Unwind di Power Electronics - 1 di 5

The efficiency of the converter

The efficiency of the converter is defined as follows:

η = Pg / PLoss

Where "Pg" is the power given by the generator "Vg". This power is delivered to the load but a little amount gets lost in the devices if they are considered as real and not as ideal. A more precise expression of the efficiency is:

η = (Pg - PLoss) / Pg

So "PLoss" represents the total power dissipated inside the converter and what happens is related to non-ideal effects. It can be estimated an expression of "PLoss" assuming non-ideal switches and non-ideal inductor.

Let's start from the inductor. A non-ideal inductor is composed by the series of an ideal inductor and a resistor with resistance "rL":

Lnon-ideal = Lideal + rL

So this is the non-ideal inductor, while this is a totally ideal inductor. In the schematic of the buck converter is represented an ideal inductor, but when the losses have to be considered, that ideal inductor must be replaced with a non-ideal inductor (composed by...)

The series of an ideal inductor and a resistor “rL”). The instantaneous power dissipation is named “pL(t)”, while the average power dissipation is named “PL”. The definition of “PL” is: Per f sf.tdtt.is.

Another non-ideal component is the primary switch. The mosfet can be approximated by the series of an ideal switch and a resistor “rON”: esi 20N. In this case, the instantaneous power dissipation and the average power dissipation are respectively named “pS(t)” and “PS”, where: Psa fedepst a.

The last non-ideal element is the diode that can be represented as a series of a full-ideal diode (whose voltage drop is 0) in series with a voltage drop that is the voltage drop presented by the diode in the ON state, that is VON. Note that power diodes are very different from small signal diodes. Small signal diodes exhibit a voltage drop of about 0.6V or 0.7V, but power diodes can exhibit a voltage drop of 1V or 1.1V (more or less).

The instantaneous power dissipation and the average power dissipation are named respectively "pD(t)" and "PD", where:
Pisa f t.fsfsf.edu

So "PLoss" is defined as follow:
PsPost plRoss

The goal is to obtain an expression of the efficiency. To reach it, it can be used a (strong) approximation that extremely simplify the calculation. In this approximation, the inductor current is considered constant, so the ripple is neglected:
tèil TOii iIl

Let's start with the calculation of "PL":
tiPer riide IEt 1ef fr il Et IL ZIO

Let's evaluate now "PS":
is ilt tTom 22am 0NIlIonper offo DTSPsa de teptpst fronted2aDiIONI

And at the end, let's evaluate "PD" (remember that "ON" and "OFF" are defined for the primary switch):
0N0ef a ètalloniVenis Von ILt OFF

So the expressions of the 3 powers lost in the converter are:
Pi IE It's evident that "PS" and "PD"

depends by the duty cycle, and2La this happens due to the fact that larger is the time that theiPs primary switch is closed more is the current that flows throughbronzia that switch and less the current that flows through thesecondary switch, in fact the diode works during the OFF time.iPd IoVonaiThe complete expression of “PLOSS” is:Pass io If IoVonDrawri D1t teIt’s so clear that for a given duty cycle the more it’s increased the output current, load the more is the powerdissipated.The relation of “PLOSS” with the output current is a mix of linear (Io) and quadratic (Io^2). This result isobtained because 2 devices are modelled with resistors and the power dissipated by a resistor depends by thesquare value of the current through it.In general “PLOSS” of a converter can be seen as a sum of 3 terms:

• A constant term (in respect of the output current ‘Io);
• A term that depends linearly by the output current and has a

effect of the constant term on the efficiency, the effect of the linear term (in respect of “Io”) on the efficiency and the effect of the quadratic term (in respect of “Io”) on the efficiency, it can be assumed:k1=k2=0. In this case: VIo Ioµ iPeonIo stintA possible sketch of this function is:MIO O If “PConstant” is considered 0, the efficiency is 100%Kirke1 for every value of “Io”. Considering “PConstant” atlow value of “Io” (this zone is called “light load”) theefficiency is degradated (goes to 0).The asymptote of the efficiency in respect of “Io” is 1.0 IoLet’s consider now only: k2=0. In this case: iIo In this case, the asymptote of the efficiency in respectMMIO of “Io” is not 1 (efficiency cannot be 100%), but less.0 oµ0 IoLet’s consider now all the terms at the denominator:i iIo IoµM Mo oµ0 IoIt’s clear that in this last case there

is a peak of the efficiency but for large value of “Io” the efficiency goes to 0(in fact the asymptote of the efficiency in respect to “Io” is 0).The usually plot of the efficiency is represented by more than 1 curve because power is studied under different assumptions.

On the last plot of the efficiency is represented a value of “Io” that is called “IoMAX” and represents the current after which internal power dissipation is not more acceptable. After this current, electronic devices heat up too much. This is the reason why the general plot of the converters are like the following:

iIo1 MAX IoLo

Generally the plot of the efficiency stops at “IoMAX” and the output current must be less than this value to avoid degradation of the efficiency.

Let’s see now the boost converter. This converter can be seen as a buck converter used at reverse. To analyse this converter, let’s start to consider the primary switch, the secondary

“S’” is the switch whose duty cycle increases the output voltage. For this reason, in the boost converter “S’” becomes “S” and “S” becomes “S’”. The circuit becomes:
lsD1Vo VgsDc

After the redefinition of the switches, the voltage conversion ratio becomes:
M eBoost tgi

Let’s redraw the circuit in a more reasonable way:
sii D1Vg Voai 5 eD

To analyze the circuit, it’s considered the “small ripple approximation” (SRA), so:
votavo

The voltage across the inductor is:
Vg 0NI t avg.tt Off

Considering the SRA, the voltage across the inductor becomes a piecewise constant function, here represented:
ottVg Evg.tk

The definition of “vL(t)” becomes:
Vg 0NI t a Vo OffVg

Due to the volt-second balance (VSB), the mean value of “vL(t)” must be equal to 0, so:
0Via tai aVSB

Obviously, considering that the mean value of the waveform is 0, two considerations can be done:
• The converter is in

steady state;

• These areas are equal;

Let's evaluate the expression of "VL" and let's impose it equal to 0:

Via tI vi Dtsitdtzfsvgbtstvg.roa aVg Vg VorVoa 0D DD Dti I I

Solving for the efficiency "M":

È IMBoost IYg 2 µ

This is the voltage conversion ratio of the boost converter.

A possible plot of the voltage conversion ratio of the boost converter is here represented:

M Apparently, the output voltage can go to infinite when dutyDBoost cycle goes to 1, but in reality, the plot is like this. When duty cycle approaches to 1, some non-idealities cannot be neglected.

The reason why the plot goes to 0 is that when the duty cycle approaches to 1, the secondary switch never closes and primary switch never opens, so there is nothing that can charge the output capacitance.

As shown in the plot, the voltage conversion ratio is 1 when the duty cycle is at 0% and goes to infinite when the duty cycle goes to 100%. It's clear that the boost converter is a