Hello everyone, welcome to the ZIMZIM DIY channel. Today, I will make a Subwoofer amplifier circuit or academically called Deep low or Deep bass, which is a circuit that filters only low frequencies to pass through. Therefore, our circuit will have only bass sound, no other ranges mixed in. This frequency range is the lowest that we humans can hear. That is, it is in the range between 20Hz and 60Hz. Most of these frequencies that we hear are booming sounds, impact sounds, and vibrations. These sounds will be emitted from the subwoofer speaker. For the subwoofer amplifier circuit, you can actually create your own circuit. You just may need to use the formula to calculate the CutOff frequency point a little more. I guarantee that you will have fun with it. Creating a subwoofer amplifier circuit Hair is divided into 2 types as follows: Type 1 is Passive. What is Passive? It is a circuit that does not require a power supply. It relies on the properties of L, C, and R to connect the signal path to the ground. Or to put it simply, it pulls the frequency that we do not use to the ground. Type 2 is Active. What is Active? It is a circuit that uses the properties of LCR to connect the signal path to the ground. But add an IC op-amp to amplify the signal to make it bigger. Make the signal in that range more prominent. In this clip, I will choose to make a Passive circuit first. In the next clip, I will do the Active. First, let's look at an example for different frequencies. I will use this MP3 player as an example. In the Equalizer section of the program, if we play normal music with the EQ value in the middle, friends will receive a complete sound signal in all frequency ranges. Is that right? This is like the sound that has not been filtered. It is the original, authentic sound. Now, let's do it again. I want to use the sound with a tweeter speaker, which is a treble. The treble is the high frequency range of 4,000 hz and above. So I will reduce the low frequency range in this range. Let's listen to it again. The sound that comes out will be purely high frequency. This is the High-pass filter circuit. But if I switch it up, now I will drive the bass. I still keep the low frequency, but pull the high frequency to the ground. The sound that comes out will be like this. This is the circuit that we will do today. Let's start designing the circuit. The circuit is to take the original sound signal through the circuit input. The signal will be filtered and then output. Therefore, for this circuit, I will use R to put between the input and output and C to put a coupling after R to ground. In this connection, only low frequencies will pass. Because the resistor here blocks some of the current, causing the capacitor to charge more slowly than before. When it charges slowly, C will accidentally release the signal. Now that we have the circuit, here's what our friends need to know. This is the equation to find the cutoff point, frequency, which will use this formula F = 1 / 2TT RC. Let's try to use this formula to randomly calculate the cutoff point. I set R to 10K and C to 10n. Substitute the values, the calculation comes out to 1592Hz or we estimate that it is approximately 1600Hz. Note that this formula will randomly find the cutoff point value by using CR to calculate. Now, in reverse, if I want to set the cutoff point, the cutoff frequency of the signal, with the value that I want to be around 60Hz or less, which is the standard for the subwoofer range. I have already gotten the value of F. The next part is, I will do the reverse. Find the value of R or the value of C instead. From my experience, I should set the value of R and then find the value of C. I have to rearrange the equation by moving C up to replace F. This circuit will look like this. C = 1 / 2TT * R * F That's right. This equation will find the value of C. I set R equal to 1k, substitute C = 1 / 2TT * 1000 * 60 = 2.65uf. But since the value of 2.65 uf capacitor is unlikely to be available for sale, I will find a close value of C, which is the one that I already have, which is 3.3uF. When the value of C changes, the R part must be moved down to 800 ohms to respond to the frequency. The cutoff point will be at 60.31 hz. But in reality, R 800 is also not available. I will use R 680 ohms and R100 ohms, which are commonly available, to connect in series. The value will be around 780 ohms, losing 20 ohms, and the cutoff point will move up another 1 Hz, which is okay. I am not serious about 1hz - 2hz. I can barely hear it anyway. In conclusion, we have R780 ohms and C 3.3 uf to use at a frequency of 61 Hz. The ideal cutoff point will be a graph like this. But in reality The frequency will gradually drop like this. It doesn't disappear suddenly like the graph a moment ago. And this is the real circuit that I have made. Go listen to the sound. Some people are not satisfied. The sound is still booming, not vibrating enough. I am one of