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(an intuitive explanation by using hydraulic analogy)
IMO it would be interesting to see the correspondence below between me and a curious web reader about the phase shift phenomenon in capacitors driven by AC. Circuit dreamer ( talk, contribs, email) 09:38, 26 June 2011 (UTC)
QUESTION: Im having a really hard time trying to understand what the phase shifts mean in real life in a circuit (I know it means current lags source voltage by X amount..but how). I understand them mathematically but not in physicality, mainly because I cant find any explanation.
ANSWER: Let's explain for now the 90 deg phase shift between current and voltage in a capacitor. I recommend to you to think "hydraulically" ("electrical current - water flow" and "voltage - water level") to understand intuitively the phase shift idea.
Well, imagine you fill (sinusoidally) a vessel with water and you picture graphically this process. Choose the half of the maximum water height as a zero level (ground). So, you first open and then close (sinusoidally) the supply faucet. But note no matter you close the faucet (in the second part of the process) the level of the water will continue rising; it is strange that you close the faucet but the water continue rising. Finally, you have completely closed the faucet (zero current), but the level of the water will be maximum (maximum voltage).
Now, at this point, you have to change the flow (current) direction to make the water level decrease. For this purpose, you open and then close another faucet at the bottom to draw the water (now you draw current from the capacitor). But again, no matter you close the faucet the level of water will continue falling; it is strange again that you close the faucet but the water continue falling. Finally, you have completely closed the faucet (zero current), but the level of the water will be maximum negative (maximum negative voltage).
So, the basic idea behind all kind of such storing elements (named integrators) is:
The sign of the output pressure-like quantity (voltage, water level, air pressure, etc.) can be changed only by changing the direction of the input flow-like quantity (current, water flow, air flow, etc.); it cannot be changed by changing the magnitude of the flow-like quantity. At this point, the current is zero but the voltage is maximum; this gives the 90 phase shift on the graph.
Circuit dreamer ( talk, contribs, email) 09:38, 26 June 2011 (UTC)
QUESTION. ...However, I full understand the phase shift INSIDE the capacitor. That I understand 110%. At the begining there is max current flowing through the capacitor as there is no opposing voltage built up across it. As time goes on the current charges up the capacitor creating a voltage accross it which will then oppose the current. So at max current theres no voltage and at max voltage across it there is no current through it. And thus 90 degree lag....
BUT, what im trying to understand is something slightly different...the bigger picture....total impedance.
So for example, we have a capacitor with reactance of 10 Ohms and a resistor in series of 2 Ohms and an AC source of 10V. The total impedance in the circuit is going to be - Squareroot(10 squared + 2 squared) = 10.198 Ohms at a phase of - inverse tan(-10/2) = -78 degrees. So total impedance = 10.198 Ohms -78 degrees.
So what im trying to work out is WHAT causes this OVERALL lag in the circuit of 78 degrees.
I understand fully the lag inside the capacitor or inductor but when it comes to the overall lag in the circuit, what analogy can be used to discribe THIS lag. What is causing it to be -78 the physical happening of it.
ANSWER. Be patient:)! We will reach the so desired RC phase-shift expalanation (I managed to do this in the early 90's) but let's first explain this phenomenon in the case of the bare capacitor.
As I can see, you have learned very well your lessons about AC supplied capacitors since this is the classic textbook explanation of the phase shift between current and voltage in a "voltage-supplied capacitor". Maybe, it is repeated hundred of thousands or millions times in textbooks through years but it is not the best explanation since this arrangement (AC voltage source driving a capacitor) is just incorrect... You know why - because the current will be unlimited and we have not such huge voltage sources producing current with infinite magnitude. That is why, I have chosen the dual arrangement - a "current-supplied capacitor", to explain why there is an exactlty 90 degree lag. This is the real situation since even real voltage sources have some internal resistance. And what is more important, we will use this arrangement to explain the overall lag (0 - 90 degree) in an RC circuit. So, I suggest to you to repeat my explanations but only in terms of electricity.
Well, imagine we drive the capacitor by a sinusoidal current source. "Current source" means that it produces and passes a sinusoidal current in spite of all. No matter what the voltage (drop) across the capacitor is - zero (empty capacitor), positive (charged capacitor) or even negative (reverse charged capacitor), our current source will pass the desired current with desired direction through the capacitor. This is the difference with your explanation - here the voltage across the capacitor does not impede the current (it impedes but the current source compensates it).
So, until the input current is positive (imagine the positive half-sine wave) it enters the capacitor and its voltage continously increases in spite of the current's magnitude (only the rate of change varies)... Imagine... the current rapidly increases -> slows down -> rapidly decreases... and finally becomes zero. At this moment there is a maximum voltage (drop) across the capacitor.
So, as you have said, "at max voltage across it there is no current through it"...Now the current changes its direction and begins rapidly increasing again -> slows down -> rapidly decreases... and becomes zero again... and again and again and again...
So, in this arrangement, the phase shift is constant and exactly 90 degree because of the ideal input current source that compensates somehow the voltage drop (losses) across the capacitor.
If you have nothing against Wikipedia, I would like to place our conversation there again. My idea is to provoke wikipedians to discuss this so interesting topic and then to add some intuitive explanations in the intro of the article... to make it more human friendly...
Circuit dreamer ( talk, contribs, email) 20:30, 26 June 2011 (UTC)
Let's now consider the ubiquitous RC circuit. First, let's build it. We have already realized that it is incorrect to drive a capacitor directly by a voltage source; we have to drive it by a current source. For this purpose, let's connect a resistor between the voltage source and the capacitor to convert the input voltage to current; so, the resistor acts as a voltage-to-current converter. From another viewpoint, we have built a current source by the input voltage source and the resistor. Let's now consider the circuit operation (I will do it electrically but the hydraulic analogy of communicating vessels is an impressive way to do it as well).
Imagine how the input voltage VIN changes in a sinusoidal manner. In the beginning, the voltage rapidly increases and a current I = (VIN - VC)/R flows from the input source through the resistor and enters the capacitor; the output voltage begins increasing lazy. After some time, the input voltage approaches the sine peak and then begins decreasing. But until the input voltage is higher than the voltage across the capacitor the current continues flowing in the same direction. As above, it is strange that the input voltage decreases but the capacitor voltage continues increasing. Figuratively speaking, the two voltages move against each other and finally meet. At this instant, the two voltages become equal; the current is zero and the capacitor voltage is maximal. The input voltage continues decreasing and becomes less than the capacitor voltage. The current changes its direction, begins flowing from the capacitor through the resistor and enters the input voltage source. It is very interesting that the capacitor acts as a voltage source that "pushes" a current into the input voltage source acting as a load. Before the source was a source and the capacitor was a load; now, the source is a load and the capacitor is a source...
So, the moment where the two voltages become equal and the current changes its direction is the moment of the maximal output voltage. Note it depends on the rate of changing (the frequency) of the input voltage: as higher the frequency is, as low the maximum voltage across the capacitor is... as later the moment is... as bigger the phase shift between the two voltages is... At the maximal frequency, the voltage across the capacitor cannot move from the ground and the moment of current direction change is when the input voltage crosses the zero (the situation is similar to the arrangement of a current-supplied capacitor).
So, in this arrangement, the phase shift varies from zero to 90 degree when the frequency varies from zero to infinity because of the imperfect input current source that cannot compensate the voltage drop (losses) across the capacitor. Circuit dreamer ( talk, contribs, email) 03:29, 5 July 2011 (UTC)
We can draw a final conclusion about the AC supplied RC circuit:
Until the input voltage is higher than the voltage across the capacitor the current flows from the input source to the capacitor and the voltage across the capacitor continuosly increases no matter if the input voltage increases or decreases.
Now we can use the acquired here knowledge to explain in such an intuitive way the odd RC "amplifying" circuit above. Circuit dreamer ( talk, contribs, email) 10:55, 5 July 2011 (UTC)
Circuit dreamer ( talk, contribs, email) 15:00, 5 July 2011 (UTC)
I assume you mean the phast shift from input to output in a series RC circuit with the output taken from across the capacitor. You can have a phase shift of exactly -90 degrees in theory only because it would take one value (w, R, or C) to be infinite. You can get close with a large value for C. For example, with w=1, R=1 and C=1000 the phase shift is -89.94 approximately. In this example the larger the value of C the closer the phase fhit will be to -90 degrees. You can also make either R and/or w large to get the same effect. The problem however is that as C, w, or R gets larger the amplitude gets closer to zero so it may do no good to get close to -90 degrees. Oscillator circuits use RC networks that have phase shifts very different than -90 degrees, for example -45 degees. The phase shift can be calculated from ph=-atan(w*R*C), where w=2*pi*f with f in Hertz, R in Ohms, and C in Farads. MrAL Gx ( talk) 17:43, 21 March 2021 (UTC)
I've removed the following section from the article. It's inserted without context and has a lot of stylistic and editorializing issues, but more than that, it is only tangentially related to the article topic.
This is the
talk page for discussing improvements to the
RC circuit article. This is not a forum for general discussion of the article's subject. |
Article policies
|
Find sources: Google ( books · news · scholar · free images · WP refs) · FENS · JSTOR · TWL |
Archives:
1Auto-archiving period: 720 days
![]() |
![]() | This article is rated C-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | ||||||||||
|
![]() | This article may be too technical for most readers to understand.(September 2010) |
(an intuitive explanation by using hydraulic analogy)
IMO it would be interesting to see the correspondence below between me and a curious web reader about the phase shift phenomenon in capacitors driven by AC. Circuit dreamer ( talk, contribs, email) 09:38, 26 June 2011 (UTC)
QUESTION: Im having a really hard time trying to understand what the phase shifts mean in real life in a circuit (I know it means current lags source voltage by X amount..but how). I understand them mathematically but not in physicality, mainly because I cant find any explanation.
ANSWER: Let's explain for now the 90 deg phase shift between current and voltage in a capacitor. I recommend to you to think "hydraulically" ("electrical current - water flow" and "voltage - water level") to understand intuitively the phase shift idea.
Well, imagine you fill (sinusoidally) a vessel with water and you picture graphically this process. Choose the half of the maximum water height as a zero level (ground). So, you first open and then close (sinusoidally) the supply faucet. But note no matter you close the faucet (in the second part of the process) the level of the water will continue rising; it is strange that you close the faucet but the water continue rising. Finally, you have completely closed the faucet (zero current), but the level of the water will be maximum (maximum voltage).
Now, at this point, you have to change the flow (current) direction to make the water level decrease. For this purpose, you open and then close another faucet at the bottom to draw the water (now you draw current from the capacitor). But again, no matter you close the faucet the level of water will continue falling; it is strange again that you close the faucet but the water continue falling. Finally, you have completely closed the faucet (zero current), but the level of the water will be maximum negative (maximum negative voltage).
So, the basic idea behind all kind of such storing elements (named integrators) is:
The sign of the output pressure-like quantity (voltage, water level, air pressure, etc.) can be changed only by changing the direction of the input flow-like quantity (current, water flow, air flow, etc.); it cannot be changed by changing the magnitude of the flow-like quantity. At this point, the current is zero but the voltage is maximum; this gives the 90 phase shift on the graph.
Circuit dreamer ( talk, contribs, email) 09:38, 26 June 2011 (UTC)
QUESTION. ...However, I full understand the phase shift INSIDE the capacitor. That I understand 110%. At the begining there is max current flowing through the capacitor as there is no opposing voltage built up across it. As time goes on the current charges up the capacitor creating a voltage accross it which will then oppose the current. So at max current theres no voltage and at max voltage across it there is no current through it. And thus 90 degree lag....
BUT, what im trying to understand is something slightly different...the bigger picture....total impedance.
So for example, we have a capacitor with reactance of 10 Ohms and a resistor in series of 2 Ohms and an AC source of 10V. The total impedance in the circuit is going to be - Squareroot(10 squared + 2 squared) = 10.198 Ohms at a phase of - inverse tan(-10/2) = -78 degrees. So total impedance = 10.198 Ohms -78 degrees.
So what im trying to work out is WHAT causes this OVERALL lag in the circuit of 78 degrees.
I understand fully the lag inside the capacitor or inductor but when it comes to the overall lag in the circuit, what analogy can be used to discribe THIS lag. What is causing it to be -78 the physical happening of it.
ANSWER. Be patient:)! We will reach the so desired RC phase-shift expalanation (I managed to do this in the early 90's) but let's first explain this phenomenon in the case of the bare capacitor.
As I can see, you have learned very well your lessons about AC supplied capacitors since this is the classic textbook explanation of the phase shift between current and voltage in a "voltage-supplied capacitor". Maybe, it is repeated hundred of thousands or millions times in textbooks through years but it is not the best explanation since this arrangement (AC voltage source driving a capacitor) is just incorrect... You know why - because the current will be unlimited and we have not such huge voltage sources producing current with infinite magnitude. That is why, I have chosen the dual arrangement - a "current-supplied capacitor", to explain why there is an exactlty 90 degree lag. This is the real situation since even real voltage sources have some internal resistance. And what is more important, we will use this arrangement to explain the overall lag (0 - 90 degree) in an RC circuit. So, I suggest to you to repeat my explanations but only in terms of electricity.
Well, imagine we drive the capacitor by a sinusoidal current source. "Current source" means that it produces and passes a sinusoidal current in spite of all. No matter what the voltage (drop) across the capacitor is - zero (empty capacitor), positive (charged capacitor) or even negative (reverse charged capacitor), our current source will pass the desired current with desired direction through the capacitor. This is the difference with your explanation - here the voltage across the capacitor does not impede the current (it impedes but the current source compensates it).
So, until the input current is positive (imagine the positive half-sine wave) it enters the capacitor and its voltage continously increases in spite of the current's magnitude (only the rate of change varies)... Imagine... the current rapidly increases -> slows down -> rapidly decreases... and finally becomes zero. At this moment there is a maximum voltage (drop) across the capacitor.
So, as you have said, "at max voltage across it there is no current through it"...Now the current changes its direction and begins rapidly increasing again -> slows down -> rapidly decreases... and becomes zero again... and again and again and again...
So, in this arrangement, the phase shift is constant and exactly 90 degree because of the ideal input current source that compensates somehow the voltage drop (losses) across the capacitor.
If you have nothing against Wikipedia, I would like to place our conversation there again. My idea is to provoke wikipedians to discuss this so interesting topic and then to add some intuitive explanations in the intro of the article... to make it more human friendly...
Circuit dreamer ( talk, contribs, email) 20:30, 26 June 2011 (UTC)
Let's now consider the ubiquitous RC circuit. First, let's build it. We have already realized that it is incorrect to drive a capacitor directly by a voltage source; we have to drive it by a current source. For this purpose, let's connect a resistor between the voltage source and the capacitor to convert the input voltage to current; so, the resistor acts as a voltage-to-current converter. From another viewpoint, we have built a current source by the input voltage source and the resistor. Let's now consider the circuit operation (I will do it electrically but the hydraulic analogy of communicating vessels is an impressive way to do it as well).
Imagine how the input voltage VIN changes in a sinusoidal manner. In the beginning, the voltage rapidly increases and a current I = (VIN - VC)/R flows from the input source through the resistor and enters the capacitor; the output voltage begins increasing lazy. After some time, the input voltage approaches the sine peak and then begins decreasing. But until the input voltage is higher than the voltage across the capacitor the current continues flowing in the same direction. As above, it is strange that the input voltage decreases but the capacitor voltage continues increasing. Figuratively speaking, the two voltages move against each other and finally meet. At this instant, the two voltages become equal; the current is zero and the capacitor voltage is maximal. The input voltage continues decreasing and becomes less than the capacitor voltage. The current changes its direction, begins flowing from the capacitor through the resistor and enters the input voltage source. It is very interesting that the capacitor acts as a voltage source that "pushes" a current into the input voltage source acting as a load. Before the source was a source and the capacitor was a load; now, the source is a load and the capacitor is a source...
So, the moment where the two voltages become equal and the current changes its direction is the moment of the maximal output voltage. Note it depends on the rate of changing (the frequency) of the input voltage: as higher the frequency is, as low the maximum voltage across the capacitor is... as later the moment is... as bigger the phase shift between the two voltages is... At the maximal frequency, the voltage across the capacitor cannot move from the ground and the moment of current direction change is when the input voltage crosses the zero (the situation is similar to the arrangement of a current-supplied capacitor).
So, in this arrangement, the phase shift varies from zero to 90 degree when the frequency varies from zero to infinity because of the imperfect input current source that cannot compensate the voltage drop (losses) across the capacitor. Circuit dreamer ( talk, contribs, email) 03:29, 5 July 2011 (UTC)
We can draw a final conclusion about the AC supplied RC circuit:
Until the input voltage is higher than the voltage across the capacitor the current flows from the input source to the capacitor and the voltage across the capacitor continuosly increases no matter if the input voltage increases or decreases.
Now we can use the acquired here knowledge to explain in such an intuitive way the odd RC "amplifying" circuit above. Circuit dreamer ( talk, contribs, email) 10:55, 5 July 2011 (UTC)
Circuit dreamer ( talk, contribs, email) 15:00, 5 July 2011 (UTC)
I assume you mean the phast shift from input to output in a series RC circuit with the output taken from across the capacitor. You can have a phase shift of exactly -90 degrees in theory only because it would take one value (w, R, or C) to be infinite. You can get close with a large value for C. For example, with w=1, R=1 and C=1000 the phase shift is -89.94 approximately. In this example the larger the value of C the closer the phase fhit will be to -90 degrees. You can also make either R and/or w large to get the same effect. The problem however is that as C, w, or R gets larger the amplitude gets closer to zero so it may do no good to get close to -90 degrees. Oscillator circuits use RC networks that have phase shifts very different than -90 degrees, for example -45 degees. The phase shift can be calculated from ph=-atan(w*R*C), where w=2*pi*f with f in Hertz, R in Ohms, and C in Farads. MrAL Gx ( talk) 17:43, 21 March 2021 (UTC)
I've removed the following section from the article. It's inserted without context and has a lot of stylistic and editorializing issues, but more than that, it is only tangentially related to the article topic.