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CSD16301Q2: Power Calculation of CSD16301Q2

Part Number: CSD16301Q2

Hi i am applying 5V as VDS and VGS as 3.3V and how can i calculate ID, and RDS(ON) with power dissipation. I am connecting a load of 2W. While i am trying to calculate ID as per given RDS (ON) in datasheet (0.027A) @VGS=3V i am getting 185A, is this right. Can you please provide me the information of RDS(ON)(TJMAX). What is he maximum allowable Chip temperature.

  • Good Morning Sujatha,
    Thanks for the questions. Please refer to figure 3 in the CSD16301Q2 datasheet. This shows drain-source current as a function of VGS with VDS = 5V at 3 temperatures. You can calculate rds(on) using the maximum specified value from the datasheet and multiply by the temperature factor shown in figure 8 of the datasheet. The max specified Tj for this device is 150degC and the factor from figure 8 is 1.5. For VGS = 3V, the max rds(on) is 34mOhm @ 25degC. For 150degC, the calculated max rds(on) is 34 x 1.5 = 51mOhm.
  • Hi John Wallace,

    Thank you for your reply. 

    The below are my calculation. Please correct me if i am wrong. I am trying to calculate PD with the Rth.

    In one application note i observed PD= ((Tch(MAX)-25C)/Rth(ch-c). In Figure1 of datasheet  i observed transient thermal impedance, but how to derive the Rth along with  channel temperature. 

    regards,

    Sujatha.T

  • Good Morning,
    The calculation in your previous note is used to determine maximum power dissipation for a particular device/package given the thermal impedance, maximum channel or junction (TJmax) temperature and ambient (25C). The thermal impedance values for junction to case, Rth(j-c), and junction to ambient, Rth(j-a) are given in table 5.2 on page 3 of the datasheet. Please review and let me know if you have any additional questions.
  • Hi John Wallace,

    Thanks for your reply.

    As you said to apply in the formula PD= ((Tch(max)-25C)/Rth(Ch-C)). In this i am assuming these values as my junction temperature of MOSFET  CSD16301Q2 may go up to 60C .

    Tch(max)=150C

    Rth(ch-c)=65-25=40

    Then PD=(150-25)/40=3.1W is this right. Please correct me if i am wrong.

    Thank You for your patience.

    regards,

    Sujatha.T

  • Hi Sujatha,
    Can you please explain why you calculate Rth(ch-c) as 65 - 25? From the datasheet, Rth(j-c) = 8.4degC/W and Rth(j-a) = 65degC/W. It is my understanding that ch (channel) and j (junction) are interchangeable. I would use Rth(j-a) in the PD calculation if the ambient temperature is known. Otherwise, you can use Rth(j-c) but you would have to measure actual case temperature of the device.

    PD(max) = (150-25)/65 = 1.92W, this is for 25degC ambient. If your ambient temperature is higher, you should use that instead of 25.

    This PD calculation is not the actual power dissipated in the FET. It is the maximum allowable dissipation for a given junciton temperature. In practise, the maximum operating junction temperature in your system is derated from 150degC for reliability purposes.

    The actual power dissipation in the FET depends on the application. For static switching (i.e. load switch), the steady state dissipation is I^2 x rds(on). For dynamic switching (i.e. switch mode power conversion) there are conduction and switching losses that need to be calculated.
  • Hi John Wallace,

    Thank you for the quick reply, the reason Why i have taken Rth(ch-c) as 65 - 25 is i assumed that "ch-c" is channel temperature minus Ambient temperature. Wrong assumption of "-". Thanks for the clarification given.

    Coming to my application, I have a power of 5V supply and i am trying to switch on the high power LED's(Peak forward current 2.5A) by controlling the Gate with 3.3V, and source connected GND. For this application can i use steady state dissipation calculation? Then it will give the power dissipation as((2.5*2.5)/0.027)=231W Is this correct? Please correct where i am going wrong?

    Thanks & regards,
    Sujatha.T
  • Hi Sujatha,
    In this case, you can use the conduction loss calculation. In your application, the FET is either fully enhanced when you apply 3.3V to the gate or off when the gate is pulled LOW. I don't think you're trying to regulate the LED current by modulating the gate-source voltage in a closed loop system. Correct me if I'm wrong. A couple of things to keep in mind for the power dissipation calculation: (1) 27mOhm is typical rds(on) @ VGS =3V and Ta = 25degC. (2) You should use the max rds(on)= 34mOhm from the datasheet and also take into account the positive temperature coefficient of rds(on). Let's assume you operate at a max case temperature of 100degC. From Figure 8 in the datasheet, the normalized rds(on) scales by a factor of 1.3 for Tcase = 100degC. Now the power dissipation looks like this:

    PD = ((2.5*2.5) * 34mOhm * 1.3) = 276mW
  • Hi John Wallace,

    Thanks for the explanation. As you said we are not regulating the current.

    So i will consider my total power dissipation of the CSD16301Q2 as PD=((2.5*2.5)*0.034*1.1)=233mW.

    regards,

    Sujatha.T