# Capacitors - How do they work

Discussion in 'Homework Help' started by forumboards360, Feb 1, 2013.

1. ### forumboards360 Thread Starter New Member

Feb 1, 2013
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Hey guys, new here and have a few questions that are really bugging me about caps.

1. I was under the impression, that in a simple RC circuit (voltage source in series with a cap in series with a resistor), that when you turn on the voltage source and supply a voltage difference, that the capacitor would charge itself, until the voltage across the cap (can also be said as, the differential voltage on the other pin of the cap than the voltage source) would be 0, meaning that after an RC time constant, the output of the cap would be the same voltage as the input DC voltage source.

2. Pertaining to Q1 above, I also understand that in my circuit design classes we talk about decoupling caps which block ALL dc voltage. This means that on the voltage source side of the cap, the voltage is DC, and the other side of the cap is at 0V, allowing only AC voltage to pass through. However, this cannot happen if Q1 happens... Either DC voltage is blocked, or the output voltage rises from 0V to DC at an RC time constant. Both cannot occur, obviously, so which is correct?

3. Why does a cap act as a short at high frequencies?

4. Why does a cap act as a open at low frequencies.

5. If Q2 is correct, how does a RC circuit work in DC? Seeing as how all the dc voltage is blocked by the cap, the voltage across the resistor would be 0, flowing no current through the rest of the circuit.

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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The current flow through capacitor only when we charging or discharging the capacitor.
So when capacitor is full charge to supply voltage no current will flow.
And this is why we say that capacitor blocked the DC voltage.

Capacitor current is proportional the rate of voltage change across it (proportional to how quickly the voltage across capacitor is changing).
The faster the voltage change (frequency of a AC signal is high) the large the current flow through capacitor (short at high frequencies).

I = C * dV/dt

This means that to sustain current through a capacitor the applied voltage must change. The more rapidly voltage changes the larger the current. On the other hand if voltage is kept constant no current will flow no matter how large the voltage.
Likewise if the current through a capacitor is found to be zero, this means that the voltage across it must be constant, not necessarily zero.

3. ### forumboards360 Thread Starter New Member

Feb 1, 2013
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In Circuit 2, Vo would always be 0 if the circuit wasn't switching, represented by the solid grey line. If the switched opened, the cap would not be able to discharge its stored voltage of DC.

However, in circuit 1, if the cap acts as an open, Vo would be 0, dropping 0 volts across R2. Now, consider if we removed R1. Vo would still always be 0 unless the the switched was flipping.... This doesn't make sense to me. How can the cap act as an open and still drop voltage over R2 in a high pass filter setup if there is no DC voltage left after the cap..

There is either DC voltage which passes through a cap at 0Hz to supply R2 or there isn't. Which is it?

4. ### WBahn Moderator

Mar 31, 2012
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I think your big problem is your notion that the voltage across the capacitor will be zero after things settle down (which is several time constants, generally five is the number used). In point of fact, the capacitor will be charged to whatever voltage is consistent with no current flowing through it. So in your circuit, there will be no current in either resistor, hence V1 will be Vdc, Vo and Vc will be zero, and the voltage across the cap will be V1-Vc = Vdc, thus satisfying KVL.

5. ### forumboards360 Thread Starter New Member

Feb 1, 2013
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This helps, thank you.

My problem is that it is hard for me to imagine the voltage at the point Vc going to 0V after 5time constants.

I get that the cap charges to vdc across its terminals; I understand that. I just don't fully get why after a cap fully charges, the DC voltage doesn't continue after the cap so that after 5time constants Vc = VDC and the cap is fully charged. (obviously going against KVL) I don't know it just seems weird to me that only AC voltage passes and there is no DC. Essentially I find it weird that in DC a cap kills everything after it which is in series.

6. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
It's not "killing" the stuff after it any more than it is "killing" the stuff before it. There is no current in EITHER resistor as a result of the cap.

You keep talking as though Vc was the voltage on the cap. It isn't. The voltage on the cap is the difference between the voltage on the positive terminal (or whichever terminal you declare as being positive) and the voltage on the negative terminal. The voltage on the cap is (V1-Vc). If Vc somehow continued to go to VDC, then there would be VDC appearing across the output (Vo and Vc are the exact same voltage at all times) and thus more current flowing in R2 than there was when the capacitor was uncharged and the current was limited by R1+R2. So where would this current be coming from?

7. ### forumboards360 Thread Starter New Member

Feb 1, 2013
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0
I don't understand the last part of your post but I do get that it killed the resistor in front of it as well.

The only thing I can think to say for the last part of your post is that when the switch is opened, VDC appears across R2... Why would it be impossible in another instance?

So essentially if you have a Voltage source in series with a resistor, but in parallel with a cap and again in parallel with an inductor, all the current (VDC/R) would flow through the inductor acting as a short in DC operation correct? But if we took the cap and put it in series with the resistor, leaving the inductor in parallel with the 3components, the cap would give 0 current to the inductor, and there would be no voltage drop across the resistor, correct?

8. ### antonv Member

Nov 27, 2012
149
27
Back to your thoughts about AC passing: remember what was said about capacitor current - it is proportional to the rate at which the voltage changes, and so when you have AC across a capacitor the voltage is always changing and so you have AC current flowing through the cap for as long as the AC voltage is applied. (Current and voltage at the cap are not in phase though - when the sine wave is changing most rapidly, going through zero, the capacitor current is highest and when the sine voltage is at its peak but briefly not changing at all, then the capacitor current is 0).

It is helpful to think of AC and DC situations separately.

Also, when used as a blocking cap the DC level after the cap is not necessarily 0, it will be whatever DC level is set up on that side of the cap.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Let as examine how this simply RC circuit work.

We have a switch in "A" position. So we apply a voltage to the circuit.
At the beginning of the charging phase the capacitor is empty, and so Vc1 = 0V
So immediately after we connect the supply voltage a current will begin to flow. And all voltage from power supply appears across R1 resistor, VR1 = Vsupply. Because II Kirchhoff's law
Vsupply = Vc1 + VR1 so since Vc1 = 0V( empty capacitor) VR1 must be equal to Vsupply.
So we can say that empty capacitor act just like a short. Because there is no voltage across capacitor.
As the capacitor begins to charge, the voltage across the capacitor begins to increase. At the same time the current flow begins to decrease. Why? Because the voltage across the capacitor begins to increase.
So voltage across resistor must decrease (Vsupply = Vc1 + VR1).
And this is why charging current begins to decrease.
The capacitor eventually charges to the full voltage source potential. When this happens, all circuit current will cease because Vc1 = Vsupply so VR1 = 0V If so no current can flow (Ohms law I = V/R = 0V/R = 0A).
And Vc1 in red show voltage across capacitor during the charging phase.
And VR1 in red show resistor voltage during the charging phase.
Notice that, when we have a full charged capacitor no current flow in the circuit.

Now see what will happens when we flipped the switch to the "B" position.
In this situation C1 capacitor was previous charged to supply voltage.
So we have a voltage across capacitor. And we connect this voltage (the capacitor) directly to the R1 resistor. This will cause that discharge current will start to flow (charged capacitor act as a voltage source).
And again at the beginning the discharge current is "larger".
I_discharge = Vc1/R1
The voltage across the capacitor begins to decrease. At the same time the current flow begins to decrease. And current will stop flow at time when we full discharge the capacitor, Vc1 = 0V.
Vc1 in green show voltage across capacitor during the discharging phase.
VR1 in green show voltage across resistor during the discharging phase.

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10. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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If I am visualizing what you are explaining correctly, the circuit with capacitor and inductor in parallel is a filter or oscillating tank at AC frequencies.

When the frequency is very low (relative to inductor value), including DC, current will go through the inductor to ground, while little or no current will go through the capacitor. Once the capacitor is charged, it will appear as an open circuit to DC, but will pass some AC based on the value of the capacitor.

When the frequency is very high, the inductance will block the rapidly changing current (dependent on value of inductor), and will not pass current. The capacitor, however, will pass current (bigger capacitor means lower frequency passed, larger inductor means lower frequency blocked).

When the AC frequency is a the resonant frequency of L and C combined, they will "take turns" charging and discharging each other, converting the current to either a magnetic field (inductor) or electrostatic field (capacitor). The oscillation will damp out due to parasitic resistance and other losses, so an amplifier and feedback is needed to continue to the oscillation.