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How to calculate efficiency of the dc-dc converter, including both thermal loss and resistive losses?

0 votes
In the demo models 'Buck Converter with Thermal Model', the efficiency has been calculated just by 1-(Ploss/Pin), where Ploss is just the switch conduction and switching loss calculated in the thermal domain. In this, why in the denominator, Ploss has not been added? As the Pin reflects only the electrical loss (resistive losses), and not the switch loss (conduction+switching), but ultimately the switch loss also comes from the Pin. So, if the switch loss has been subtracted from the Pin separately in the numerator, then it should have been added in the denominator as well. Basically, as per my understanding, it should have been, (Pin-Ploss_switch)/(Pin+Ploss_switch). Can you please clarify this point?

So, If I connect the parasitic resistors along with the inductor and capacitor in my simulation, I observe that Pin-Po gives me the resistive losses only (not the switch loss). That means Pin=Po+resistive loss. But Pin doesn't include switch loss. So, how should I calculate efficiency in this case? Should it be, (1-(Ploss_resistive+Ploss_Switch)/Pin) or (Pin-Ploss_switch)/(Pin+Ploss_switch)?
asked Oct 2 by Kavita_iitb (12 points)

1 Answer

0 votes

This is a good question and fundamental to how PLECS works with ideal switching and thermal losses based on lookup table data. As the switching losses are inserted as energy impulses in (zero) time, PLECS does not introduce the thermal losses back to the electrical domain, but only uses the latest junction temperature information for computing new thermal losses. Therefore, the only correct calculation for efficiency when using the thermal domain is to draw the thermal losses from the source power. It is true that if you include ohmic losses in the switch elements (on-resistances), and try and measure the output power having also included all electrical losses due to resistances in passive elements that you can expect a mismatch between a purely electrically-calculated efficiency versus the accepted-as-most-appropriate method that we provide in our demo model. By the way, you can also have a heatsink cover resistors to include ohmic losses in your thermal calculations. But in conclusion, yes, Pin doesn't include switch loss, but we know the total losses of the switching cell from the thermal calculations and should subtract those from the input power (and any other ohmic losses you want to include in the calculation, e.g., filters, winding losses, etc.). Let me know if this is clear.

answered Oct 2 by Kris Eberle (966 points)
Hi Kris. Thanks for understanding the depth of my problem. But still, I am having a doubt. Can you provide me a way to calculate the efficiency, if I don't extend my heat sink to other ohmic components (like parasitic winding resistances etc), and keep it only around the Switches? Will the formula still remain the same as 1-(P_switchloss/Pin) or will it change?
No, it will still remain the same. You are just trying to to see the difference between the input power and losses from the thermal domain as a percentage of the input losses: (Pin - Psw)/Pin. You can still add all of ohmic components you want, such as to see their affect on output voltage, for example, but if you were to directly probe the output power it will not include the thermal losses and therefore not give an accurate representation for efficiency calculations.
Thanks Kris. But my confusion is this:

What I see from the simulation in plecs is: Pin=Po+ Pohm   (Pohm=ohmic losses in winding resistors etc), and Psw is calculated separately in thermal domain.

So, technically, Psw is also drawn from the source, i.e., it should have been: Pin=Po+Pohm+Psw, which is not the case in plecs if we give thermal model.

Efficiency should be equal to 1-(Ploss/Pin), where Ploss should accomodate all the loss, i.e., Pohm+Psw.

Therefore, as per the plecs model, the efficiency should be= 1-((Psw+Pohm)/(Pin+Psw)). Am I correct? Then it doesn't match the formula given in the manual.
Hi again,

Sorry for the confusion. And you are generally correct. I wrote a long explanation and then realized this may be best explained by saying that we are promoting the efficiency of an otherwise ideal circuit where you only need to calculate the losses of the switching cell. Of course, you can add ohmic losses into the efficiency formula. Typically the power semiconductors contribute the most substantial portion of the losses so that's at minimum, your staring point. But as long as you don't use only the output power to calculate efficiency, you can choose the level of fidelity for the rest of your model and calculate further losses accordingly. Is that more clear?


Yes Kris. Thank you. Now it is more clear..