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What is the formula for finding the peak to peak AC ripple voltage across the load resistor for this LC full wave bridge rectifier circuit?

The rectifier output is a pulsating DC voltage which has both DC and AC components. The LC filter is to remake the rectified pulsating DC voltage to a pure DC voltage with some small AC ripple voltage across the load resistor.

The DC component across the resistor is found by the formula V(DC)=2V(p)/pi where V(p) is the peak rectified voltage, but I am unable to find the formula for finding the filtered peak to peak AC voltage across the resistor.

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    \$\begingroup\$ Why is the inductor so large and the capacitor so small? \$\endgroup\$
    – AnalogKid
    Aug 18, 2023 at 14:04

2 Answers 2

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This should be as this simulation ... C1 is not important enough.
So remain L/R filter and ondulation proportional to \$ R / (j*2*pi*f*L+R)\$ ... (f=100).

Here, you get about 20 V when input voltage is about 100 V peak, 50 Hertz.

EDIT: "I" is the "i" needed by MapleSoft to specify it is a "complex" operation.

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  • \$\begingroup\$ What is j in the formula? \$\endgroup\$
    – Alex
    Sep 5, 2023 at 7:44
  • \$\begingroup\$ "j" is also "i" ... These are used for complex numbers. \$\endgroup\$
    – Antonio51
    Sep 5, 2023 at 8:18
  • \$\begingroup\$ So I can ignore this "i" if I am interested in only calculating the magnitude of peak to peak voltage? \$\endgroup\$
    – Alex
    Sep 5, 2023 at 8:53
  • \$\begingroup\$ And the formula is now R/(R+2pifL) if I ignore "i"? \$\endgroup\$
    – Alex
    Sep 5, 2023 at 8:55
  • \$\begingroup\$ @Alex theoretically, one can't ignore the "i". The "real" value of this function is : R / (sqrt(R^2 + (2*pi *f *L) ^2) ) ("complex" numbers operation) \$\endgroup\$
    – Antonio51
    Sep 5, 2023 at 9:20
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Well, I'll provide an output expression for your filter. First, let's recall the Fourier series of the absolute sine-wave.

The function \$x\mapsto f(x):=|\sin x|\$ is even and \$\pi\$-periodic; therefore \$f\$ has a Fourier series of the form:

$$f(x)=\frac{\text{a}_0}{2}+\sum_{\text{k}\space\geq\space1}\text{a}_\text{k}\cos\left(2\text{k}x\right)\tag1$$

With:

$$\text{a}_\text{k}={2\over\pi}\int\limits_0^\pi f(x)\cos\left(2\text{k}x\right)\space\text{d}x={2\over\pi}\int\limits_0^\pi \sin\left(x\right)\cos\left(2\text{k}x\right)\space\text{d}x\tag2$$

It follows that:

$$\eqalign{\text{a}_\text{k}&={1\over\pi}\int\limits_0^\pi\left(\sin\bigl((1+2\text{k})x\bigr)+\sin\bigl((1-2\text{k})x\bigr)\right)\space\text{d}x\cr\\&={1\over\pi}\left({\cos\bigl((2\text{k}-1)x\bigr)\over 2\text{k}-1}-{\cos\bigl((2\text{k}+1)x\bigr)\over 2\text{k}+1}\right)\Biggr|_0^\pi\cr\\&={2\over\pi}\left({1\over 2\text{k}+1}-{1\over 2\text{k}-1}\right)\cr\\&=-{4\over\pi(4\text{k}^2-1)}\cr}\tag3$$

Therefore we have:

$$\left|\sin\left(x\right)\right|={2\over\pi}-{4\over\pi}\sum_{\text{k}\space\geq\space1}{\cos(2\text{k}x)\over 4\text{k}^2-1}\tag4$$

The transfer function of your circuit is given by:

\begin{equation} \begin{split} \text{Y}\left(\text{s}\right)&=\frac{\displaystyle\text{R}\space\text{||}\space\frac{\displaystyle1}{\displaystyle\text{sC}}}{\displaystyle\text{sL}+\left(\text{R}\space\text{||}\space\frac{\displaystyle1}{\displaystyle\text{sC}}\right)}\\ \\ &=\frac{\displaystyle\frac{\displaystyle\text{R}\cdot\frac{\displaystyle1}{\displaystyle\text{sC}}}{\displaystyle\text{R}+\frac{\displaystyle1}{\displaystyle\text{sC}}}}{\displaystyle\text{sL}+\frac{\displaystyle\text{R}\cdot\frac{\displaystyle1}{\displaystyle\text{sC}}}{\displaystyle\text{R}+\frac{\displaystyle1}{\displaystyle\text{sC}}}}\\ \\ &=\frac{\displaystyle\frac{\displaystyle\text{R}}{\displaystyle1+\text{sCR}}}{\displaystyle\text{sL}+\frac{\displaystyle\text{R}}{\displaystyle1+\text{sCR}}}\\ \\ &=\frac{\displaystyle\text{R}}{\displaystyle\text{R}+\text{sL}\left(1+\text{sCR}\right)}\\ \\ &=\frac{\displaystyle\text{R}}{\displaystyle\text{R}+\text{Ls}+\text{CLRs}^2} \end{split}\tag5 \end{equation}

Where \$\displaystyle\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$.

Taking the inverse Laplace transform, we can see:

\begin{equation} \begin{split} y\left(x\right)&=\mathscr{L}_\text{s}^{-1}\left[\text{Y}\left(\text{s}\right)\right]_{\left(x\right)}\\ \\ &=\mathscr{L}_\text{s}^{-1}\left[\frac{\displaystyle\text{R}}{\displaystyle\text{R}+\text{Ls}+\text{CLRs}^2}\right]_{\left(x\right)}\\ \\ &=\frac{2}{\sqrt{\text{L}\left(\text{L}-4\text{CR}^2\right)}}\cdot\exp\left(-\frac{x}{2\text{CR}}\right)\cdot\sinh\left(\frac{\sqrt{\text{L}\left(\text{L}-4\text{CR}^2\right)}}{2\text{CLR}}\cdot x\right) \end{split}\tag6 \end{equation}

We can write the voltage after the full bridge rectifier as follows:

\begin{equation} \begin{split} \text{v}_\text{i}\left(t\right)&=\left(\hat{\text{u}}_\text{i}-2\text{V}_\text{d}\right)\left|\sin\left(2\pi\text{f}t\right)\right|\\ \\ &=\left(\hat{\text{u}}_\text{i}-2\text{V}_\text{d}\right)\left({2\over\pi}-{4\over\pi}\sum_{\text{k}\space\geq\space1}{\cos\left(4\pi\text{k}\text{f}t\right)\over 4\text{k}^2-1}\right) \end{split}\tag7 \end{equation}

Where \$\hat{\text{u}}_\text{i}\space\left[\text{V}\right]\$ is the amplitude of the input voltage, \$\text{V}_\text{d}\space\left[\text{V}\right]\$ is the voltage drop across one diode.

Now, using the convolution property of the Laplace transform, we can write the output voltage:

\begin{equation} \begin{split} \text{V}_\text{o}\left(t\right)&=\int\limits_0^t\mathscr{L}_\text{s}^{-1}\left[\text{V}_\text{i}\left(\text{s}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\text{Y}\left(\text{s}\right)\right]_{\left(t-\tau\right)}\space\text{d}\tau\\ \\ &=\int\limits_0^t\text{v}_\text{i}\left(\tau\right)\cdot y\left(t-\tau\right)\space\text{d}\tau\\ \\ &=\int\limits_0^t\left(\hat{\text{u}}_\text{i}-2\text{V}_\text{d}\right)\left({2\over\pi}-{4\over\pi}\sum_{\text{k}\space\geq\space1}{\cos\left(4\pi\text{k}\text{f}\tau\right)\over 4\text{k}^2-1}\right)\cdot y\left(t-\tau\right)\space\text{d}\tau\\ \\ &=\left(\hat{\text{u}}_\text{i}-2\text{V}_\text{d}\right)\left({2\over\pi}\int\limits_0^t y\left(t-\tau\right)\space\text{d}\tau-\frac{4}{\pi}\sum_{\text{k}\space\geq\space1}\frac{1}{4\text{k}^2-1}\int\limits_0^t\cos\left(4\pi\text{k}\text{f}\tau\right)y\left(t-\tau\right)\space\text{d}\tau\right) \end{split}\tag8 \end{equation}

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    \$\begingroup\$ I appreciate your effort and accept your answer but really what you wrote is beyond my comprehension. I just needed a simple algebric expression that relates the peak to peak ac ripple voltage to the known quantities like V(DC) or peak value of the pulsuating DC voltage or ac input voltage to the rectifier etc. Hope you will write such an expression for the people like me who are much less smart than you! @Jan Earland \$\endgroup\$
    – Alex
    Aug 18, 2023 at 19:01

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