LC filter circuit parameter design verification

shaorc

Design and select LC filter circuit for Class D power amplifier chip output. Now we know that the bandwidth of audio analog signal is 100Hz-16kHz, and the minimum load of load speaker is 2 ohms (now calculated as 2 ohms) My output topology is as shown in Figure 1 Figure 1 Now calculate L=15uH, C=10uF [1] But according to the formula f=1/2*pi*root LC, the cutoff frequency is calculated to be 13kHz, not 16kHz. Do we need to reselect the value of LC? [2] Since the channel power is 40W and the effective value of analog audio signal is 8V, the current of inductor L is 5A. So when selecting L, how large should the selected inductor current be? In addition to considering the volume and cost of the inductor, do we need to look at the ripple voltage or current through simulation? In addition, what other parameters should be considered when selecting the inductor? Here, [3] Capacitor C is a film capacitor. The withstand voltage of the capacitor should only be greater than the voltage amplitude of my analog signal (1 times), 1.414*8=12V. However, when I see the capacitor C of the LC filter circuit of the same chip from TI, the withstand voltage of the film capacitor is 250V (as shown in Figure 2). Why is this the case? Do all film capacitors have a larger withstand voltage? Figure 2

maychang

Changed to 100Hz-16kHz again? Can your speakers produce 16kHz sound? Speakers that can work up to 16kHz are much more expensive than your Class D power amplifier chip.

shaorc

maychang posted on 2019-4-29 16:15 Changed to 100Hz-16kHz again? Can your speakers produce 16kHz sound? Speakers that can work up to 16kHz are better than your Class D power amplifier chip...

You can play the playback frequency of this speaker first and then select L and C according to theoretical calculations. Please answer the questions I posted for the first time.

maychang

shaorc posted on 2019-4-29 16:18 You can first play the playback frequency of this speaker and then select L and C according to theoretical calculations. Please answer the questions I posted for the first time

Let's not talk about LC filtering, let's talk about current calculation. Your speaker impedance is 2 ohms, can the power amplifier output 40W? It should be noted that the output power of the power amplifier is not fixed. You can find out this by checking the datasheet.

maychang

The withstand voltage of film capacitors is generally large, because the withstand voltage is related to the thickness of the film, and the thicker the film, the higher the withstand voltage. However, it is difficult to manufacture a film that is too thin.

shaorc

maychang published on 2019-4-29 16:51 The withstand voltage of thin film capacitors is generally large, because the withstand voltage is related to the thickness of the film, and the thicker the film, the higher the withstand voltage. However, it is difficult to manufacture a film that is too thin.

"The withstand voltage of thin film capacitors is generally large, because the withstand voltage is related to the thickness of the film, and the thicker the film, the higher the withstand voltage. However, it is difficult to manufacture a film that is too thin." According to your logic, shouldn't it be the thinner the film, the higher the withstand voltage? Otherwise, how can the withstand voltage of thin film capacitors be so large?

shaorc

maychang posted on 2019-4-29 16:49 Let's not talk about LC filtering, let's talk about the calculation of current. Your speaker impedance is 2 ohms, can the power amplifier output 40W? It should be noted that the output power of the power amplifier is not...

Yes, it can

maychang

shaorc posted on 2019-4-29 17:00 "The withstand voltage of film capacitors is generally large, because the withstand voltage is related to the thickness of the film. The thicker the film, the higher the withstand voltage. However, it is difficult to manufacture a film that is too thin." Press...

"According to your logic, shouldn't the thinner the film, the higher the withstand voltage?" That's impossible. You can find out by looking up the relationship between the withstand voltage of air at a certain pressure (air gap breakdown) and the distance.

maychang

shaorc posted on 2019-4-29 17:22 Yes, it is possible

So, how did you get the 40W? You can't control your Class D power amplifier chip to output exactly 40W of power, because you can't control the input signal (microphone, recorder, etc.). What if the person using the microphone shouts hard?

topwon

【1】But according to the formula f=1/2*pi*root LC, the cut-off frequency is calculated to be 13kHz, not 16kHz. Does this mean that we need to reselect the value of LC? -------Are you using the TI calculation table I uploaded? The cut-off frequency of 13KHz should be the f0 I mentioned in yesterday's post, and the cut-off frequency fc should be the frequency of the intersection of the speaker curve and the horizontal axis (-3dB) in the right graph (the Q value calculated by your parameters is greater than 0.707, so the cut-off frequency will be higher than 13KHz).

shaorc

topwon posted on 2019-4-29 18:22 【1】But according to the formula f=1/2*pi*root LC, the cutoff frequency is calculated to be 13kHz, not 16kHz. Does this mean that we need to reselect the value of LC? -- ...

Thank you, I don't understand here. Can you illustrate fc and fo, and how to see that the cutoff frequency you said is higher than 13khz? I remember that teacher Maychang said that these two values are not the same for the second-order filter circuit

topwon

f=1/2*pi*root LC, where f should be the resonant frequency f0 of the LC circuit, which should also be called the natural frequency (or center frequency?). The cutoff frequency fc is generally the frequency point corresponding to -3dB, which can be found on the amplitude-frequency response curve (the two orange circles on the right side of the figure, the rightmost is a partial enlargement, and the horizontal axis is the -3dB line). The Ti calculation table uses cut-off to describe f0, which I personally think is wrong. If you choose a suitable load resistance value (such as 1.733 ohms) so that the Q value is equal to 0.707, the two will be equivalent (the curve corresponding to 1.733 ohms in the picture will intersect with -3dB at 13KHz).

topwon

In view of your specific design requirement of 100Hz-16kHz, I think the first thing to consider is to make the inductance value as small as possible (the volume cost can be reduced, and the current can be easily increased). In addition, the accuracy range of the components needs to be considered. The inductance is generally 20%, the film capacitor is considered 5%, and the load resistance will be more complicated (because the speaker is not a pure resistive load in fact). Let's assume it is 10%. You need to ensure that the fc calculated within the accuracy range of the LCR component can be greater than the 16KHz you require. In fact, for this kind of actual product, if I make it myself, I will make fc above 20KHz, because the main function of this LC low-pass filter is to remove the switching frequency of the class D amplifier (equivalent to the carrier frequency), which is generally several hundred KHz. In this way, even if fc is 4 or 50KHz, there should be no problem (in fact, many products that do not consider EMI issues can save this LC and directly use the low-pass filter characteristics of the human ear to remove the carrier frequency). In this way, the inductance and capacitance values can be smaller, which is convenient for selecting suitable materials (volume, cost, supply, etc.).

topwon

3. The general withstand voltage of film capacitors is 50V-63V-100V-250V DC. The higher the withstand voltage, the larger the price and size should be. Generally, it should be OK to take a value higher than PVDD (it should not be compared with the audio analog voltage amplitude, because LC filters the square wave of PVDD amplitude, and some overshoot margin should also be considered). For example, there should be no problem with a 50V withstand voltage film capacitor for a 24V-36V power supply. I have not tried using a 63V withstand voltage for a 48V power supply, but my personal analysis shows that it is OK.

topwon

【2】Since the channel power is 40W and the effective value of the analog audio signal is 8V, the current of the inductor L is 5A. So when selecting L, how large should the selected inductor current be? ----40W@2 ohms, why is it 8Vrms? It should be about 8.9Vrms, right? How did you calculate the 5A current of the inductor? 40W/8V (actually 40W/8.9V) ?This is the average value of the inductor current. The nominal current value of the inductor should be greater than the average current + the peak value of the ripple current. The peak value of the ripple current should have a corresponding calculation formula for each power amplifier chip.

topwon

maychang posted on 2019-4-29 17:27 So, how did you get your 400,000? You can't control your Class D power amplifier chip to output exactly 40W of power, because the input signal (microphone, recording...

When designing this specifically, it should be ensured that when the input signal amplitude is the largest, the maximum power value of 40W can be output. The amplitude range of the input audio signal seems to be quite confusing at present (the standards of audio sources are not uniform). I personally consider it as 0.2Vrms-2Vrms, and add a limiting circuit to ensure that a larger signal will not damage the subsequent stage. Some high-end power amplifiers will have input sensitivity adjustment capabilities to more conveniently match the amplitude of various audio source signals. His product is used in specific occasions, and the selection and compatibility of the pre-stage audio source will probably be relatively easier to handle.

gmchen

What the 13th poster said makes sense. It is a bit like trying to find a sword by carving a mark on the boat to find a sword. If you want to learn LC filter design, it is best to read some relevant books. Many design software on the Internet only give results. If you don't know the principle, you will be confused by it. Take the design of the first post as an example. I roughly verified that it is a Chebyshev type low-pass filter with an in-band fluctuation of 0.25dB. The -3dB bandwidth is indeed around 16kHz.

maychang

topwon posted on 2019-4-30 10:18 When designing this specifically, it should be ensured that when the input signal amplitude is the largest, it can output a maximum power value of 40W. The amplitude of the input audio signal...

The original poster used a Class D power amplifier chip, but did not specify the supply voltage of the chip. For the chip, the supply voltage determines the maximum output power, and the original poster may not be aware of this. If the maximum output power of the amplifier is much greater than the rated power of the speaker, then the speaker is dangerous when working, and the original poster does not seem to be aware of this.

shaorc

gmchen posted on 2019-4-30 10:37 What the 13th floor said makes sense. So it's a bit like trying to find a sword by carving a mark on a boat to find a sword. If you want to learn LC filter design...

Thank you for your advice. I am also reading books to check information, but the books always like to use the 5th order LC low-pass filter as an example.

gmchen

The cutoff frequency of the LC filter is not simply calculated as 1/(2*pi*sqrt(LC)). This is because the impact of the load and signal source internal resistance on the Q value of the LC loop must be considered, and for filters higher than the second order, there is also the mutual impact of the previous and next stages.