What is the principle of RC filter circuit? What is going on?

小太阳yy

What is the principle of RC filter circuit? What is going on? [Copy link]

 

The above circuit converts the PWM voltage into a DC voltage through RC filtering. The principle is that the capacitor will charge and discharge. When the PWM power supply is low, the capacitor will discharge. So what is the use of this R? And how is the converted voltage calculated?

Why is VOUT equal to VCC*DUTY? What is the relationship between the PWM frequency and the cutoff frequency of the RC filter? I know that the capacitor has a cutoff frequency, but what about the cutoff frequency of the RC? For those who have just started designing, these questions are hard to understand after thinking for a long time.

maychang

"For those who have just started designing, it takes a long time to figure these things out."

Nothing to think about.

The cut-off frequency of RC filtering is shown in the first-order circuit. At this frequency, the capacitive reactance of the capacitor is exactly equal to the resistance value of the resistor, so the voltage across the capacitor is equal to the voltage across the resistor (but not half!). For signals with a lower frequency than this, the capacitive reactance of the capacitor is larger than the resistance value of the resistor, and the signal voltage divided on the capacitor is larger than the voltage divided on the resistor. For signals with a higher frequency than this, the capacitive reactance of the capacitor is smaller than the resistance value of the resistor, and the signal voltage divided on the capacitor is smaller than the voltage divided on the resistor. Therefore, the RC in the figure is a low-pass filter (first order).

maychang

"Also, how is the converted voltage calculated?"

As mentioned above, RC forms a first-order low-pass filter. The DC voltage at the Vout terminal is the average value of the PWM voltage, that is, VCC*DUTY. DUTY is the duty cycle.

maychang

"I know that the capacitor has a cutoff frequency."

Which book says this? Who said this?

maychang

"What is the relationship between the PWM frequency and the cutoff frequency of the RC filter?"

PWM is a rectangular wave. In addition to DC and fundamental components, it also contains a large number of higher harmonics (secondary, third, etc.). Now you want to remove the DC component and remove the fundamental component and higher harmonic components. Obviously, the fundamental frequency should be much higher than the RC cutoff frequency. Otherwise, the fundamental component will appear at the Vout end. In fact, multiple first-order low-pass cascades are often used to filter out the AC components (fundamental component and higher harmonic components) as thoroughly as possible.

maychang

{What is this R used for? }

If there is no R (short circuit), the waveform at the Vout end will always be exactly the same as the PWM end (no filtering effect).

小太阳yy

maychang posted on 2022-4-11 18:01 "And how is the converted voltage calculated? " As mentioned earlier, RC forms a first-order low-pass filter. The DC voltage at the Vout end is the PWM voltage...

What I mean is why is it calculated like this, won't there be any loss of charge through the resistor R?

小太阳yy

maychang published on 2022-4-11 18:07 "What is the relationship between the frequency of PWM and the cutoff frequency of RC filtering? PWM is a rectangular wave. In addition to the DC and fundamental components, it also contains a large number of high-order harmonics...

I don't quite understand this. Can you explain the principle of capacitor charging and discharging?

maychang

小太阳yy posted on 2022-4-11 18:38 I mean why is it calculated like this? Won't there be any loss of charge passing through the resistor R?

"I mean why is it calculated like this?"

You can look at how to calculate the capacitive reactance of capacitors and the inductive reactance of inductors, and then look at first-order circuits. For example: "Circuits 5th Edition" by Qiu Guanyuan, Chapter 6 and Chapter 7.

maychang

小太阳yy posted on 2022-4-11 18:42 This is a bit confusing. Teacher, can you explain it from the principle, the charging and discharging principle of capacitors?

"I don't quite understand this. Teacher, can you explain it in terms of the principle of capacitor charging and discharging?"

The charging and discharging of a capacitor is a qualitative judgment. Calculation, even if it is an approximate calculation, is a quantitative analysis.

Rectangular waves contain a large number of high-order harmonics, which require Fourier expansion and cannot be explained "from the perspective of charging and discharging."

maychang

Xiaotaiyangyy posted on 2022-4-11 18:42 This is a bit confusing. Teacher, can you explain the principle of capacitor charging and discharging?

PWM signal is a non-sinusoidal periodic signal. Non-sinusoidal periodic function is decomposed into Fourier series, see Qiu Guanyuan's "Circuit Fifth Edition", Chapter 13.

chunyang

小太阳yy posted on 2022-4-11 18:38 I mean why is it calculated like this? Won't there be any loss of charge passing through the resistor R?

It is the loss of the resistor that can form a low pass. That is, the higher the frequency of the signal, the smaller the capacitance reactance, and thus the greater the loss on the resistor, thus achieving a "low pass". To understand the principle and specific calculation, you might as well use Ohm's law. You should be able to calculate the voltage division of two resistors, right? It's just that the capacitive reactance of the capacitor (equivalent to a resistor at a certain frequency) is a function of frequency. In the Ohm's law voltage division resistor calculation formula, the capacitive reactance formula can be used to replace the grounded resistor.

chunyang

小太阳yy posted on 2022-4-11 18:42 This is a bit confusing. Teacher, can you explain it from the principle, the charging and discharging principle of capacitors?

The thinking direction is wrong. The key concept here is "capacitive reactance", that is, at a certain frequency, the capacitor can be equivalent to a resistor. After understanding the capacitive reactance, you can use Ohm's law to understand the working principle of the RC low-pass circuit, and then you can use the capacitive reactance formula combined with Ohm's law to derive the calculation formula of the RC low-pass circuit.

maychang

Xiaoshangyy published on 2022-4-11 18:38 What I mean is why is it calculated like this? Won't the charge be lost when passing through the resistor R?

Chunyang is right. You can understand RC low-pass by using Ohm's law and think of RC as two resistors dividing the voltage (in fact, it is the resistor and capacitor components dividing the voltage). However, the resistance value of the resistor does not change with the frequency, but the capacitive reactance of the capacitor does. Therefore, when two resistors divide the voltage, the ratio of the output-input voltage does not change with the frequency. When resistors and capacitors divide the voltage, the ratio of the output-input voltage changes with the frequency and is a function of the frequency.

Fred_1977

This is a simple first-order filter circuit, not for DC conversion.

redsip

It can be understood that this R will limit the charge and discharge current of the capacitor, thereby adjusting the charge and discharge time of the capacitor. When R is large enough, the capacitor starts charging again before it is fully discharged, which is reflected in the appearance of a DC component at the Vout end.

se7ens

Is there any load connected to Vout?

R can limit the charging and discharging current of the capacitor

mingplus

zhangdaoyu

Small questions lead to big knowledge. These are the basics. You may not be able to use them even if you have learned them. It is so valuable to apply what you have learned. I am here to learn.

PowerAnts

maychang posted on 2022-4-11 19:13小太阳yy posted on 2022-4-11 18:38 I mean why is it calculated this way? Won't the charge be lost through the resistor R? c ...

Assuming RL is infinite, PWM level is 3.3V, duty cycle is 0.5, frequency is 10KHZ, R=1K, C=1uF, can the ripple level on C be calculated using "capacitive reactance" and "Ohm's law"?