I am looking for effective voltage values for calculating power in a AC sine wave supplied circuit with a resistor under 2 conditions : full wave rectification; half wave rectification. I have searched the internet for the effective value of a half wave rectified voltage and have found nothing. I think, I have figured it out, but find puzzling the lack of this information.

Effective voltage of a rectified full wave sine is Vpeak/square root of 2 Effective and RMS are the same, average voltage has no use.

Effective voltage of half wave rectified sine wave is not RMS, but rather a RMS of the original sine wave /2. Or Vpeak/ 2 x square root of 2 RMS and average voltage have no use. RMS of a half wave rectified voltage is Vpeak/2 and this is not an effective voltage. What did I get right or wrong? thank you

  • \$\begingroup\$ I see my mistake, I was dividing by 2 the voltage instead of the power. The V RMS of half wave rectification = V peak/square root of 2/square root of 2 = V peak/2 The V RMS is the effective DC equivalent voltage. Could I say that this is the effective dc component of the voltage, and the AC component only modulates the dc, and doesn’t do any work? For example if I mix 1VAC (1.414 V peak) and 1.414 VDC, the effective voltage would be 1.414V, and the AC portion would not do any work. If I am modulating DC voltage with an AC voltage, am I increasing the effective voltage or it stays the same? \$\endgroup\$ – sparky Al Jan 2 '15 at 15:40
  • \$\begingroup\$ If you put 1 Vrms sine wave on top of a 1.414 VDC, you will get a steady power draw from the DC and a varying power draw averaging to the equivalent of 1 VDC from the AC. This simply adds because of superposition. So you would draw equivalent power to a 2.414 DC voltage. \$\endgroup\$ – caveman Jan 2 '15 at 15:52
  • \$\begingroup\$ Thank you for the reply. I have found a formula for mixing AC and DC. ... V RMS = square root of(VDC^2 + (VAC peak^2)/2) The above example would then be 1.73V. (the VAC peak is relative to its 0V before mixing). It appears that they don’t just add together, but adding the AC increases the overall V RMS. I think, that at the instant of the AC being -1.414V, and the DC +1.414, voltage would be 0V, and no current. I would be interested is there is a simple way to calculate the AC portion of rectified voltage. \$\endgroup\$ – sparky Al Jan 2 '15 at 20:03
  • \$\begingroup\$ Ah, you are right. I messed up in the same way you did earlier. You will get the sum of the powers of the two AC and DC. So once you cancel the R out: Vdc^2 + Vac^2 = Veq^2. \$\endgroup\$ – caveman Jan 2 '15 at 21:37
  • \$\begingroup\$ The DC portion is just the average value. When you subtract that off at every point, you get the AC. If you want to know the frequencies that exist in the AC portion, you have to use Fourier analysis. \$\endgroup\$ – caveman Jan 2 '15 at 21:39

Let's clarify what "effective" voltage means. First of all let's note that this is only really used to describe time-varying periodic voltages. We can imagine taking that voltage and driving across some resistor, R. Since this is a periodic wave it will on average consume some set amount of power.

If we take the same resistor and drive a particular DC voltage, we will consume the same amount of power. This voltage is the Effective Voltage. It is independent of the resistor used. Different resistors will give different power consumption, but for a given periodic voltage input, the effective voltage is the same.

If we go to work the problem out for determining what this effective voltage is for any wave, it effectively comes down to performing the following steps. I recommend working this out from V^2/R to understand why these steps are correct.

  1. Divide the period up into a bunch of small samples.
  2. Square the voltage value at each sample.
  3. Average up all these squared values.
  4. Take the square root of this average to get an approximation of the effective value.

As you do this with more and more samples (i.e. smaller time between each sample), you approach the effective value. This sequence is called the root-mean-square, or RMS. In other words, RMS Voltage is Effective Voltage.

Now, here is where people get confused. If you go and do the math, you can calculate analytically that for a sine wave $$V_{RMS} = \frac{V_{peak}}{\sqrt 2}$$ This is only true for a sine wave. For most other waves, you simply have to go do the math.

However, for some waves, like a full- and half-wave rectified sine waves, you can use the definition to determine the RMS voltage more simply.

For a full wave rectified wave driven across a resistor, it can be realized that a resistor doesn't care if a voltage is positive or negative, just what the magnitude is. The power is the same as a regular sine wave, therefore the RMS voltage is the same.

For a half wave rectified wave driven across a resistor, again, it can be worked out that the power consumed is 1/2 of the typical sine wave. The power is 1/2 the regular sine wave, but voltages are squared to get power, so:

$$\frac{P_{sine}}{P_{half}} = 2 = \frac{{V_{{RMS}_{sine}}}^2}{{V_{{RMS}_{half}}}^2}$$ which gives $$\frac{V_{{RMS}_{sine}}}{V_{{RMS}_{half}}} = \sqrt 2 => {V_{{RMS}_{half}}} = \frac{V_{{RMS}_{sine}}}{\sqrt 2} = \frac{V_{peak}}{2}$$

A bit more complicated, but it's easier than calculus.

  • \$\begingroup\$ I see my mistake, I was halving the voltage instead of the power. Thank you \$\endgroup\$ – sparky Al Jan 2 '15 at 15:44

If by "effective" you mean the voltage (and current) which you can put in the equation:

$$ P = IV $$

or derivatives, which can be obtained by substituting Ohm's law for \$I\$ or \$V\$:

$$ P = I^2 R\\ P = V^2 / R $$

and get electrical power, then RMS is that "effective" value.

Calculating the power of a sine waveform that has all its positive (or negative) peaks lopped off is easy: it's half, because it's doing the same work but only half of the time. So, if you have an AC voltage that makes 1W of power in a resistor, and then you remove all the negative voltages (by half-rectification), you will then get 0.5W of power. Assuming an ideal rectifier, of course.

Voltage being proportional to the square of power (\$P=V^2/R\$), then the RMS voltage of a half-rectified sine wave is less than an ordinary sine wave by a factor of \$\sqrt{2}\$.

  • \$\begingroup\$ RMS of half wave is V peak/2 \$\endgroup\$ – sparky Al Jan 2 '15 at 3:56
  • \$\begingroup\$ RMS of half wave is V peak/2, I don't think, that is an effective voltage. Effective voltage is a half of full wave rectification, so the formula for half wave would be V peak/2 x square root of 2. \$\endgroup\$ – sparky Al Jan 2 '15 at 4:03
  • \$\begingroup\$ I see my mistake, I was halving the voltage instead of the power. Thank you \$\endgroup\$ – sparky Al Jan 2 '15 at 15:44

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