This article is original by the author, please indicate the source if reprinted. This question has troubled the author for more than 5 years, and one product lost millions due to this issue.Most MCU enthusiasts do not understand why a 22pF capacitor needs to be connected on both sides of the MCU crystal, as this capacitor can sometimes be omitted. Many books mention it only as a stabilizing component or load capacitor, but do not provide a deep theoretical analysis.The problem is that many enthusiasts do not care about these two capacitors; they think it is enough to follow the reference design. I was the same until one time a mobile phone project had issues because of this capacitor, resulting in a loss of millions, which made me seriously consider the role of this capacitor. In fact, the MCU oscillation circuit is called a “three-point capacitor oscillation circuit”. Please refer to the images in the webpage.Y1 is the crystal, equivalent to the inductor in the three-point circuit, C1 and C2 are the capacitors, and 5404 and R1 form an NPN transistor. You can refer to the three-point capacitor oscillation circuit in high-frequency books. Next, let’s analyze this circuit. The 5404 needs a resistor; otherwise, it will be in the saturation cutoff region, not the amplification region. R1 serves as the bias for the transistor, allowing the 5404 to be in the amplification region, making it an inverter. This achieves the function of an NPN transistor, which is also an inverter in common emitter configuration. Now, let’s explain the working principle of this three-point oscillation circuit in simple terms; you can also check the book directly. Everyone knows that the condition for a sine wave oscillation circuit to oscillate is that the system gain must be greater than 1, which is easy to achieve, and the phase must satisfy 360°. Next, we mainly explain the phase issue: since the 5404 is an inverter, it achieves a 180° phase shift. Therefore, C1, C2, and Y1 need to achieve a 180° phase shift, which happens when C1, C2, and Y1 resonate. The simplest way to understand this is to take the ground as a reference. When resonating, the current through C1 and C2 is the same, with the ground in between, so the voltages are opposite, achieving a 180° phase shift. When C1 increases, the amplitude at C2 increases, and when C2 decreases, the amplitude also increases. Sometimes, C1 and C2 can still oscillate without soldering; this does not mean there are no C1 and C2, but rather due to the distribution capacitance of the chip pins, since C1 and C2 do not need to be very large, so this point is very important. Next, let’s analyze the impact of these two capacitors on oscillation stability. The feedback voltage of the 7404 relies on C2; if C2 is too large, the feedback voltage is too low, which is also unstable. If C2 is too small, the feedback voltage is too high, storing too little energy, making it susceptible to external interference, which can also radiate and affect the outside. The role of C1 is exactly the opposite of C2. When we design PCBs, if it’s a double-sided board and relatively thick, the influence of distribution capacitance is not significant. However, in high-density multi-layer boards, distribution capacitance must be considered, especially for oscillation circuits like VCOs, where distribution capacitance should be taken into account. For some industrial control projects, it is recommended not to use crystal oscillation methods, but rather to directly connect an active crystal oscillator. Many times, people use a 32.768K clock crystal for timing instead of using the MCU’s crystal frequency division to create a clock. Many people do not understand this reason; it actually relates to the stability of the crystal. The higher the frequency of the crystal, the harder it is to achieve a high Q value, leading to lower frequency stability. The 32.768K crystal has good stability and other aspects, forming an industrial standard that is easier to achieve high performance.END
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