When I have either a rectified signal (like the unfiltered rectified sine wave out of a diode bridge) or say a fast pulsed signal feeding a LED and resistor, in which both the DC component and the RMS component are high, which value RMS or DC should I use to make my calculations?

In some cases Ive read books which use the DC (mean) values to analyze voltage or current values, yet they use the RMS value to calculate power, and in other cases they use the mean (DC) values of voltage and current to calculate power.

Im confused as to when should I use DC values for current, voltage and power and when should I use RMS when both are present and with great magnitude. I understand the definitions of both RMS and Average and their integral formulas, in which RMS is the mean squared and produces the same "heat" as a DC signal with the same value. The problem is not the definition, the problem is when to apply RMS values or DC values when I have both present in a signal.

Let me give you an example to clarify my question, say Im pulsing an LED in series with a resistor with a fast signal (0 to 5v pulse, 20% duty cycle). The RMS voltage and current values of the LED may be 1.5V and 6mA, and the DC voltage and current say 1.7V and 4.5mA, which should I consider the voltage and current of each LED: 1.5V and 6mA, or 1.7V and 4.5mA?, what about if I want to calculate the power dissipated on both the LED and the resistor, should I use the RMS voltage and current values or the DC values?


Short answer: in most cases RMS values should be considered to calculate power in a component, however if there is a need to calculate power supplied by a DC source, then the mean or DC components should be used.

An important distinction should be made: When I first asked this question I wrongfuly thought that a Multimeter set to AC volts or amps displayed the RMS value of a signal regardless of whether DC was present or not, so when both DC and AC were present, I was confused on which value to use for example to calculate power, instead, when set to AC, a multimeter displays the RMS value of the AC component of the signal only, however, if you want the RMS value of a signal in which both DC and AC are present, then you should measure both the AC and DC component in a multimeter and \$V_{RMS}=\sqrt{V_{DC}^2+V_{RMS_{AC}}^2}\$ should be used. It is obvious that if there is no DC present, the mean value would be zero and the value displayed by the multimeter set to AC is in fact the RMS value of the signal, .

The RMS value of a signal is

\$RMS=\sqrt{\frac{1}{T}\int_{0}^{T} f(t)^2dt}\$

This is the value that should be used, for example in a rectified signal through an LED.

The contribution of both the DC and AC components can be easily seen if the analysis is focused on harmonics, then, power is calculated as:

$$P=V_{DC}I_{DC}+\Re \{\frac{1}{2}\sum_{n=1}^\infty V_nI_n^*\}$$


\$V_{DC}\$ and \$I_{DC}\$ are the DC voltage and current


\$V_n\$ and \$I_n\$ are phasors and include the peak voltage and current of the nth harmonic along with its phase.

In the case where only one frequency is present, then \$P\$ is simply

$$P=V_{DC}I_{DC}+\Re \{\frac{1}{2} V_pI_p^*\}$$

Thus, the power in for example a resistor, is due to both the DC + AC component.

When calculating the power being supplied by a DC source, the DC voltage of the source and current through the source must be considered to calculate the power being delivered by the source, same thing happens with an AC source, but in that case the AC voltage and AC current should be considered.

Regarding current, the RMS value is



\$I_{DC}\$ is the DC component and \$I_n\$ is the peak value of the nth harmonic, again if only the fundamental is present, the equation reduces to:


The RMS voltage is calculated in a similar way, thus, in general, in order to calculate power in a component in which both the DC component and the AC component are present, we must consider the RMS value.

Consider the following example of 2 resistors in series, there is also a 10V AC component on top of a 12V DC component feeding the circuit, I also added a power meter and a current-voltage probe.

enter image description here

The Peak voltage is clearly half of the peak to peak voltage, so


The DC voltage is


The RMS voltage is:


Which agrees with the value displayed in the yellow box in the picture

The current can be calculated the same way, its value is

$$I_{RMS}=6.95 mA$$

The power is simply \$P=V_{RMS}I_{RMS}=48.3mW\$ which agrees with the power meter, (Note: I have noticed that in Multisim the voltage and current values displayed by the probes are not 100% accurate, as opposed to the values displayed by the Multimeter which are more precise, this is why theres a slight difference between the calculated power and the power displayed by the power meter)

Note that the power could have been computed using \$P=V_{DC}I_{DC}+\Re \{\frac{1}{2} V_pI_p^*\}\$, and the results would be the same.

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In order to calculate average power, you need to integrate the instantaneous power (instantaneous voltage times instantaneous current) over some time interval, which yields total energy for that interval, and then divide by the time interval.

The details of doing this depend on the circuit you're considering. For example, in a purely resistive circuit, the instantaneous power is proportional to the square of the voltage (or the square of the current), and so the average power can be computed directly from the RMS voltage (or current).

In nonlinear circuits (including those with diodes), some part of the power dissipation might be proportional to current only; in this case, you could use the average value of the current to compute the average power.

But in general, you need to work out the details for your specific circuit and then determine which simplifications apply.

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  • \$\begingroup\$ So basically what you are suggesting is that the fool-proof method of calculating power is to just integrate the peak voltage vs peak current over a period T, and forget about if should use RMS or DC or whatever. Fair enough, but then what about the regular voltage and current calculations, when should one use the DC or RMS value? \$\endgroup\$ – S.s. Jun 8 '14 at 21:02
  • \$\begingroup\$ No, not the peak values, but the actual moment-to-moment values (waveforms) of the voltage and current. This is the definition of average power. Anything else is a simplification that applies in certain circumstances. For example, we know that the RMS value of a sinewave it 0.707 times its peak value because we use sinewave power in so many ways. \$\endgroup\$ – Dave Tweed Jun 8 '14 at 21:32
  • \$\begingroup\$ yes sorry, I meant the v(t) and i(t) \$\endgroup\$ – S.s. Jun 8 '14 at 21:39

In the case of the pulsed current to the LED, the power dissipated in the resistor will be \$I_{RMS}^2 \cdot R\$ the (RMS current) squared, multiplied by the resistance.

More generally, in a linear or nonlinear circuit without reactance (no capacitors or inductors), where current is in phase with voltage (whatever the waveforms), the mean power is \$I_{RMS} \cdot V_{RMS} \$, and that is the case for the LED as well as for the resistor.

If the circuit has inductor or capacitors (and I think if it is not time-invariant) then you have to integrate the instantaneous voltages and currents to find the power. Many oscilloscopes can perform that calculation, but you have to be careful to integrate over an integral number of periods or you may get a correct answer (the mean power over the integration time you happened to pick) that does not reflect the mean power over many cycles.

I don't think you ever want to use average current or voltage to calculate power, excepting trivial situations where it is constant.

Average current is important in a rapidly (much more than about 10Hz) pulsed LED because it provides an indication of the perceived light output (by a human eye). Since the LED voltage increases with current, and output is usually more-or-less proportional to current, you usually get maximum brightness for a given power dissipation by running the LED at DC.

In other words, an LED run at 0.1A with 20% duty cycle will appear about as bright as the same LED run at 20mA with 100% duty cycle, but the first one will get hotter (and will not last as long).

Average voltage may also be important if you want to interpret what a non-true-RMS meter is reading (such meters measure the average voltage or current, but display a voltage or current reading that is higher by a factor \$ \frac{\pi}{\sqrt 8}\$, so they show the correct value for a sinusoidal wave). They can be very wrong for other waveforms.

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