The square wave generator is powered by a single power supply. The voltage division method is used to create a virtual ground. How to analyze and obtain an approximate reference value?

PunkGibson

The square wave generator is powered by a single power supply. The voltage division method is used to create a virtual ground. How to analyze and obtain an approximate reference value? [Copy link]

When designing a square wave generator, a single power supply is used, so a virtual ground needs to be created. Since only one op amp is allowed, the virtual ground here is created by a simple resistor voltage division method (a more reliable method should be to use a voltage follower to generate VCC/2 to supply the next stage op amp as a virtual ground).
But I don't quite understand how to calculate a more accurate reference value using this simple voltage division method to create a virtual ground.

The specific description is as follows:

The above picture shows what I call a single-power square wave generator. The reference value is generated by the Vcc voltage divider. In theory, R6, R7, R1, and R2 all share the VCC voltage, so the reference value generated at the Up terminal of the op amp through the superposition principle should be Vcc * (R7||R1||R2)/[ (R7||R1||R2) + R6]=Vcc/3=5/3V. The Up value is further used to calculate that the square wave frequency is 20khz.
The calculation process is as follows:

The measurement found that the square wave frequency is indeed around 20khz (as shown in the figure below), which means there is no problem in approximating Up according to this idea.

However, from the waveform of the output square wave (the red waveform in the figure below, channel A), we can see that the average value of the square wave voltage is (1.541+3.469)/2=1.5V. This reference value is inconsistent with the 5/3 just calculated.

So I don't quite understand how the reference value should be calculated in this case, and whether the voltage divider of R2 should be considered. If it is considered, then the theoretically calculated frequency value is accurate, but the calculated reference value will contradict the actual value, and vice versa.

wenyangzeng

If the output is DC isolated, all problems will be solved.

PunkGibson

wenyangzeng posted on 2024-6-30 09:43 Output DC isolation solves all problems

Why do we need to output DC isolation? I don't quite understand. Can you explain it to me?

maychang

[I don't quite understand how this reference value should be calculated in this case, and whether the voltage divider of R2 should be considered]

Of course it should be considered.

However, [if it is taken into account, then the theoretically calculated frequency value is accurate, but the calculated reference value will contradict the actual value]. The theoretical calculation error of the frequency of this multivibrator is quite large, and you can see it by looking at the voltage swing amplitude at the output end of the op amp of the simulation waveform. The high level of the op amp output is about 1.5~2V lower than the positive end of the power supply, and the low level of the output is about 1.5~2V higher than the negative end of the power supply. Each chip is different, and the output swing amplitude has a considerable impact on the frequency. Therefore, it is difficult to accurately output the frequency of this oscillation circuit.