Core Mathematics & equations for SMPS design




Much of the mathematics of this document can be found in the brief book:

“Understanding the mathematics of electronics”

by F.A.Wilson

Babani Electronics Books (1997)



Many of the equations of this document can be found in the brief book:

“Practical Electronics Calculations and Formulae”

by F.A.Wilson

Babani Electronics Books (1979 & 1981)





General Equations:


V = I.R

Q = I.t

P = V.I

Energy = P.t = V.I.t

Energy = Joules

Power = joules/ second



a.(b+c) = a.b + a.c

(b+c)/a =  b/a + c/a

a/b = (c.a)/(c.b)

(a/b)^c = (a^c)/(b^c)


(a/b) /  (b/c) = (ac/b^2)


“a + (b/c)”  does  NOT equal “(a +b)/c”

…remember “BODMAS” (see google)


Resistor Equations.


Two impedances in parallel =


Impedance = Z1.Z2/(Z1+Z2)


Inductor equations.


V = L.(di/dt)

V = N.[d(phi)/dt]


Energy = (0.5) *L * I^2


Inductive reactance = 2.pi.f.L


Complex inductive reactance  = jwL

(where w = 2.pi.f)



Capacitor equations.

Capacitive reactance = 1/ (2.pi.f.C)

(where w = 2.pi.f)


Complex  capacitive reactance  = 1/(jwC) = -j/wC



Q = C.V

i(t) = C.(dv/dt)

Energy = (0.5)*C*V^2


Capacitor charging from constant voltage source (V) via resistor:

v(t) = V.[1- exp (-t/RC)]

{where exp = 2.71828}


Capacitor discharging from Voltage V through Resistor R,

v(t) = V.[exp (-t/RC)]



Inductor,Capacitor equations


Frequency of resonance = 1/  { 2.pi.[SQRT(L.C)] }


0.5Ohms = SQRT(L/C) , -for critical damping of an LC circuit. (no overshoot)


Frequency Response:


RC circuit:

Be able to calculate vout/vin for an RC circuit as follows, with a sinusoidal voltage source input, and be able to do this at any frequency between say 0.1Hz and 200KHz.

Use complex numbers to do this.










..then calculate 20*Log(Vout/Vin) and know that this gives the gain, at whatever frequency.


Know how to make out the complex impedance for the above series RC circuit…ie R-j[1/(wc)]



Opamp circuit:

Be able to calculate vout/vin for an opamp circuit as follows, with a sinusoidal voltage source input, (eg sine input voltage = 1Vpk at zero degrees phase) and be able to do this at any frequency between say 0.1Hz and 200KHz.

Use complex numbers to do this.















..then calculate 20*Log(Vout/Vin) and know that this gives the gain, at whatever frequency.



Transformer equations.


Vp/Vs = Np/Ns


Np.Ip = Ns.Is

  • - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

(Lenz’s Law…………………….Faraday’s Law

V = L.di/dt;                                 V = N.[d(phi)/dt]

{where phi = flux in Webers; N = number of turns}


L.(di/dt) = N.[d(phi)/dt]


d(phi) = L. di / N                 


now:    phi = B.A

(where A = Area)



B = L. di / [A.N]

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


AL = L(nH) / N^2

{AL value is  quoted in ferrite datasheets}


L = N^2/Rel

(where Rel = Reluctance)



Reluctance = l/(u(0).u(r).A)


 u(0) = 4.pi*10e-7

l = length of the uniform magnetic circuit,

A = area of the magnetic core}



{by “mu” I mean the greek letter mu}


N.I = H.dl (Ampere’s Law)


    N = Number of turns

    I = Current

    H = Magnetising force (A/m)

    dl  = magnetic path length of core.


Waveform equations (typical SMPS waveforms)
















(a) Pulse Waveform:


DC = D*V1

AC = SQRT(RMS^2 - DC^2)

(b) Triangle Waveform:

RMS = SQRT((I1^2 + I1*I2 + I2^2)/3)

DC = (I1 + I2)/2

AC = SQRT(RMS^2 - DC^2)

(c) Trapezoidal Waveform:

RMS = SQRT(D*(I1^2 + I1*I2 + I2^2)/3)

DC = D*(I1 + I2)/2

AC = SQRT(RMS^2 - DC^2)


(d) Discontinuous Boost converter input current:





If you wish, the above can be treated as two separate trains of discontinuous right angle triangles (back to back).

The right angle triangle waveform can be taken as a trapezoid waveform with one side zero.

Then, find the RMS current of each right angled triangle train.

Then RMS current = SQRT{    [I1(rms)]^2 + [I2(rms)]^2    }


Sine Wave:


v(t) = A.sin(2.pi.f.t + theta)

{where A = Peak value; f = frequency; pi = 3.14159; t = time, theta = phase difference from reference phase}


RMS of sine wave of  peak value A = A/ [SQRT(2)]

(remember that this concerns a whole number of half periods)


Average value of sine wave of  peak value A, from zero degrees to 180 degrees = A*0.637

{i.e. (2.Vpk)/pi}


If you have a sine wave of  frequency 1KHz, and zero DC offset, how much time is between the rising zero crossing and the first positive peak encountered therafter



Be able to calculate the frequency of a sine wave given the time for one period.


Given two sine waves of the same frequency but different phase, be able to assign one as the zero phase reference, and then calculate the phase difference between the two sine waves.


For a three phase supply, with each phase being a sine wave with Vrms = 230V,  be able to calculate that the peak of the line to line voltage is 563V. Also know how to check this with a simulation.


If see a drawing of  2 sine waves of the same frequency, on the  same time scale, be able to recognise that they are either in_phase or out_of_phase.


Be able to basically draw two sine waves, with  roughly 60 degrees of phase difference between them.


Understand lagging and leading phase of sine waves.


Know that pi radians is 180 degrees.


Know that all waveforms can be considered to be composed of an (infinite) sum of sine waves of different magnitudes, phases and frequencies. (Fourier analysis)



Know the equations for sin cos and tan for a right angled triangle.

Eg sin (theta) = Opp/Hypotenuse


For a right angled triangle, know that hyp = SQRT[opp^2 + adj^2]




Log(a.b) = Log(a) + Log(b)

{logs all to same base}

Noting that logarithms can turn multiplication operations into simpler addition operations.


If log(100) = 2 then 10^2 = 100   {log to the base 10}





e^(-t/RC) = 10


take natural logs (ln) of both sides in order to get “t” on its own, say…



-t/RC= ln(10)

-t = ln(10) * R * C

t = -ln(10) * R * C


Know eg  that 1nF  = 1*10^(-9) Farads.

See that 3*10^(-3) * 2*10^(-6) = 6*10^(-9) add the exponents


if dividing, then subtract the exponents….

eg [3*10^(-3)] / [6*10^(-9)] = 0.5 * 10^6


know that 1 = 1*10^(0)


know that eg {[10^2]^3} = 10^6 multiply the exponents.


Know eg that 10mA* 12kOhms = 120 Volts




V=IR is Ohms Law

Know that V is proportional to I, and that R is the constant of proportionality


Differentiation and Integration



Know that differentiation involves  the rate of change. current into a capacitor = C * dv/dt, where “dv/dt”is the rate of change of voltage across the capacitor

dv is a  range of voltage, or the change in voltage.

dt is an interval of time



Know that Integration involves finding the total value of some quantity by splitting it up into lots of equal , small pieces, and then adding all these up to find the total.


The purpose of doing an integration is often to find an average of some value that changes with time. So the integration finds the total value, and then uses that to find the average of the value over some period of time, say, (often one second, the unit of time).


The area under a graph of current against time, is the amount of charge delivered.


(The area under a graph of   velocity against time is the distance covered…..since if you travel at 50mph for one hour, then you will go 50 miles)


There is reading material on  integration from page 198  of the book “Understanding the mathematics of electronics” by F.A.Wilson, Babani publishers.


Ultimately, the RMS value of periodical electrical waveforms is calculated by integration. In this document, there are equations presented which mean you don’t have to perform the integral every time for those waveforms. (it would take too long to do that in a fast working engineering environment)


Instantaneous values    notation


Know that v(t) means “the voltage at the instant of time, t”


…this distinguishes it from other values for “v” which could be eg an average value of v, or an RMS value of v.


Note that finding the RMS value of a periodic voltage waveform involves identifying a whole period of the voltage waveform, then over this period, one should square every single instantaneous value , v(t) , of voltage, and then find the average of these multiple squared values, and then the square route of the resulting value is taken to give the RMS value.


Here is the above expressed in mathematical notation….



…note that the ‘squigley’ line is the summation line, and means that  every u(t) value between 0 and T (one whole period), should be squared, and then each of these squared values should be added together cumulatively.


 [in the above expression “u(t)” is used instead of “v(t)”]


….Note that over a period of a voltage waveform,  there are obviously an infinite number of  instantaneous values of v(t), and squaring each one, and then adding those squared values up cumulatively sounds like a daunting proposition!….however, as you know, there are standard  methods available for doing this.


In most hard-switched SMPS’s, the method of integration does not have to be carried out to find eg RMS values, because there are standard expressions available for use, these expressions are given in this document…(they have been checked by doing the integration.)




Gain (dB) = 20 * log (Vout/Vin)

{log to base 10}


Complex numbers

1/(jwC) = -j/(wC)

j^2= -1

s = jw


Complex number representation of resistors, capacitors and inductors…also, RCL circuits and calculating the overall complex impedance.


SMPS equations.


DC transfer functions (Continuous Conduction Mode)



{( Vout + Vd) / Vin} = 1/(1-D)

-where Vd =Output diode forward voltage



(Vout +Vd) = D.Vin


BuckBoost & SEPIC:

(Vout +Vd)/Vin = D/(1-D)


Two Transistor Forward:

(Vout +Vd)/Vin = N.D

where N = Ns/Np



(Vout +Vd)/Vin = N.D/(1-D)

where N = Ns/Np


Full Bridge:

Consider as transformer isolated Buck & work it through.


Half Bridge:

Consider as transformer isolated Buck & work it through

..Remember “Vin” is effectively the input voltage divided by two.



Continuous mode hard switched equations:

D = t(ON)/T

t(ON) = D.T                            (FET on time)

t(OFF) = T(1-D)                     (FET off time)


{where “t(ON)” = FET on time; T = switching period}




Vout/Vin for the following…….


a…Non-inverting opamp   (Vout/Vin = {[Rfb/Rin[ + 1})

b…Inverting Opamp        -Rfb/Rin

c…Differential opamp



Complex numbers

Be able to calculate modulus and argument of a complex number, or at least know about it and know where to look it up.

Generally, the math software you use will actually do it for you…. in excel, Modulus of 1+j2 = IMABS(1,2)

also, Argument of 1+j2 = IMARGUMENT(1,2)

…know how to divide and multiply complex numbers in excel with IMDIV and IMPRODUCT (unless you are using a different math software)



mm^2to m^2…….


eg 10mm^2 = 10 {[m*10^(-3)]^2} = 10 m^2 * 10^(-6) = 10*10^(-6) m^2

…which equals 1*10^(-5) m^2, ie 10^(-5) m^2


Convert  radians to degrees (..multiply by 180/pi)




u(0) = permeability of free space = 4.pi*10^(-7) Henries per metre

{where “u(0)” = ‘mew’(0)…the Greek letter “mew”


permitivity of free space = 8.854e-12





Eg for graph of  i(t) against t, know that area beneath curve is the charge.


Know how to calculate the gradient of a straight line.

Know that an increasing gradient is positive, and a decreasing gradient is negative.


Know that  y = mx + c is the equation of a straight line of gradient m, and intercept on y axis, c

(the equation of a straight line can come in useful when finding out your delta B losses in non linear inductors. This is because the permeability gives the gradient of the B,H curve, so you find permeability at H=0, then draw your straight line up from there, to the next point, and so on, and so on)



Functions such as…..



Scientific Notation

This must be used.

Eg, if typing 1nF   into an excel spreadsheet, we must type “1e-9”, and not “0.000000001”.

….typing the latter will take far too long, and result in typo’s where the wrong number of zero’s is typed in.


so “1e-9” mains  “1 times 10 to the  minus 9” , or     “1 x 10^(-9)”, to see what it means,  first write number “1.”, remembering to  put the decimal point after it, then “jump” the decimal point 9 places to the left, because it is to the power “minus” 9. Put a zero under each jump, and you get 0.000000001




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