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I have tested a boost converter under different duty cycles. I know the voltage is related to the input voltage by V = Vin/(1-D) where D is the duty cycle. But for some reason, as I increase this duty cycle with my input voltage kept constant, the power efficiency of the converter goes down.

I am using a function generator to make the MOSFET switching, and a DC voltage supply for input. (something like this: http://shop.rabtron.co.za/catalog/bench-power-supply-c-46_374.html)

Can someone tell me why the power efficiency of a boost converter reduces as the duty cycle is increased?

EDIT: I put a circuit to make the question more clear. I actually did not bother to check the type of diode or MOSFET, and it seems like that would've been important...

  • \$\begingroup\$ Hint: What happens to V as D increases? What happens to the MOSFET switch current as D increases? \$\endgroup\$ Commented Oct 2, 2014 at 18:09
  • \$\begingroup\$ Without a circuit this cannot be definitely and sensibly answered. \$\endgroup\$
    – Andy aka
    Commented Oct 2, 2014 at 19:23
  • \$\begingroup\$ So I know that V should increase with D, and I think that the MOSFET would be conducting current for a longer time as D increases. So my best guess is that the MOSFET losses are increasing (or the diode losses are decreasing). Is this the proper way to think about it? \$\endgroup\$
    – AAC
    Commented Oct 2, 2014 at 21:58
  • \$\begingroup\$ Diode duty cycle will decrease as your MOSFET duty cycle increases, but so does your peak current. Also, the reverse recovery time of the diode is becomming a larger and larger part of the diodes total on-time as this happens so the switching losses increase expoentially here. \$\endgroup\$
    – winny
    Commented Feb 5, 2018 at 10:22

5 Answers 5


The main power losses in a boost converter can be summarized as follows:

  1. Power switch switching losses (e.g. MOSFET, BJT. Hereafter I will refer to the Power switch as the MOSFET)
  2. MOSFET conduction losses.
  3. DIODE switching losses.
  4. DIODE conduction losses.
  5. Other conduction losses (e.g. inductor resistance)

The efficiency of a converter is given by: eff = Po/Pin = (Pin - Plosses)/Pin.

As the losses change the efficiency therefore changes.

One can not make a blanket statement as to why the efficiency reduces or losses increase as the duty cycle increases because then one would need to know all five loss parameters as a function of current, voltage and switching frequency.

However, a simplified explanation of this phenomena is that the MOSFET conduction losses are unequal to the diode conduction losses. As the duty cycle increases, the MOSFET will conduct for a longer period and the diode for a shorter period. This in turn alters the power losses in the circuit. If the DIODE happens to have higher conduction losses than the MOSFET for example, then as the duty cycle changes, causing the DIODE to conduct for a relatively longer period than the MOSFET, then the efficiency will decrease. This is a simplified explanation, but the main principle is that as you change the duty cycle, the operating conditions for each element in the circuit change. Since the losses for each device depend on it's specific operating point, then changing the duty cycle changes the losses.


Additional to other answers:

A point perhaps not covered by others -

Vout = Vin + converted power as the inductor "stands on" Vin.

The Vin component is "100% efficient" by definition. As Vout gets increasingly > Vin the converted component gets larger and efficiency falls.

eg say conversion efficiency was 50%.
Assume ideal diode and zero resistance inductor to make picture easier.

At Vin = 5, Vout = 5, efficiency = 100%
At Vout = 10V Z = (100+50)/2 = 75%.
At Vout = 100V Z = (5 x 100 + 95 x 50)/100 = 52.5%

The above drop in efficiency is with the actual converter efficiency remaining constant.
If the converter losses increase as the conversion fraction increases (as they generally do) then the effect is even greater (but actual conversion loss will not usually be 50% :-) ).


Can you explain a little more what you mean by "converted power as the inductor stands on Vin"?

Consider the diagram.
Vout = Vin + Vinductor - Vdiode_drop.
If diode is perfect and inductor is zero Ohms resistance then if Vout = Vin ALL of output comes from input directly at 100% efficiency.
As Vout climbs the Vin part remains and is added to by converted energy at < 100% efficiency(Z).
The net Z is the mix of 100% Vin and lower % converted.

  • \$\begingroup\$ Can you explain a little more what you mean by "converted power as the inductor stands on Vin"? \$\endgroup\$
    – AAC
    Commented Oct 2, 2014 at 22:10
  • \$\begingroup\$ @AAC Look at the diagram. Vout = Vin + Vinductor - Vdiode_drop. If diode is perfect and inductor is zero Ohms resistance then if Vout = Vin ALL of output comes from input directly at 100% efficiency. As Vout climbs the Vin part remains and is added to by converted energy at < 100% efficiency(Z). The net Z is the mix of 100% Vin and lower % converted. \$\endgroup\$
    – Russell McMahon
    Commented Oct 3, 2014 at 21:27
  • \$\begingroup\$ @AAC Vin is like a "relative ground" for the inductor. \$\endgroup\$
    – MOSFET
    Commented Mar 28 at 16:29

The output of a boost converter is not a linear function of the duty cycle. For any one output to input voltage ratio, there is a optimum duty cycle. Both higher and lower will cause less output.

Think about it. There are two competing requirements pushing the duty cycle in opposite directions. As the duty cycle increases, you get higher and higher inductor current, which allows more to be transferred to the output. However, this transferring happens during the off time. As the duty cycle increases, the off time (1 - duty cycle) decreases, and leaves less and less time to dump any current onto the output.

The point of maximum power transfer is a tradeoff between these requirements. Increasing the duty cycle past this point will result in less output power, but will still keep increasing the input power, so efficiency goes down. This is probably what you saw. If you start at 0 duty cycle, you will see a rise in the output voltage to some point, then a decrease after that.

  • \$\begingroup\$ For a perfectly-efficient boost converter operated at a high enough frequency that current never had time to change significantly, the ratio of input voltage to (output minus input voltage) would essentially coincide with the duty cycle (the limiting case as frequency increased would have the ratios coincide). Inefficiencies cause the ratios to differ. \$\endgroup\$
    – supercat
    Commented Oct 3, 2014 at 22:40

What I think happens physically in the converter is that with the switch in the ON position there is no transfer of energy from the input to the load. During the Toff interval (when the switch is off) the input is connected to the load and the energy stored in the inductor is transferred to the output and the boost action occurs.

So intuitively it can be seen as a process where when we increase the duty cycle 'D' to 1 the switch is ON for the entire switching period. All the power is dissipated in terms of conduction loss in the Rds of the MOSFET and so no power transferred and efficiency ultimately reaches zero when we operate at maximum duty cycle.


Actually it's quite easy to explain while looking at the circuit.

Clearly at \$\delta=0\$, \$U_{out}=U_{in}-U_{diode}\$ and the only loss is that in the diode due to conduction.

When you want to increase the output voltage of the boost converter, you periodically switch on \$M_1\$, which then periodically will 'short' \$L_1\$ leading to an increased current amplitude through it plus a periodic current through \$M_1\$ which until then carried no current at all, i.e. had no losses associated to it.

Further, when you switch off \$M_1\$, \$D_1\$ is forced into conduction where it was blocking before that. So the separation charge in the die of \$D_1\$ has to be eliminated, which represents a (fixed) loss, but also the current, which until then was continuous, now is put through in the shape of a pulse.

Now if we assume constant average output current, a pulsed current through \$D_1\$ with the same average value will have a higher RMS value, and therefore will cause higher losses during the conduction, both in \$M_1\$ as well as \$D_1\$.

Now if we increase the duty cycle in order to get a higher output voltage, all those loss mechanisms, except for the switch-on losses in \$D_1\$ will increase. The time that \$L_1\$ is 'shorted' increases, hence its current and loss, and also the losses in \$M_1\$ because both the period the current flows and its amplitude increases, and the RMS current through \$D_1\$ also increases. The reduction in conduction time of \$D_1\$ however will reduce the conduction loss caused by the fixed voltage drop (apart from the resistive one), so there is a little caveat, but most probably the resistive losses dominate as this is a 'power' converter. Where the fixed voltage losses dominate it could be better to select a synchronous rectification instead of a simple diode.


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