average voltage

Average Voltage Tutorial

Having looked at the RMS Voltage value of an alternating waveform in a previous tutorial, we can now look at calculating another value using either the mid-ordinate rule or analytical rule to find a waveforms “average” or mean voltage.

The process used to find the Average Voltage of an alternating waveform is very similar to that for finding its RMS value, the difference this time is that the instantaneous values are not squared and we do not find the square root of the summed mean.

The average voltage (or current) of a periodic waveform whether it is a sine wave, square wave or triangular waveform is defined as: “the quotient of the area under the waveform with respect to time”. In other words, the averaging of all the instantaneous values along time axis with time being one full period, (T).

For a periodic waveform, the area above the horizontal axis is positive while the area below the horizontal axis is negative. The result is that the average or mean value of a symmetrical alternating quantity is zero because the area above the horizontal axis (the positive half cycle) is the same as the area below the axis (the negative half cycle) and cancel each other out in the sum of the two areas as a negative cancels a positive producing zero average voltage.

Then the average or mean value of a symmetrical alternating quantity, such as a sine wave, is the average value measured over only half a cycle since over a complete cycle the average value is zero regardless of the peak amplitude.

The electrical terms Average Voltage and Mean Voltage or or even average current, can be used in both an AC and DC circuit analysis or calculations. The symbols used for representing an average value are defined as: VAV or IAV.

Average Voltage Graphical Method

Again consider only the positive half cycle from the previous RMS voltage tutorial. The mean or average voltage of a waveform can be found again with a reasonable amount of accuracy by taking equally spaced instantaneous values.

The positive half of the waveform is divided up into any number of “n” equal portions or mid-ordinates. The width of each mid-ordinate will therefore be no degrees (or t seconds) and the height of each mid-ordinate will be equal to the instantaneous value of the waveform at that point along the x-axis of the waveform.

The Graphical Method

average voltage graphical method


Each mid-ordinate value of the voltage waveform is added to the next and the summed total, V1 to V12 is divided by the number of mid-ordinates used to give us the “Average Voltage”. Then the average voltage (VAV) is the mean sum of mid-ordinates of the voltage waveform and is given as:

average voltage definition


and for our simple example above, the average voltage is therefore calculated as:

average or mean voltage


So as before lets assume again that an alternating voltage of 20 volts peak varies over one half cycle as follows:

Voltage 6.2V 11.8V 16.2V 19.0V 20.0V 19.0V 16.2V 11.8V 6.2V 0V
Angle 18o 36o 54o 72o 90o 108o 126o 144o 162o 180o

The Average voltage value is therefore calculated as:

average voltage formula


Then the Average Voltage value using the graphical method is given as: 12.64 Volts.

Average Voltage Analytical Method

As said previously, the average voltage of a periodic waveform whose two halves are exactly similar, either sinusoidal or non-sinusoidal, will be zero over one complete cycle. Then the average value is obtained by adding the instantaneous values of voltage over one half cycle only. But in the case of an non-symmetrical or complex wave, the average voltage (or current) must be taken over the whole periodic cycle mathematically.

The average value can be taken mathematically by taking the approximation of the area under the curve at various intervals to the distance or length of the base and this can be done using triangles or rectangles as shown.

Approximation of the Area

approximation of the area under a curve


By approximating the areas of the rectangles under the curve, we can obtain a rough idea of the actual area of each one. By adding together all these areas the average value can be found. If an infinite number of smaller thinner rectangles were used, the more accurate would be the final result as it approaches 2/π.

The area under the curve can be found by various approximation methods such as the trapezoidal rule, the mid-ordinate rule or Simpson’s rule. Then the mathematical area under the positive half cycle of the periodic wave which is defined as V(t) = Vp.cos(ωt) with a period of T using integration is given as:

area under a curve by integration


Where: 0 and π are the limits of integration since we are determining the average value of voltage over one half a cycle. Then the area below the curve is finally given as Area = 2VP. Since we now know the area under the positive (or negative) half cycle, we can easily determine the average value of the positive (or negative) region of a sinusoidal waveform by integrating the sinusoidal quantity over half a cycle and dividing by half the period.

For example, if the instantaneous voltage of a sinusoid is given as: v = Vp.sinθ and the period of a sinusoid is given as: , then:

average voltage by integration


Which is therefore given as the standard equation for the Average Voltage of a sine wave as:

Average Voltage Equation

average voltage equation

Then the average voltage (VAV) of a sinusoidal waveform is determined by multiplying the peak voltage value by the constant 0.637, which is two divided by pi (π). The average voltage, which can also be referred to as the mean value, depends on the magnitude of the waveform and is not a function of either the frequency or the phase angle.

Referring to our graphical example above, the peak voltage, (Vpk) was given as 20 Volts. Using the analytical method the average voltage is therefore calculated as:

VAV = Vpk x 0.637 = 20 x 0.637 = 12.74 volts

Which is the same value as for the graphical method.

To find the peak value from a given average voltage value, just rearrange the formula and divide by the constant. For example, what is the sinusoidal peak value, Vpk if the average value is 65 volts.

Vpk = VAV ÷ 0.637 = 65 ÷ 0.637 = 102 volts

Note that multiplying the peak or maximum value by the constant 0.637 ONLY applies to sinusoidal waveforms.

Average Voltage Summary

Then to summarise. When dealing with alternating voltages (or currents), the term Average value is generally taken over one complete cycle, whereas the term Mean value is used for one half of the periodic cycle.

The average value of a whole sinusoidal waveform over one complete cycle is zero as the two halves cancel each other out, so the average value is taken over half a cycle. The average value of a sine wave of voltage or current is 0.637 times the peak value, (Vp or Ip. This mathematical relationship between the average values applies to both AC current and AC voltage.

Sometimes it is required to be able to calculate the value of the direct voltage or current output from a rectifier or pulse type circuit such as a PWM motor circuit because the voltage or current, although not reversing, is changing continuously. Since there are no phase reversals the average value is used and the RMS (root-mean-square) value is unimportant for this type of application.

The main differences between an RMS Voltage and an Average Voltage, is that the mean value of a periodic wave is the average of all the instantaneous areas taken under the curve over a given period of the waveform, and in the case of a sinusoidal quantity, this period is taken as one-half of the cycle of the wave. For convenience the positive half cycle is generally used.

The effective value or root-mean-square (RMS) value of the waveform is the effective heating value of the wave compared to a steady DC value and is the square root of the mean of the squares of the instantaneous values taken over one complete cycle.

For a pure sinusoidal waveform ONLY, both the average voltage and the RMS voltage (or currents) can be easily calculated as:

Average value = 0.637 × maximum or peak value, Vpk

RMS value = 0.707 × maximum or peak value, Vpk

rms and average voltage


One final comment about using Average Voltage and RMS Voltage. Both values can be used to represent the “Form Factor” of a sinusoidal alternating waveform. Form factor is defined as being the shape of an AC waveform and is the RMS voltage divided by the average voltage (form factor = rms value/average value).

So for a sinusoidal or complex waveform the form factor is given as: ( π/(2√2) ) which is approximately equal to the constant, 1.11. Form factor is a ratio and therefore has no electrical units. If the form factor of a sinusoidal waveform is known, then the average voltage can be found using the RMS voltage value and vice-versa.


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  • F

    It is a good platform for learning electronics for beginners

  • m

    In graphite which bonding is present
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    c) vanderwall
    d) both b and c

  • m

    which conductor has highest conductivity
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    b Aluminium
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    • M

      If pure silver is used then its conductivity is 105%.
      But copper has 100%conductivity,Aluminum hass 61%.

      This percentage are taken with respect to copper.

  • m

    The band gap of a semiconductor is 1.43ev . Its cutoff wavelength is

    Please solve by giving the formula

  • m

    what is the average value of triangular wave

    • Wayne Storr

      If the triangular waveform is symmetrical around the zero axis, then the average value measured from peak-to-peak is zero because over one full period the positive and negative areas cancel in the sum of the two areas. It may be periodic, but they are still two triangles and the area of a triangle is 1/2 x base x height. The RMS value however is calculated as Vpeak divided by (root 3) which equals 0.577Vp

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    Ankit Choudhury

    What is the effective value of the current represented by the equation
    I = 2 + 3 sinwt ?

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    Muhammad waseem mukhtar

    what is the average value of the alternating signal
    a) zero
    b) half
    c) both
    d) none of above

    • Wayne Storr

      a), the average value of an alternating sinusoidal wave taken over one full cycle is zero, as it has equal and opposite positive and negative areas.

  • S

    b) The Voltage Measuring Circui:
    The mains voltage across the load (low voltage side of 11kV/415V transformer) is divided by a factor 253 (115 % of rated voltage 220V) and shifted to a DC level of 2.5 V. This shifting enables to measure positive and negative voltages. Fig. 2, shows the voltage measuring circuit. Resistor R1 and R3 (which are equal 62.51 KΩ, 4.7 KΩ respectively) bydefault produces a 2V AC signal when the line voltage reaches 115% of nominal voltage as shown in “(1),”. VMax = 1.15 VRated (N2 / N1) R3 / (R3 + R1) (1) The 253V limit is there to leave 15% margin for overvoltages.
    The DC bias ladder R2, R3 (both are 4.7K Ω) positions the AC signal halfway up the analog to digital converter (ADC) voltage reference to enable measuring positive and negative voltages as shown in “(2),”. VDC = VCC R2 / (R2 + R3) (2) Then the sumation of two voltages is fed to an ADC input (AN1, AN2, and AN3 for phase 1, 2, 3 respectively) of the PIC18f4550.

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    Faisal masood


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