How do I calculate the output voltage here?

小太阳yy

How do I calculate the output voltage here? [Copy link]

How to calculate the output voltage?

maychang

If the input impedance of the next stage marked with text is infinite, then the voltage at the junction of the four 10 kilo-ohm resistors is easy to calculate. The voltage at this point is affected by four voltages: PWM high or low level, 5V, output voltage (that is, the voltage across the capacitor on the right), and zero (that is, ground), so the left capacitor is charged (discharged when PWM is low). After obtaining the expression for the voltage across the left capacitor, the expression for the voltage across the right capacitor can also be obtained. In this way, the voltage ripple across the right capacitor can also be calculated.

Approximate calculation ignores the change of output voltage (regarded as the average voltage on the left capacitor), and only calculates the average value of the three resistors, upper, left, and lower, in one cycle of PWM high and low levels. The result of this calculation is that the voltage on the right capacitor is considered to be equal to the average voltage on the left capacitor.

小太阳yy

maychang published on 2023-10-17 12:01 If the input impedance of the subsequent stage marked with text is infinite, then the voltage at the junction of the four 10 kilo-ohm resistors is easy to calculate. The voltage at this point is affected by four voltages: PW ...

Teacher, how do I calculate the average voltage in this pwm cycle? Do you think I am doing it right? When the pwm output of the MCU is low, the two 10K resistors below are in parallel, and the equivalent resistance is 5K. The output voltage is the voltage divided by the 10K and 5K above. When the Pwm output is high, it is just the opposite, which is equivalent to 5K above and 10k below. In this way, two voltage values are obtained, one Max value and one MIN value. Finally, the average voltage is equal to the Max value multiplied by the Pwm duty plus the min value multiplied by 1 minus the pwm duty. Is this correct?

maychang

小太阳yy posted on 2023-10-17 13:04 Teacher, how should I calculate the average voltage in this pwm cycle? Do you think I am right? When the pwm output of the MCU is low, the following two 1 ...

[Do you think I am doing this right?]

Yes, that's how it's calculated.

If the PWM high level is not 5V or the low level is not 0V, or the resistors are not all the same value, the calculation is the same. It's just a little more troublesome to calculate them separately.

Nubility

The calculation method is the same as that of the resistor series-parallel network, except that the resistor is replaced by a capacitor, and the angular velocity ω of the signal waveform and the imaginary unit j are added to the calculation formula.

吾妻思萌

maychang published on 2023-10-17 12:01 If the input impedance of the subsequent stage marked with text is infinite, then the voltage at the connection point of the four 10 kilo-ohm resistors can be easily calculated. The voltage at this point is affected by four voltages: PW...

Teacher, how should the average voltage within this pwm cycle be calculated? Do you think my calculation is correct? When the pwm output of the MCU is low, the two 10K resistors below are in parallel, and the equivalent resistance is 5K. The output voltage is the voltage value obtained by dividing the 10K and 5K above. When the Pwm output is high, it is just the opposite, which is equivalent to 5K above and 10k below. In this way, two voltage values are obtained, a Max value and a MIN value. Finally, the average voltage is equal to the Max value multiplied by the Pwm duty plus the min value multiplied by 1 minus the pwm duty. Is this correct?

Points

wang12011201

It is said that there is no DC noise effect due to the isolation of C2 capacitor. At high gain, even a small DC voltage at the amplifier input will affect the dynamic range of the op amp and may even cause output saturation. The input bias current flows through the coupling capacitor, charging it until it exceeds the rated common-mode voltage of the amplifier input circuit or exceeds the output limit.