Please, if you could explain me why on step-up (boost) converter the sign of the t-off voltage part is positive in the following equation:

Vd * tOn + (Vd - Vo) * tOff = 0

I thought that the difference between (Vd - Vo) is positive (otherwise the diode would be off) and this justifies the above equation but if (Vd - Vo) is positive, the current in the inductor during the tOff period would increase and not decrease.

Thank you in advance.

Source: https://en.wikipedia.org/wiki/Boost_converter Book: Mohan, Ned; Undeland, Tore M.; Robbins, William P. (2003). Power Electronics. Hoboken: John Wiley & Sons, Inc. ISBN 978-0-471-42908-1.

  • 4
    \$\begingroup\$ While the boost converter is a pretty standard circuit, it would be helpful if you drew a diagram and labelled what you mean by each symbol. \$\endgroup\$ – Hearth May 28 '19 at 15:46
  • 1
    \$\begingroup\$ Assuming you mean Vd is the supply voltage (Vi in the article you reference), (Vd-Vo) is negative (or it wouldn't be a boost converter). The voltage across the inductor (due to the decaying field) makes up the difference and puts the diode into conduction. \$\endgroup\$ – Phil G May 28 '19 at 15:57
  • \$\begingroup\$ Webers are the equivalent of magnetic charge (comparable to Coulombs for capacitors.) Just as John and Phil say, the sum of the Webers during the ON time plus the Webers during the OFF time must sum to zero. If they didn't sum to zero, then the "magnetic charge" would either walk continuously more positive or negative and eventually saturate the core (unless it's an air-core, I suppose.) As Phil says, Vd-Vo is negative for the boost converter. So all this works out okay. \$\endgroup\$ – jonk May 29 '19 at 4:51

For the boost to be in steady-state the inductor current delta (ramp up) during the on time has to equal the inductor current delta (ramp down) in the off time.

Let's assume Vd is the input voltage and Vo the output voltage, and the diode is ideal.


simulate this circuit – Schematic created using CircuitLab

Rearrange your equation

$$Vd * tOn + (Vd - Vo) * tOff = 0$$

Like this:

$$Vd * tOn = (Vo - Vd) * tOff$$

Using the relation $$V=L*di/dt$$ and $$di = V*dt/L$$ you can see that the above equation balances the inductor up-slope during the on time with the down slope during the off time. The L cancels out on both sides of the equation.

The waveforms look like this, (From Here) enter image description here

| improve this answer | |
  • \$\begingroup\$ Isn't Figure 3 in the wiki the OP mentioned, continuous mode boost, correct? (Maybe I'm just turned around, tonight, and things will clear up for me in the morning.) \$\endgroup\$ – jonk May 29 '19 at 4:43
  • \$\begingroup\$ @jonk Yes, it seems correct to me, I didn't intend to imply that it wasn't. In the drawing I linked the top waveform was incorrect as drawn but if we consider it the switch node voltage not the switch state it's OK. Good point about the equations being for CCM. In DCM things are somewhat different, though of course the energy added each cycle has to balance the energy delivered to the load in the same way. \$\endgroup\$ – John D May 29 '19 at 14:57
  • \$\begingroup\$ @jonk Edited to make the diagram more clear. \$\endgroup\$ – John D May 29 '19 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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