Many friends who are new to the industry will ask, what is the load capacitance used by a single-chip crystal? In fact, this is easy to remember because the circuit capacitance of a single-chip crystal is either 22pf or 30pf. Why is this? This article will explain to everyone why the value should be taken this way through an example. What is the principle of single chip value?

The example in this article uses an ATMEGAL16 single-chip temperature acquisition system. The fact is not easy after welding. Even if a multimeter is used, no errors can be found. Afterwards, it is suspected that the single-chip is locked. The situation has not improved after replacing several single-chip computers. . After careful investigation, it was found that two 0.1uf capacitors were welded to the crystal circuit, which caused the crystal to fail to vibrate, so the entire circuit behaved abnormally, and then replaced with a 22pf capacitor and it was normal immediately.

So what exactly caused the exception?

In fact, the real name of the oscillation circuit of the microcontroller and some other ICs is "three-point capacitor oscillation circuit", as shown in Figure 1.

Y1 is a crystal, which is equivalent to a three-point inductor. C1 and C2 are capacitors. The 5404 NOT gate and R1 implement an NPN transistor. Next analyze this circuit.

5404 must require a resistor, otherwise it will be in the saturation cutoff region instead of the amplification region. R1 is equivalent to the biasing effect of the transistor, so that 5404 is in the amplification region, then 5404 is an inverter, which realizes the role of the NPN transistor. The NPN transistor is also an inverter when the common emitter is connected.

的 A sinusoidal oscillation circuit needs to oscillate if the system magnification is greater than 1. This is easy to achieve, the phase satisfies 360 degrees, and a small oscillation with the same oscillation frequency as the crystal is amplified.

The following mainly explains the phase problem:

Because 5404 is an inverter, that is to say, 180 ° phase shift is achieved, then C1, C2, and Y1 need to achieve 180 ° phase shift. Just when C1, C2, and Y1 form resonance, 180 phase shift You can solve the equations, etc., and treat Y1 as an inductor. You can also use the characteristics of the capacitor inductance, such as the capacitor voltage is 90 ° behind the current, the inductor voltage is 90 ° ahead of the current for analysis. When C1 increases, the amplitude at C2 increases, and when C2 decreases, the amplitude also increases.

Sometimes C1 and C2 can be vibrated without soldering. This is not to say that there is no C1 or C2, but it is caused by the distributed capacitance of the chip pins. Because C1 and C2 do not need to be very large, this is important. Next, analyze the influence of these two capacitors on the oscillation stability.

Because the voltage feedback of 5404 depends on C2, assuming C2 is too large, the feedback voltage is too low. Assume that C2 is too small, the feedback voltage is too high, the stored energy is too small, it is easy to be interfered by the outside world, and it will also affect the outside world by radiation. The effect of C1 is exactly the opposite of C2. Because when the board is laid out, it is assumed that the double-sided board is relatively thick, so the effect of distributed capacitance is not great. It is assumed that when high-density multilayer boards are used, distributed capacitance needs to be considered.

It needs to be explained to everyone that when it is used in industrial control design, it is recommended not to use a passive crystal oscillator to start, but directly connect to an active crystal oscillator. Because industrial control projects have higher requirements for stability, passive crystal oscillators cannot start directly. Crystals with higher frequencies have lower stability, so lower speed crystals will be used when speed requirements are not high.