Shouldn't the charging curve of a capacitor be inversely proportional to its discharging curve?

QWE4562009

Shouldn't the charging curve of a capacitor be inversely proportional to its discharging curve? [Copy link]

 

Shouldn't the charging curve of a capacitor be inversely proportional to its discharging curve?

maychang

“Green: Discharging, Red: Charging”

No. The green curve is the voltage across RC in series, and the red curve is the voltage across C. The relationship between the red and green curves in the simulation is completely correct.

littleshrimp

The red signal is the charging curve when the green signal is at a high level, and the low level is the discharging curve. Is it true that when the capacitor is just charged, the voltage difference is large, the current is large, and the charging speed is fast, and when it is fully charged, the voltage difference is small, the current is small, and the charging speed is slow? Discharging and charging are exactly the opposite.

QWE4562009

littleshrimp posted on 2019-12-14 09:27 The red signal is the charging curve when the green signal is at a high level, and the low level is the discharge curve. Is it true that when the capacitor is just charged, the voltage difference is large, the current is large, and the charging speed is fast...

Not the other way around.

QWE4562009

maychang posted on 2019-12-14 08:51 “Green: discharge, red: charge” is wrong. The green curve is the voltage across the RC after the series connection, and the red curve is the voltage across C. Simulation...

Why do I feel that the waveforms are not inverse?

QWE4562009

maychang posted on 2019-12-14 08:51 “Green: discharge, red: charge” is wrong. The green curve is the voltage across the RC after the series connection, and the red curve is the voltage across C. Simulation...

Why is the discharge a square wave?

maychang

QWE4562009 posted on 2019-12-14 11:28 Why is the discharge a square wave?,,,

“Why is the discharge a square wave?”

The green curve is the voltage curve at the left end of R1. The voltage at this point has only two values: either equal to V1 or zero, which is of course a square wave.

maychang

QWE4562009 posted on 2019-12-14 11:28 Why is the discharge a square wave?,,,

“Why is the discharge a square wave?”

Who said the green curve is the discharge curve? It seems there is no such statement.

maychang

QWE4562009 posted on 2019-12-14 11:28 Why do I feel that the waveforms are not inverted?

“Why do I feel like the waveforms are not in anti-phase?”

It is definitely not an inverse relationship.

However, who said that it must be an inverse relationship? It seems that there is no such saying.

PowerAnts

The owner should dig out the discharge part, mirror it vertically, and compare it with the charging part.

QWE4562009

PowerAnts posted on 2019-12-14 12:58 You should dig out the discharge part, mirror it vertically, and compare it with the charging part. It is the same

Why do we need a mirror image?

QWE4562009

maychang posted on 2019-12-14 12:00 "How can the discharge be a square wave?" Who said that the green curve is the discharge curve? It seems that there is no such saying.

What is that green curve?

maychang

QWE4562009 posted on 2019-12-14 13:47 What is the green curve?

"What is that green curve?"

I said in the 7th post: "The green curve is the voltage curve at the left end of R1". This is consistent with the simulation diagram in the first post. In the first post, oscilloscope channel A is green and channel B is red.

chunyang

The original poster has misunderstood. The charge and discharge curve of the capacitor is the red one, that is, the voltage change curve across the capacitor. The rising section is charging, and the falling section is discharging.

maychang

QWE4562009 posted on 2019-12-14 13:47 Well, why do we need a mirror? ?

The red curve is the capacitor charging and discharging curve.

Flip the rising segment of the red curve upside down, and you get the falling segment of the red curve. The rising segment after flipping completely overlaps with the falling segment that has not been flipped. This is called "mirror image".

QWE4562009

maychang posted on 2019-12-16 10:46 The red curve is the charging and discharging curve of the capacitor. Flipping the rising section of the red curve upside down is exactly the falling section of the red curve. The rising section after flipping is exactly the same as...

That's not the green line.

maychang

QWE4562009 posted on 2019-12-16 11:35 That doesn’t mean there is no green line.

"That means there is no green line."

How can there be “nothing to do with the green line”?

The green line is the input signal of this RC circuit, or the excitation signal. If the green line is not a rectangular wave with a valley bottom of zero but other waveforms (such as a triangle wave, a sine wave, etc.), the red curve will not be the shape of the first post. In a sense, the green line is the "cause" and the red line is the "effect".