Switching losses calculation in resistive load and inductive load are different. For a full-bridge DC-DC topology with a resistive load at the output, should I calculate my switching losses like I have a resistive load or should I consider the transformer as a load to switches and calculate the losses like I have an inductive load?

  • \$\begingroup\$ What calculations would you use, and under what assumptions do they apply? \$\endgroup\$ May 14, 2022 at 14:57
  • 1
    \$\begingroup\$ If it's a DC-DC then the load will not matter in terms of \$\cos{\varphi}\$. Whether it has a reactance, or a negative incremental resistance, those will influence the control loop, but the power will be DC. \$\endgroup\$ May 22, 2022 at 16:52
  • \$\begingroup\$ Are you trying to calculate the losses in the converter? Or are you trying to calculate losses through the load? \$\endgroup\$
    – soup
    May 22, 2022 at 21:08

1 Answer 1


The transformer will indeed present as an inductive load (two of them, actually) to whatever is driving the primary. if there is a resistive load across the secondary, then there will also be a resistive load on the primary.

What your MOSFETs will actually see as a load is comprised of 3 components:

  1. Reflected load - this is an effective resistance that is seen on the primary due to a given resistive load on the secondary. The resistive load will be 'transformed', however, depending on turns ratio: $$ R_{reflected} = R_{load} \left ( \frac{N_{pri}}{N_{sec}}\right )^{2}$$ where \$ R_{load} \$ is the resistance across the secondary, and \$ N_{pri} \$ and \$ N_{sec} \$ being the number of turns on the primary and secondary windings respectively.

  2. Mutual inductance - Conveniently, tightly coupled inductors will generally be seen as a single effective inductor that is connected in parallel with the reflected load. The mutual inductance is simply: $$ M = k\sqrt{L_{pri}L_{sec}}$$ where \$ k \$ is the coupling coefficient of the transformer (a good transformer will have a coefficient greater than 0.99) and \$ L_{pri},L_{sec} \$ being the self-inductances of the primary and secondary windings.

    Note: if you attempt to measure the self-inductance of the windings with the other winding open circuit, you will actually be measuring that winding's self-inductance plus leakage inductance, though this will typically be very small compared to the self-inductance. More on that later.

    You can also measure the mutual inductance directly by connecting the primary and secondary windings in series (such that they are in phase) and measuring the inductance as if it was one single winding. You will measure an inductance that is greater than the two windings' self-inductances added together. If you subtract each winding's self-inductance from that measured inductance and divide it by 2, that is the mutual inductance of the transformer.

  3. Leakage inductance - this is the one that can cause some real trouble for your MOSFETs, not only in the form of additional losses, but also potentially destructive voltage spikes that will need to be dealt with (using some kind of snubber usually). This is something that typically needs to be empirically measured using the actual physical transformer. Fortunately, this is easily done. Simply short the secondary winding and measure the inductance across the primary winding. That value is your leakage inductance.

The resistive load on the secondary as well as the entire transformer itself can be represented (ignoring capacitive parasitics which are beyond scope here), at least from the point of view of what is actually loading your MOSFETs, like this using these 3 things:

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The inductance seen by the MOSFETs will only be the mutual inductance and leakage inductance, you do not need to factor the windings' self-inductances (or rather, you already have). They are accounted for in the form of the mutual and leakage inductances.


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