# average calculation vs rms calculation in rectifiers

In rectifier $$\R_{load}\$$ we calculate $$\P_{load}=V_{avg}\cdot I_{avg}\$$ why cant we take $$\V_{rms}\$$ and $$\I_{rms}\$$ as we are taking like in electrical calculations and also what is the use of calculating rms value in rectifier $$\R_{load}\$$? • I think you are taking something out of context. Are you trying to find average voltage output of a capacitor smoothed rectifier? Apr 4, 2021 at 2:20
• kollipara, I'm just projecting for a moment. Bear with me. Are you talking about a situation where there is a bridge diode rectifier and a resistor as the load?
– jonk
Apr 4, 2021 at 2:31
• I'm saying that load dc power calculation in rectifier r-load is vavg *iavg why couldn't we take rms values like as in electrical calculations i.e 1phase ac supply with rload Apr 4, 2021 at 2:41
• @kollipararajkumar What exactly is the meaning of "rectifier r-load" to you? Edit your question (you can do that) and use the schematic editor that is available. Draw out the schematic. Yes, I understand "1phase ac supply" and I understand "rectifier" and I understand the words themselves. I just don't understand how you put them all together. It's not working in English. Best is to shift to pictures. That will bypass any language issues. A useful answer will be based upon the details. Your question is right to ask in many circumstances. But in a few, there is a different answer.
– jonk
Apr 4, 2021 at 2:49
• @kollipararajkumar If you want, I can completely ignore your situation (because you've not communicated it sufficiently well) and propose for your consideration a case where a "rectifier r-load" and "1phase ac supply" would, in fact, find that "pload=vavg*iavg" as an acceptable approximation (though not perfect.)
– jonk
Apr 4, 2021 at 3:05

In the case of resistive loads (loads that cannot temporarily store energy), the power is the product of its current and the voltage across its terminals. For cases where this current and terminal voltage is steady over time, this is easy since they don't vary and you can just multiply them out.

If you are only measuring one of them, whether just terminal voltage or just current through the resistor, then you have to infer the other facet. That's done easily enough and it results in using either the voltage squared or else the current squared. That's based upon various re-arrangements of the very familiar Ohm's Law.

For cases where the current and terminal voltage varies over time, you have to keep track of all the little product-bits over time (and divide by that time) to get the average (mean) power.

There is an instantaneous power, too. But it's for only an instant of time. In continuous systems, we are almost always interested in how that averages out for the period of interest. (In the case of fuses there is more of an interest in those instantaneous values.)

Note that for cases where the device's current/voltage varies over time and where we want to calculate the average power that results, we can't use the average of the device's terminal voltage or its current in order to get its average power.

For example, suppose the $$\10\:\Omega\$$ resistor's terminal voltage is $$\1\:\text{V}\$$ for $$\1\:\text{s}\$$, then $$\3\:\text{V}\$$ for $$\2\:\text{s}\$$, and then $$\1\:\text{V}\$$ for $$\1\:\text{s}\$$ more. We want the average power for those $$\4\:\text{s}\$$. The average voltage is exactly $$\\frac{1\:\text{V}\cdot 1\:\text{s}+3\:\text{V}\cdot 2\:\text{s}+1\:\text{V}\cdot 1\:\text{s}}{4\:\text{s}}=2\:\text{V}\$$. But the average power is $$\\frac{\left(1\:\text{V}\right)^2\cdot 1\:\text{s}+\left(3\:\text{V}\right)^2\cdot 2\:\text{s}+\left(1\:\text{V}\right)^2\cdot 1\:\text{s}}{4\:\text{s}\cdot 10\:\Omega}=500\:\text{mW}\$$. If we wanted a single value to represent the changing terminal voltage, where we could just use $$\P=\frac{V^2}{R}\$$ then we'd need $$\V=2.236068 \:\text{V}\$$!

Where would that come from??

RMS voltage and RMS current is really about creating just that value -- a value that somehow matches up with the DC-equivalent heating that results from a device's varying voltage/current. To compute the power, you want the DC-heating equivalent. In the above example, that's $$\V=2.236068 \:\text{V}\$$.

So let's just define $$\V_\text{RMS}\$$ as being the DC-heating equivalent of some $$\V_\text{DC}\$$, with the only caveat being that when we say RMS we are implying that things are ever-changing and probably not DC. But for the purposes of average power computation with a resistive load, $$\V_\text{RMS}\equiv V_\text{DC}\$$. In the case of DC, then these two things are exactly the same thing.

In general, this DC-heating equivalent, for voltage, is:

$$V_\textrm{RMS}=\sqrt{\frac1{T}\int_0^TV_\textrm{t}^2\:\textrm{d} t}$$

In the earlier example, we can compute this:

$$V_\textrm{RMS}=\sqrt{\frac1{4\:\text{s}}\cdot\left[\left(1\:\text{V}\right)^2\cdot 1\:\text{s}+\left(3\:\text{V}\right)^2\cdot 2\:\text{s}+\left(1\:\text{V}\right)^2\cdot 1\:\text{s}\right]}=\sqrt{5}\:\text{V}$$

(Note that the integral here was just the sum of each part.)

In the case of a sinusoidal AC voltage and a single diode rectifier (so that half of the cycles are missed), we find that $$\V_\text{AVG}=\frac1{2\pi}\int_0^\pi V_\text{PK}\sin x\:\textrm{d}x\$$. (The diode drop is considered zero here.) The goal for analysis is to pick a convenient period of time that makes the calculations easier. In this case, I chose one cycle of exactly $$\2\pi\$$ duration (timeless, but the idea is the same.) But the integral only goes from 0 to $$\\pi\$$, as you can see, because half the cycles are skipped. It turns out that the solution for this is $$\V_\text{AVG}=\frac{V_\text{PK}}{\pi}\$$.

So you really do have the right answer for the average voltage. But by now you also realize that, like a stopped clock that is right twice per day, the average voltage is almost never the right one to use with time varying signals in order to compute the power, directly. You should use the RMS values, instead.

Also in the case of a sinusoidal AC voltage and a single diode rectifier (so that half of the cycles are missed), we find $$\V_\textrm{RMS}=\sqrt{\frac1{2\pi}\int_0^\pi \left[V_\text{PK}\sin x\right]^2\:\textrm{d} x}=\sqrt{\frac{V_\text{PK}^2}{2\pi}\int_0^\pi \sin^2 x\:\textrm{d} x}=\sqrt{\frac{V_\text{PK}^2}{2\pi}\cdot\frac{\pi}{2}}\$$ and that's obviously just $$\V_\text{RMS}=\frac{V_\text{PK}}{2}\$$. (In this particular case! Do not project this into other cases.)

So you also really do have the right answer for the RMS voltage. And this is the value to use for average power computations.

• Thank you sir i understand it clearly lastly i also have some doubt on average voltage,average current i know its meaning but i want to know where it used Apr 4, 2021 at 8:04
• @kollipararajkumar The average voltage or average current can be used where computations do not involve powers. So you can use the average voltage along with time to compute Webers for an inductance in some situations, for example. You'll find these times when you do find them. I cannot provide a glossary of every possible case.
– jonk
Apr 4, 2021 at 8:10

In rectifier $$\R_{load}\$$ we calculate $$\P_{load}=V_{avg}⋅I_{avg}\$$

That's incorrect. Consider the case without using an ideal diode rectifier. The power will be: -

$$I_{RMS}^2\cdot R_L$$

So, with a rectifier, only half the power is delivered to the load hence, the power will be: -

$$\dfrac{I_{RMS}^2}{2}\cdot R_L$$

Where the above RMS current value is the equivalent continuous RMS current should the diode be shorted.

If we use peak values instead we would have: -

$$\dfrac{I_{PK}^2}{(\sqrt2)^2\cdot2}\cdot R_L = \dfrac{I_{PK}^2}{4}\cdot R_L$$

This tells us that the true RMS value of a rectified sinewave current is $$\\dfrac{I_{PK}}{2}\$$

In other words, the power delivered to a load resistor via a rectifier diode has got nothing to do with average values (unless of course you then change those average values back to RMS or peak values of the applied sinewave).