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?
Best Answer
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:
$$P=V_{DC}I_{DC}+\Re \{\frac{1}{2}\sum_{n=1}^\infty V_nI_n^*\}$$
Where:
\$V_{DC}\$ and \$I_{DC}\$ are the DC voltage and current
and
\$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
$$I_{RMS}=\sqrt{I_{DC}^2+\frac{1}{2}\sum_{n=1}^{\infty}I_n^2}$$
Where
\$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:
$$I_{RMS}=\sqrt{I_{DC}^2+\frac{1}{2}I_p^2}$$
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.
The Peak voltage is clearly half of the peak to peak voltage, so
$$V_p=9.94/2=4.97V$$
The DC voltage is
$$V_{DC}=6V$$
The RMS voltage is:
$$V_{RMS}=\sqrt{6^2+\frac{1}{2}4.97^2}=6.95V$$
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.