# How to find the average value of the half rectified sine wave

The question is as follows:  I get part (a), but how do they get that answer for part (b)? The answers are stated below the question? How do they get the average value of the half rectified sine wave to be (1/pi)Vs-Vd/2 ?

• Did you obtain the analytical expression for the output instantaneous voltage of the rectifier? Did you attempt finding the average value using integrals? Here in Electronics Stack Exchange, you should post your work. Sep 21, 2021 at 22:31

The analytical expression of the average value isn't that complicated, nor it is hard to obtain, so I'm not sure why the textbook showed an approximated expression.

I'll derive the exact expression using the piecewise-linear model of the diode, which is more general than the simplified model and the ideal model. You can read about those models/approximations as well as Shockley's/exponential model here. In this model, the blocking/OFF/reverse-bias region is for $$\v \lt V_\gamma\$$, while the conducting/ON/forward-bias region is for $$\i \gt 0\$$; the diode is mathematically modeled by the following simultaneous equations:

$$\v_\text{D}(t) = V_\gamma + R_\text{D} \, i_\text{D}(t) \iff i_\text{D}(t) \ge 0 \tag*{}\$$

$$\v_\text{D}(t) \le V_\gamma \iff i_\text{D}(t) = 0 \tag*{}\$$

For a given diode, $$\V_\gamma > 0\$$ and $$\R_\text{D} > 0\$$ are known constants. If the real diode you’re modeling is a silicon diode, then $$\V_\gamma = 0.7 \text{ V}\$$, while if it's a germanium diode, then $$\V_\gamma = 0.3 \text{ V}\$$.

The circuit diagram of the rectifier is: Figure 1. Circuit diagram of single-phase uncontrolled rectifier, with a purely resistive load, and a sinusoidal instantaneous input voltage. Image source: own.

The output instantaneous voltage is:

$$\v_\text{out}(t) = \begin{cases} 0 \text{ V} & \text{, if k T + \le t \le k T + t_1} \\ \dfrac{R}{R + R_\text{D}} (v_\text{in}(t) - V_\gamma) & \text{, if k T + t_1 \le t \le k T + t_2} \\ 0 \text{ V} & \text{, if k T + t_2 \le t \le k T + T} \end{cases} \tag 1\$$

where $$\k\$$ is any integer number, and:

$$\v_\text{in}(t) = V_\text{m} \sin{(\omega t)}, \tag 2\$$

$$\t_1 = \dfrac{\sin^{-1}{(V_\gamma/V_\text{m})}}{\omega}, \tag 3\$$

$$\t_2 = \dfrac{\pi - \sin^{-1}{(V_\gamma/V_\text{m})}}{\omega}, \tag 4\$$

$$\T = \dfrac{2 \pi}{\omega}, \tag 5\$$

with $$\\omega\$$ and $$\V_\text{m} > V_\gamma\$$ known constants.

Note that what I call $$\V_\text{m}\$$ and $$\V_\gamma\$$ is what your textbook calls $$\V_\text{S}\$$ and $$\V_\text{D}\$$.

If you want a proof of the above, read this.

Now let's find the average value $$\V_\text{out,avg}\$$ of $$\v_\text{out}\$$. We start with:

\\begin{align} V_\text{out,avg} &= \dfrac{1}{T} \displaystyle\int_0^T v_\text{out}(t) \, \mathrm dt \\ &= \dfrac{1}{T} \left( \displaystyle\int_0^{t_1} 0 \, \mathrm dt + \left( \dfrac{R}{R + R_\text{D}} \right) \displaystyle\int_{t_1}^{t_2} (v_\text{in} - V_\gamma) \, \mathrm dt + \displaystyle\int_{t_2}^T 0 \, \mathrm dt \right) && \text{substituting (1)} \\ &= \dfrac{1}{2 \pi/\omega} \dfrac{R}{R + R_\text{D}} \left( V_\text{m} \displaystyle\int_{t_1}^{t_2} \sin{(\omega t)} \, \mathrm dt - V_\gamma \displaystyle\int_{t_1}^{t_2} \, \mathrm dt \right) && \text{substituting (5) and (2)} \\ &= \dfrac{\omega}{2 \pi} \dfrac{R}{R + R_\text{D}} \left( V_\text{m} \left[ -\dfrac{\cos{(\omega t)}}{\omega} \right|_{t_1}^{t_2} - V_\gamma \left[ t \right|_{t_1}^{t_2} \right) \\ &= \dfrac{\omega}{2 \pi} \dfrac{R}{R + R_\text{D}} \left( -\dfrac{V_\text{m}}{\omega} \left[ \cos{(\omega t_2)} - \cos{(\omega t_1)} \right] - V_\gamma \left[ t_2 - t_1 \right] \right) \\ &= \dfrac{\omega}{2 \pi} \dfrac{R}{R + R_\text{D}} \left( -\dfrac{V_\text{m}}{\omega} \left[ \cos{ \left( \omega \dfrac{\pi - \sin^{-1}{(V_\gamma/V_\text{m})}}{\omega} \right) } - \cos{ \left( \omega \dfrac{\sin^{-1}{(V_\gamma/V_\text{m})}}{\omega} \right) } \right] - \right. \cdots \\ &\cdots \left. - V_\gamma \left[ \dfrac{\pi - \sin^{-1}{(V_\gamma/V_\text{m})}}{\omega} - \dfrac{\sin^{-1}{(V_\gamma/V_\text{m})}}{\omega} \right] \right) && \text{substituting (3) and (4)} \\ &= \dfrac{\omega}{2 \pi} \dfrac{R}{R + R_\text{D}} \left( -\dfrac{V_\text{m}}{\omega} \left[ \cos{ \left( \pi - \sin^{-1}{[V_\gamma/V_\text{m}]} \right) } - \cos{ \left( \sin^{-1}{[V_\gamma/V_\text{m}]} \right) } \right] - \right. \\ &\cdots \left. - \dfrac{V_\gamma}{\omega} \left[ \pi - \sin^{-1}{(V_\gamma/V_\text{m})} - \sin^{-1}{(V_\gamma/V_\text{m})} \right] \right) \\ &= \dfrac{1}{2 \pi} \dfrac{R}{R + R_\text{D}} \left( -V_\text{m} \left[ \cos{ \left( \pi - \sin^{-1}{[V_\gamma/V_\text{m}]} \right) } - \cos{ \left( \sin^{-1}{[V_\gamma/V_\text{m}]} \right) } \right] - V_\gamma \left[ \pi - 2 \sin^{-1}{(V_\gamma/V_\text{m})} \right] \right) \tag 6 \end{align}\

Recalling the following trigonometric identities (source, source [reflection of cosine in π/2], source):

\\begin{align} \sin^{-1}{(x)} + \cos^{-1}{(x)} &= \dfrac{\pi}{2} \\ \implies \cos^{-1}{(x)} &= \dfrac{\pi}{2} - \sin^{-1}{(x)} \\ \implies \dfrac{\pi - 2 \sin^{-1}{(x)}}{2} &= \cos^{-1}{(x)} \\ \implies \pi - 2 \sin^{-1}{(x)} &= 2 \cos^{-1}{(x)}, \tag 7 \end{align}\

$$\\cos{(\pi - \theta)} = -\cos{(\theta)}, \tag 8\$$

$$\\cos{(\sin^{-1}{[x]})} = \sqrt{1 - x^2}, \tag 9\$$

and substituting equations (7) to (9) in (6), we get:

\\begin{align} V_\text{out,avg} &= \dfrac{1}{2 \pi} \dfrac{R}{R + R_\text{D}} \left( -V_\text{m} \left[ -\cos{ \left( \sin^{-1}{[V_\gamma/V_\text{m}]} \right) } - \sqrt{1 - (V_\gamma/V_\text{m})^2} \right] - V_\gamma \left[ 2 \cos^{-1}{(V_\gamma/V_\text{m})} \right] \right) \\ &= \dfrac{1}{2 \pi} \dfrac{R}{R + R_\text{D}} \left( -V_\text{m} \left[ -\sqrt{1 - (V_\gamma/V_\text{m})^2} - \sqrt{\dfrac{V_\text{m}^2 - V_\gamma^2}{V_\text{m}^2}} \right] - V_\gamma 2 \cos^{-1}{[V_\gamma/V_\text{m}]} \right) && \text{substituting (9)} \\ &= \dfrac{1}{2 \pi} \dfrac{R}{R + R_\text{D}} \left( -V_\text{m} \left[ - 2 \dfrac{\sqrt{V_\text{m}^2 - V_\gamma^2}}{V_\text{m}} \right] - V_\gamma 2 \cos^{-1}{[V_\gamma/V_\text{m}]} \right) \tag*{} \end{align}\

$$\\implies \boxed{V_\text{out,avg} = \dfrac{1}{\pi} \dfrac{R}{R + R_\text{D}} \left( \sqrt{V_\text{m}^2 - V_\gamma^2} - V_\gamma \cos^{-1}{[V_\gamma/V_\text{m}]} \right) \quad \text{, piecewise-linear model}} \tag {10}\$$

If instead we use the simplified model (i.e. we set $$\R_\text{D} = 0 \text{ } \Omega\$$), then equations (1) and (10) simplify to:

$$\v_\text{out}(t) = \begin{cases} 0 \text{ V} & \text{, if k T + \le t \le k T + t_1} \\ v_\text{in}(t) - V_\gamma & \text{, if k T + t_1 \le t \le k T + t_2} \\ 0 \text{ V} & \text{, if k T + t_2 \le t \le k T + T} \end{cases} \tag {11}\$$

$$\\boxed{V_\text{out,avg} = \dfrac{1}{\pi} \left( \sqrt{V_\text{m}^2 - V_\gamma^2} - V_\gamma \cos^{-1}{[V_\gamma/V_\text{m}]} \right) \quad \text{, simplified model}} \tag {12}\$$

If instead we use the simplified model (which your textbook seems to have used) and further assume that $$\V_\gamma \ll V_\text{m}\$$, then $$\V_\text{m}^2 - V_\gamma^2 \approx V_\text{m}^2 - 0 = V_\text{m}^2\$$ and $$\\cos^{-1}{(V_\gamma/V_\text{m})} \approx \cos^{-1}{(0/\infty)} = \pi/2\$$, thus substituting these two approximations in equation (12), we get:

\\begin{align} V_\text{out,avg} &\approx \dfrac{1}{\pi} \left( \sqrt{V_\text{m}^2} - V_\gamma \dfrac{\pi}{2} \right) \\ &= \dfrac{1}{\pi} V_\text{m} - \dfrac{\pi}{2} \dfrac{1}{\pi} V_\gamma \tag*{} \end{align}\

$$\\implies \boxed{V_\text{out,avg} \approx \dfrac{V_\text{m}}{\pi} - \dfrac{V_\gamma}{2} \quad \text{, simplified model with V_\gamma \ll V_\text{m}}} \tag {13}\$$

which is the formula your textbook shows.

If instead we use the ideal model (i.e. we set $$\R_\text{D} = 0 \text{ } \Omega\$$ and $$\V_\gamma = 0 \text{ V}\$$), then equations (1), (3), (4) and (10) simplify to:

$$\v_\text{out}(t) = \begin{cases} 0 \text{ V} & \text{, if k T + \le t \le k T + t_1} \\ v_\text{in}(t) & \text{, if k T + t_1 \le t \le k T + t_2} \\ 0 \text{ V} & \text{, if k T + t_2 \le t \le k T + T} \end{cases} \tag {14}\$$

$$\t_1 = 0 \text{ s} \tag {15}\$$

$$\t_2 = \dfrac{\pi}{\omega} = \dfrac{T}{2} \tag {16}\$$

$$\\boxed{V_\text{out,avg} = \dfrac{V_\text{m}}{\pi} \quad \text{, ideal model}} \tag {17}\$$

As a numerical example, below is shown a plot in Wolfram Mathematica, using the piecewise-linear model,of the input instantaneous voltage (eq. (2)), output instantaneous voltage (eq. (1)), average output voltage (eq. (12)) and RMS output voltage (derived here), from $$\t = 0 \text{ s}\$$ to $$\t = 2T\$$, where $$\V_\text{m} = 4 \text{ V}\$$, $$\\omega = 2 \pi \text{ rad/s}\$$, $$\R = 10 \text{ } \Omega\$$, $$\V_\gamma = 0.7 \text{ V}\$$, and $$\R_\text{D} = 0.1 \text{ } \Omega\$$. Figure 2. Waveforms of input instantaneous voltage, output instantaneous voltage, average output voltage and RMS output voltage, of a single-phase uncontrolled rectifier with a sinusoidal input instantaneous voltage and a purely resistive load. Image source: own.

Below we compare eq. (12) (simplified model) and eq. (13) (simplified model with $$\V_\gamma \ll V_\text{m}\$$), from $$\V_\text{m} = V_\gamma\$$ to $$\V_\text{m} = 480 \sqrt{2} \text{ V}\$$, where $$\V_\gamma = 0.7 \text{ V}\$$; observe the error is extremely small: Figure 3. Waveforms of average output voltage using the simplified model and average output voltage using the simplified model with $$\V_\gamma \ll V_\text{m}\$$. Image source: own.

• I'd not say that the book uses an incorrect formula, it's an approximate one. That's a very common engineering practice, the main point behind is improving comprehension of influence of involved quantities. Looking at your exact formula one hardly understands what's the role of Vg all on all. On the other hand the approximated one clearly states: "You loose half Vg in the average". Clearly approximation has an error and applicability should be stated and understood. In this case the radicals and arcocosine one gives 3.4762V while the simple one 3.4697V, Identical to all practical purposes Sep 22, 2021 at 5:29
• True; I've fixed it. But still we need to derive the equation (your comment doesn't explain the why, it only explains what the formula says), which I've also added now. Many thanks for the feedback! Sep 22, 2021 at 7:15
• Hi Alejandro :) deriving it it's plain sailing, you have half sine integrated over a full period which gives $\frac{V_\text{m}}{\pi}$ Than you have to subtract a quasi rectangle at the base of the sine, it's Vg high, nearly T/2 long and integrated over T is $\frac{V_\gamma \times T/2}{T}=\frac{V_\gamma}{2}$ The error is than the two small quasi-triangles at the beginning and end (sines near zero actually). Otherwise you might try some Taylor expansion of your exact result...(I see now you've already gonr that way) Sep 22, 2021 at 10:12

The frequency of the signal is doubled, so the period becomes $\pi$ The rectified voltage is a diode drop lower than the original sine wave.

• also a sine pumped current half wave the average V of a half wave sine is peak divided Vs/pi . to translate pi period to Vs/pi =Vavg +Vd, you have to integrate the half sine. May 28, 2017 at 20:22
• until it gets loaded by RC then it takes 8RC/T to get 10% ripple.. May 28, 2017 at 20:26
• @kva No, the frequency of the output is not doubled. The rectifier is half-wave, not full-wave. Sep 15, 2021 at 23:20