just physically on another device. 

    In the calculation sheet 

    D1 is the conduction loss through the diode in Watts which is Joules/sec, during the negative half the loss will be on the other SiC diode but still be the same average.

    Same goes for all the other components listed on the sheet. I am not sure what else I can explain better.

    -Manish 

  • 0 in reply to Manish Bhardwaj
    Prodigy 80 points

    Hi Manish,

    I completely get what you are saying.

    Only one point that still bothers me is this. 

    As you say "Losses during the other half will be similar , just physically on another device." which means over long time (not instantaneous) the average loss is going to be distributed among these devices.

    I mean instantaneous loss per device is going to be different from average loss per device since we have more devices and same average loss." during the negative half the loss will be on the other SiC diode but still be the same average."

    As you say the instantaneous loss per device is what is given in the excel sheet, but over long run, this loss is distributed among all the devices included.

  • 0 in reply to arun jeyaprakash
    TI__Mastermind 43730 points

    That is correct, i guess you might be coming at it from thermal perspective, 

    Average loss on one device will be half for the diodes, for the T connected FETs it is going to remain similar and not change.