For a project I'm working on, I've to calculate a function for the output voltage of this filter:

enter image description here

The things I calculate:

  1. The input voltage: $$\text{U}_{\text{in}}\left(t\right)=\hat{\text{u}}\sin\left(\omega t+\theta\right)$$
  2. The voltage after the rectifier: $$\text{U}_{\text{C1}}\left(t\right)=\left|\text{U}_{\text{in}}\left(t\right)\right|=\left|\hat{\text{u}}\sin\left(\omega t+\theta\right)\right|$$
  3. Using Laplace transform, I got for the output voltage: $$\frac{\text{U}_{\text{C2}}\left(\text{s}\right)}{\mathcal{L}_t\left[\left|\text{U}_{\text{in}}\left(t\right)\right|\right]_{\left(\text{s}\right)}}=\frac{1}{1+\text{L}\cdot\text{C}_2\cdot\text{s}^2}$$

Where \$\omega=2\pi\text{f}\$

Questions: 1. Why does the output voltage not depend on the value of \$\text{C}_1\$? 2.Is the amplitude of the voltage after the (not ideal) rectifier equal to: \$\hat{\text{u}}-\text{V}_{\text{d}}\approx\hat{\text{u}}-0.7\$

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    \$\begingroup\$ A fullwave rectification has two diode losses. \$\endgroup\$ Dec 17, 2016 at 21:13
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    \$\begingroup\$ @JImDearden So the amplitude drops with \$\text{V}_{\text{d}}\approx1.4\$? \$\endgroup\$
    – KlaasP
    Dec 17, 2016 at 21:15
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    \$\begingroup\$ only when current is flowing, yes (ish) \$\endgroup\$
    – user16222
    Dec 17, 2016 at 21:19
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    \$\begingroup\$ Equation 2 is not valid, the whole point of having a C1 is to fill in the valley of the absolute sine. There is no transfer function for a rectifier circuit, it is significantly non-linear. \$\endgroup\$
    – rioraxe
    Dec 18, 2016 at 0:15
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    \$\begingroup\$ @rioraxe, There is a TF for the LC network (which is what the OP actually has), and the output voltage can be determined using the LT of the rectified sine as input signal. The resultant LT is non-analytic, though! \$\endgroup\$
    – Chu
    Dec 18, 2016 at 9:10

2 Answers 2


Your transfer function doesn't depend on \$C_1\$ because after the diode bridge, \$C_1\$ acts pretty much as a source. I would think of this circuit as a DC source at \$C_1\$ and an LC network to filter out the ripple.

Ideally, you could just consider \$C_1\$ as a DC voltage source (how good of a DC source depends on the its capacitance and how much current it is being drawn by the circuitry to the right of \$C_1\$). That's what you have in the denominator of your TF on the left hand side, you're using \$V_{c1}\$ as your input.

Even though you have a transfer function that includes \$L\$ and \$C_2\$, ideally, the DC portion coming from \$C_1\$ is unaffected. That means that not even the last LC filter changes the DC portion (ideal case). That is evident if you where to evaluate your TF at \$s=0\$, which is the case for a DC signal. Now, in reality, what's going to change your output voltage/current are the ESR losses associated with the capacitors and inductor of the filter. And of course, \$L\$ and \$C_2\$ will further smooth out your output signal.

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    \$\begingroup\$ Thanks for you answer. But are my calculations true? Are they right or not? And what whould this circuit look like when it is not idealy? \$\endgroup\$
    – KlaasP
    Dec 17, 2016 at 22:19
  • \$\begingroup\$ @KlaasP Your transfer function is Ok and the output voltage is actually the the peak value of the sine wave minus two voltages drops of the diodes that are on during each half cycle. If you take into account ESR losses, then you'd have to add a series resistance to each of the elements, the inductor and the caps. \$\endgroup\$
    – Big6
    Dec 17, 2016 at 23:09

Here is a way for an approximation and I will assume ideal components:


simulate this circuit – Schematic created using CircuitLab enter image description here

Start with an abbreviation circuit on the left, assume a constant current is drawn. Uc takes on the classic scalloped-shaped rippled voltage. This would be the first part of the approximation.

If we apply the same constant current approximation to the circuit on the right, then Uc would take on the form \$u |\sin(wt)|\$. But that is a bad approximation because if there is any filter element further down, the current becomes discontinuous when the diodes are off. To get the form \$u |\sin(wt)|\$, current must flow continuously through a pair of diodes, it can be true if the load is a resistor to ground, but this is not the case here.

This part of the circuit is clearly non-linear. Couple of indications are -- the diodes switching on and off; the fundamental frequency of the signal doubled.

Now go back to the scalloped-rippled voltage, one way is to represent that with a Fourier series. Assuming decent filtering, there will be a large DC component and smaller harmonics of \$2w\$. A reasonable approximation can be taking only the DC component and the first harmonic. Equate the DC to the average, and equate the amplitude of the first harmonic to the ripple. Now that can be passing through a LC filter and be processed linearly.

One of the earlier assumption is that the current coming off the rectifier with capacitor is constant. That assumption can be checked by comparing the inductor L current to the average current.

Let's say we try to get a transfer function end to end through some kind of approximation, but no transfer function can produce the harmonics, including the dominant \$2w\$ component, from a pure \$w\$ source.


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