In the IEEE magazine 6. NO. 3. JULY 1991 A Novel Soft-Switching Full-Bridge DC/DC Converter: Analysis, Design Considerations, and Experimental Results at 1.5 kW, 1OOkHz the following converter has the active state when any of the pair of transistors S1-S4 and S2-S3 are on.

enter image description here

C1 and C2 represents the parasitic capacitance ( Snubber, Trafo, etc ).

The converter has the following equivalent circuit for the passive state, the state in which no power is transferred through the Transformer.


simulate this circuit – Schematic created using CircuitLab

All the energy needed to charge the capacitance must be accounted for and the the author came up with the following time function for voltage across C1. What was his approach to it?

$$\begin{aligned} V_{C1}(t) =& V'_O \frac{L_m}{L_m+L'_O}(1-\cos{\omega t}) \\ -& (I_m+I'_P)\sqrt{\frac{L_1+L_C+L_m||L'_O}{C_1}}\sin{\omega t} \\ +& V_{in}\cos{\omega t} \end{aligned}$$

EDIT: Franc enlightened me! Thank you!

  • 1
    \$\begingroup\$ Who was the author? What was the paper? \$\endgroup\$
    – Andy aka
    Commented Jan 7 at 20:59
  • \$\begingroup\$ To make it a better question, please edit it and tell us which parts of the function are unclear. The function is a sum of three terms, so, hopefully, you understand at least some of them, or at least some of the factors (things on either side of multiplication signs). \$\endgroup\$ Commented Jan 8 at 4:01

1 Answer 1


Assuming that the applied voltage has a sinusoidal time trend with pulsation ω and unit amplitude and that the initial conditions are zero, the calculation of the voltage across the capacitor is simple:

enter image description here

  • \$\begingroup\$ Thank you. I made sense of it. My only question is : where does "sLo" disappear in the first equation? It seems that only the parallel xL ( inductive reactance ) are computated. \$\endgroup\$
    – i33SoDA
    Commented Jan 8 at 14:04
  • \$\begingroup\$ @i33SoDA Thanks, I had missed it. \$\endgroup\$
    – Franc
    Commented Jan 8 at 16:40
  • \$\begingroup\$ Thank you again so much. You really helped me understand!! I very appreciate it! \$\endgroup\$
    – i33SoDA
    Commented Jan 8 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.