My understanding of RC circuits is broken

Question

I asked a relatively simple question. Unfortunately, the answers provoke far more questions! :-(

It seems that I don't actually understand RC circuits at all. In particular, why there's an R in there. It seems completely unnecessary. Surely the capacitor is doing all the work? What the heck do you need a resistor for?

Clearly my mental model of how this stuff works is incorrect somehow. So let me try to explain my mental model:

If you try to pass a direct current through a capacitor, you are just charging the two plates. Current will continue to flow until the capacitor is fully charged, at which point no further current can flow. At this point, the two ends of the wire might as well not even be connected.

Until, that is, you reverse the direction of the current. Now current can flow while the capacitor discharges, and continues to flow while the capacitor recharges in the opposite polarity. But after that, once again the capacitor becomes fully charged, and no further current can flow.

It seems to me that if you pass an alternating current through a capacitor, one of two things will happen. If the wave period is longer than the time to fully charge the capacitor, the capacitor will spend most of the time fully charged, and hence most of the current will be blocked. But if the wave period is shorter, the capacitor will never reach a fully-charged state, and most of the current will get through.

By this logic, a single capacitor on its own is a perfectly good high-pass filter.

So... why does everybody insist that you have to have a resistor as well to make a functioning filter? What am I missing?

Consider, for example, this circuit from Wikipedia:

What the hell is that resistor doing there? Surely all that does is short-circuit all the power, such that no current reaches the other side at all.

Next consider this:

This is a little strange. A capacitor in parallel? Well... I suppose if you believe that a capacitor blocks DC and passes AC, that would mean that at high frequencies, the capacitor shorts-out the circuit, preventing any power getting through, while at low frequencies the capacitor behaves as if it's not there. So this would be a low-pass filter. Still doesn't explain the random resistor through, uselessly blocking nearly all the power on that rail...

Obviously the people who actually design this stuff know something that I don't! Can anyone enlighten me? I tried the Wikipedia article on RC circuits, but it just talks about a bunch of Laplace transform stuff. It's neat that you can do that, I'm trying to understand the underlying physics. And failing!

(Similar arguments to the above suggest that an inductor by itself ought to make a good low-pass filter — but again, all the literature seems to disagree with me. I don't know whether that's worthy of a separate question or not.)

Answer 1

Let's try this Wittgenstein's ladder style.

First let's consider this:

^{simulate this circuit – Schematic created using CircuitLab}

We can calculate the current through R1 with Ohm's law:

$$ {1\:\mathrm V \over 100\:\Omega} = 10\:\mathrm{mA} $$

We also know that the voltage across R1 is 1V. If we use ground as our reference, then how does 1V at the top of the resistor become 0V at the bottom of the resistor? If we could stick a probe somewhere in the middle of R1, we should measure a voltage somewhere between 1V and 0V, right?

A resistor with a probe we can move around on it...sounds like a potentiometer, right?

^{simulate this circuit}

By adjusting the knob on the potentiometer, we can measure any voltage between 0V and 1V.

Now what if instead of a pot, we use two discrete resistors?

^{simulate this circuit}

This is essentially the same thing, except we can't move the wiper on the potentiometer: it's stuck at a position 3/4th from the top. If we get 1V at the top, and 0V at the bottom, then 3/4ths of the way up we should expect to see 3/4ths of the voltage, or 0.75V.

What we have made is a resistive voltage divider. It's behavior is formally described by the equation:

$$ V_\text{out} = {R_2 \over R_1 + R_2} \cdot V_\text{in} $$

Now, what if we had a resistor with a resistance that changed with frequency? We could do some neat stuff. That's what capacitors are.

At a low frequency (the lowest frequency being DC), a capacitor looks like a large resistor (infinite at DC). At higher frequencies, the capacitor looks like a smaller resistor. At infinite frequency, a capacitor has to resistance at all: it looks like a wire.

So:

^{simulate this circuit}

For high frequencies (top right), the capacitor looks like a small resistor. R3 is very much smaller than R2, so we will measure a very small voltage here. We could say that the input has been attenuated a lot.

For low frequencies (lower right), the capacitor looks like a large resistor. R5 is very much bigger than R4, so here we will measure a very large voltage, almost all of the input voltage, that is, the input voltage has been attenuated very little.

So high frequencies are attenuated, and low frequencies are not. Sounds like a low-pass filter.

And if we exchange the places of the capacitor and the resistor, the effect is reversed, and we have a high-pass filter.

However, capacitors aren't really resistors. What they are though, are impedances. The impedance of a capacitor is:

$$ Z_\text{capacitor} = -j{1 \over 2 \pi f C} $$

Where:

• \$C\$ is the capacitance, in farads
• \$f\$ is the frequency, in hertz
• \$j\$ is the imaginary unit, \$\sqrt{-1}\$

Notice that, because \$f\$ is in the denominator, the impedance decreases as frequency increases.

Impedances are complex numbers, because they contain \$j\$. If you know how arithmetic operations work on complex numbers, then you can still use the voltage divider equation, except we will use \$Z\$ instead of \$R\$ to suggest we are using impedances instead of simple resistances:

$$ V_\text{out} = V_{in}{Z_2 \over Z_1 + Z_2}$$

And from this, you can calculate the behavior of any RC circuit, and a good deal more.

Answer 2

I think some of the answers are over-complicating things. The only physics you really need to know is that the "resistance" of a capacitor goes inversely with frequency, and the famous 3-dB formula: $$f_{-3dB} = \frac{1}{2\pi RC}$$ So, presuming you're familiar with those, let's look at it like this.

Low Pass Filter

So you don't like R, eh? Well, let's say the resistor isn't there--

Oops, we can't! There is always some resistance. You don't get to imagine what happens without it. The wire will have milliohms or micro-ohms, but there is still some resistance. The smaller it is, the farther away your 3-dB point gets, according to our handy 3-dB formula--and the less "low pass" it becomes. Adding a discrete resistor lets you choose the 3-dB point, instead of it being determined for you by small wire- or trace resistance, which most of the time you don't know (and can't even measure!).

High Pass Filter

Here, we can imagine life without R. One night, you got into an argument with it, and in a fit of rage, you took it out. So now let's say it's absent.

But now look what we have; the capacitor is just a big, dumb resistor whose resistance, as you know, varies inversely with frequency.

It is still a filter in the sense that it will attenuate voltages of certain frequencies. Certainly it will block DC; in that sense, it is "low pass". But now it's terrible! Why?

For low frequencies, as I said, it's now just a "big" resistor; depending on how much current you're pulling, that means low frequencies will be attenuated somewhat: as you know, the more current you pull over an impedance, the more the voltage drops across it.

But, like in the low-pass filter case when you removed R, your circuit now depends on something you don't usually control: current. If this filter is connecting to a a high-impedance (i.e. megaohm) load, very little current will be drawn; the capacitor won't drop much voltage for most frequencies, and so it may as well not be there. You want to be able to put this filter anywhere and have it work some pre-determined way.

Let's look at some simulations. Say you have a 1uF cap, and your load is 1k:

(Ignore the phase plot, as it's irrelevant for this post). OK, we have a rolloff starting around 200Hz. That's alright, I guess, if that's what you want. But what happens when the resistor changes? I.e., what happens when your circuit wants a different amount of current?

Goodness! Our 3dB point is now around 1Hz. So our "filter" is moving all over the place whenever something in your circuit wants the current to change! It's totally unpredictable.

So you make amends with the resistor, and you put it back, and it fixes your filter for you.

Wait-- how does R fix your high pass filter, you ask? Well, with it and the capacitor, it acts as a voltage divider! If it's stiff enough--that is, if its output impedance is much lower than the input impedance driving the rest of your circuit--it insulates your filter from changes in current draw.

Answer 3

I know that you already have got many answers. Let me try my own way.

What I have to design is filter. Both low-pass and high-pass. What I have is a capacitor only.

Consider the first implementation, where all the components are ideal.

When Vout is measured using an ideal oscilloscope, What we would get is Vout = Vin.

So this circuit can not work as any filter.

Considering second implementation,

Here, there is no current through C and hence here also Vout is Vin.

So the second circuit also can not work as a filter.

So one can not implement a filter only with capacitor (at least in ideal case).

Now coming to your mental model, as you said that "Current will continue to flow until the capacitor is fully charged.."

But have you ever thought about how long will it take for a capacitor to get fully charged?

The charging time of a capacitor is decided by the capacitance value C and the current passing through it (which can be controlled by placing a resistor of appropriate value in series with C).

$$V = \frac{Q}{C} = \frac{I\times t}{C}$$ $$\Rightarrow t = \frac{V\times C}{I} \propto RC$$

In short, the charging time is decided by the product RC.

Now placing a finite resistance in series with C we can control the time taken by the capacitor to get fully charged. So with a series resistance R, the first circuit can act as a low pass filter and second circuit can act as high pass filter as shown in your question.

If R = 0 (short circuit), then the capacitor get charged instantly and it acts as open circuit for every frequency. That is what happened in first circuit.

If R = infinity (open circuit), then the capacitor never start to charge or no current flows through capacitor. And that happens in the second circuit.

Answer 4

Forget the idea of "power getting through"; power is the product of current and voltage, and the kind of applications where you'll see this configuration of components has nothing to do with the transfer of power.

In a simple AC circuit (let's start here at least) a capacitor has a characteristic called reactance. Reactance is essentially the relationship between the capacitance and the frequency of the signal involved. It's calculated using the infamous formula of 1/2πfC, where f is frequency in Hertz and C is capacitance in Farads, and reactance is measured in Ohms. Essentially, a capacitor is a frequency dependent resistor.

For reactive components, i.e. caps and inductors, their frequency-based resistance is often referred to as impedance. You'll often find circuits or devices with "input impedance" rather than resistance, implying that it can vary depending on the input signal frequency but should usually be flat(ish) over the range of frequencies the circuit/device is intended for.

Back to the mysterious inclusion of the resistor; think back to my earlier comment about the cap being a frequency controlled resistor. That means, for a given frequency, you now have two resistors forming a potential divider. If you know R and C, you can plot a graph of Vout vs frequency.

The most common place you'll find these filters is in basic/passive signal processing circuits. One would expect to see the high-pass configuration at the input to an operational amplifier (to save amplifying obnoxious low frequencies). Op amps benefit from having MASSIVE input impedances - typically terraohms - so you can't say that the parallel resistor is siphoning off the current because that's it's exact purpose: almost no current whatsoever will end up in the op amp, so a cap in series by itself will be useless.

Yes, things change a bit when you move to current amplifiers, but that's really quite a different topic. Transistor amplifiers are in their own league, and a bit beyond this question.

However, for some extra information, there are situations where power is transferred across a series resistor/parallel capacitor configuration. The winner of that category is, as the name suggests, power lines (carrying electricity across the country, etc.). Transmission Line analysis is done by modelling a power line as a series resistance plus a parallel cap and inductor, representing the resistance of the copper wire, the parasitic capacitance between the copper conductor and it's outer "ground" sheath, and the voltage induced from external factors, respectively. In such a case, these components represent the real-world imperfections, so power is indeed lost. The Lumped Transmission Model (the name may vary) will use this LRC circuit on a 'per unit distance' basis, such that several of these circuits are lumped together, one after the other, to represent a particular length power line.

Answer 5

The resistor is made to control the current. You seem to forget the voltage across a capacitor cannot change instantly, it is the result of negative charges accumulating on one plate and leaving the other, finally resulting in creating the electric field equivalent to its voltage. If this voltage can't change instantly and you apply a different voltage across it, the wires need to drop that difference of voltage and their resistance is tiny which will make a massive current flow (U=RI). There is basically nothing slowing the electrons down except the wires. The uncontrollable very high current will charge the capacitor in no time if it doesn't damage it, which renders the filter useless as it's supposed to absorb and deliver current as required. There is a duality between time domain and frequency domain - by controlling how fast the capacitor reacts to input changes via a precise value of R, you have control on the cutoff frequency of your filter.

Sometimes high reactivity is desired, for decoupling capacitors for example which do not have limiting resistors, but not in filters.

Note that if you're supplying current, you don't need a current limiting resistor however you do need a voltage limiter because the capacitor voltage will increase linearly and eventually go past the breakdown voltage. But it's not a filter anyway; you would use an inductor to filter current.

In the high pass filter/edge detector (first circuit), the resistor is there to form a voltage divider with the capacitor. Capacitors grossly said act like frequency-dependent resistors (they also phase shift the signals but let's let that slide). The resistor is there to create a voltage that depends on frequency without drawing any current: at high frequencies the impedance of the capacitor will decrease and you get more of the input out (and vice-versa). So without that resistor, if no current is drawn the input will be mirrored in the output (no voltage drop).

In the low pass filter the resistor is also there to form a voltage divider except that this time, the voltage of interest is that across the capacitor ("gets stronger with time" => low pass) and not the image of the current ("gets weaker with time" => high pass). If you short circuit the resistor the capacitor will react too quickly and will be useless as a filter, just like I mentioned at the beginning of this post.

Answer 6

Great question.

It seems to me that if you pass an alternating current through a capacitor, one of two things will happen. If the time to fully charge the capacitor is longer than the wave period, the capacitor will spent most of the time fully charged, and hence most of the current will be blocked. But if the wave period is shorter, the capacitor will never reach a fully-charged state, and most of the current will get through.

I agree with part of this analysis. If you put a current into a capacitor, you can figure out the voltage across it pretty easily by using

\$ V = \frac{1}{C}\int i(t)dt \$

However, then you start talking about a capacitor that is "fully charged". At what voltage is a capacitor fully charged? There's a voltage where the capacitor might fall apart, but I don't think that's what you're thinking of.

This doesn't really make sense anyways. Where is this current coming from? Usually, it's easier to work with voltages - I have a much easier time applying a sinusoidal voltage to a capacitor than a sinusoidal current.

So, here's my intuition:

• The current that passes through a resistor is \$I = \frac{V}{R}\$.
• The current that passes through a capacitor is \$I = C\frac{dV}{dt}\$.
• At low frequencies, \$\frac{dV}{dt}\$ is small, so there isn't much current through the capacitor; since there's low current, there's little voltage across the resistor, and most of the voltage is across the capacitor.
• At high frequencies, \$\frac{dV}{dt}\$ is big, so the capacitor can pass as much current as it wants; the resistor is the limiting factor for the current in the circuit, so most of the voltage drop is across the it.
• At medium frequencies, there's a transition from the low frequency case to the high frequency case. This happens around \$f = \frac{1}{2\pi RC}\$.
• Without a resistor, you can't tell where low frequencies and high frequencies cross over.

PS: you're right about "blocking the power" - if you want to transfer the current flowing through this filter into something else further down the line, it's going to behave differently.

Answer 7

For low-pass filter case: resistor is there to limit the current from input voltage source. In theory ideal components used, so this voltage source can deliver infinite current. If we take resistor out, there will be no filtering at all, capacitor will be charged to input volage instantly (as any current required to match voltage rate of change can be supplied), no matter what frequency signal is. That's where resistance comes into play. With any non zero value capacitor voltage start to lag behind input, and so filtering effect created. And if ideal current source connected to low pass RC filter, R actually CAN be taken out, as it have no influence on current flowing in.

Answer 8

If you try to pass a direct current through a capacitor, you are just charging the two plates. Current will continue to flow until the capacitor is fully charged, at which point no further current can flow.

The resistor answers the question "how much current?", and consequently the question how long current will continue to flow.

At any rate, "current will continue to flow until the capacitor is fully charged" is misleading. If we are talking about "direct current", the current will continue to flow until the capacitor hands in his resignation. For an electrolytic capacitor, that can be surprisingly smelly.

Now usually we don't have an ideal current source in charge. It is more common to have a voltage source and a resistor (hint hint), and the current through the resistor will decrease while the voltage across the capacitor approaches the voltage on the other side of the resistor. The ratio between this voltage difference and the charging current is determined by the resistor.

Answer 9

If you apply a CURRENT then the resistor is doing nothing and the voltage on the cap will increase linearly to infinity. However, if you apply a VOLTAGE then the resistor will 'resist' the flow of current and generate an opposing voltage drop. The capacitor will only see a part of the voltage and whatever current the resistor lets through. As the cap charges, the voltage on the cap increases and the resistor lets less and less current through. The voltage on the resistor will asymptotically approach zero.

A capacitor will no load will actually pass arbitrarily low frequencies as there will be no current path to charge or discharge through.

Answer 10

If the time to fully charge the capacitor is longer than the wave period,

But how long is that time? It turns out to take \$R \cdot C\$ seconds to charge to approximately 2/3 (charging is asymptotic). This is called the RC time constant.

If you take the resistor out of the first circuit, and have nothing at Vout, then you don't have a circuit - there is no loop round which current can flow. In reality, if you put say a meter or an audio input there then it will look like a resistor of a few megaohms. The current flows through the capacitor, through the meter, and back to the negative rail. Putting a specific resistor there gives you a predictable sensibly-sized resistance to calculate with. It doesn't divert power - in fact by ohm's law it develops a voltage across it in proportion to the alternating current flow.

In the other example, the series resistor is there otherwise Vout would always equal Vin; it delays the charging of the capacitor to a specific time constant.

An inductor on its own is called a "choke" and is indeed an effective lowpass filter. It's never entirely on its own, there is always a few picofarads of wire capacitance around..

(Your question conflates voltage, current and power carelessly, which may be confusing you)

Answer 11

If there is no actual or implicit resistor in your circuit, you are driving the capacitor with either an ideal voltage source, or an ideal current source. Putting a resistor in series with an ideal current source is pointless, so the only interesting case is the one with an ideal voltage source.

The point of the ideal voltage source is that the capacitor will follow the voltage immediately. And that means that the current into the capacitor will be \$d/dt U * C\$. A jump in voltage will result in infinite current spikes.

The usual purpose of an RC element, however, is not as a differentiator but rather as a delay element. Putting a resistor in series will limit the current and thus stop the capacitor from tracking the voltage immediately.

Answer 12

@MathematicalOrchid, thanks for the wonderful question and intuitive way of reasoning. I admire you because I myself have always tried to answer these questions in this way. I'll share only a few thoughts that would add something new to what has been said already.

Indeed, in the case of the differential CR circuit below, the resistor can be omitted if you replace it with the load itself... but the load should be low resistive enough. It is possible here since the load is connected in series to the capacitor.

In the case fo the integrating RC circuit below, it cannot be omitted since the load is connected in parallel to the capacitor. Then what is the role of the resistor in this arrangement?

The capacitor is a kind of a "container" that should be "filled" of "fluid"; so its input quantity is flow-like (current)... and its output quantity is pressure-like (voltage)... it is a device with current input and voltage output... an ideal (linear through time) integrator... a current-to-voltage integrator. You have to drive ("fill") it by a current source... but you have a voltage source. So you have to convert the voltage into a current... and this is the role of the resistor... it acts as a voltage-to-current converter...

If you combain the input voltage source and the resistor, you can think of this combination as of a simple (imperfect) current source driving a current integrator.

I have created many stories about these circuits (some of them - animated). Here are a few of them; maybe they can help your intuitive understanding:

How to make a perfect RC integrator - Wikibooks

Class exercise - my students, 2004

Op-amp RC integrator - circuit-fantasia.com (Circuit stories on the whiteboard)

Ramp generator - Circuit stories on the whiteboard

Why there is a phase shift between the current and voltage in a capacitor - Wikipedia talk page

Building an op-amp inverting integrator - Flash animated story

Answer 13

Let's do a simpler, more effective, approach...

But first:

What the hell is that resistor doing there? Surely all that does is short-circuit all the power, such that no current reaches the other side at all.

This is incorrect in two main points:

• Shorting means making two points the same voltage (in reference to ground), which is clearly not the case here: Assuming the resistor's value isn't zero, the voltage across the resistor is not zero.. unless the current through the resistor is. Since the voltage across the resistor is V=R*i. If one of the two is zero, then the voltage is zero.

• Even if it was a short circuit, there would still be a current (but no voltage, since the voltage across a "short/wire" is zero. So V=R*i. Assuming it's a short (R=0), there can be a current flowing and the voltage would still be zero...

Now...

Let me ask you a question.. In the first circuit (assuming R is not zero), what would make the voltage zero? Well, no current.

And assuming you're applying a voltage across your input (at your left), why would there not be a current?

Because the capacitor is preventing current flowing.

And in which case the capacitor would do that? In which case any component would prevent current from flowing?

Answer: When a component has an impedance of infinity..

See: V=Z*I.. So I = V/Z, right?

So if Z = Infinity, then you have a null current... In other words, your component becomes equivalent to an open-switch..

Now: When does a capacitor behave that way? In other words, when is the impedance of a capcitor infinity? Well Zc=1/(jwC)..

Assuming C is not zero.. That leaves omega = 0... In other words, what you call "DC". Frequency zero.

So, let's call "gain" the ratio between the voltage at your output and input..

G= Voutput/Vinput..

When omega = 0, capacitor behaves as an open circuit, meaning your current doesn't even "make it" to your resistor, meaning volage across R (which is Voutput) is 0..

Which means G=0/Vinput=0.

Okay.. We saw the case for omega = 0..

What about omega = infinity?

Well, the capacitor then behaves as a closed switch.. Which means: Vinput=R*I=Voutput.

Which means G=1.

So.. Our circuit's gain is 0 in the low frequencies, and 1 in the high frequencies... In other words, it lets high frequencies pass, and blocks low frequencies.. In other words: A High Pass Filter.

Can we do our second circuit?

Omega --> 0 ===> Capacitor is open circuit (remove it from your schematic). All you're left with is Vout = Vin.. So gain G = 1.

Omega --> Infinity ==> Capacitor is a short circuit, and Vout = 0, so G=0.

In other words, that circuit lets signals of Low frequencies pass, and blocks high frequency signals..

It's a Low Pass Filter..

Some remarks:

I suggest you get a solid understanding on the basics, first. Really understand how each of these components works individually.

Chapter 1 (Foundations) of The Art of Electronics would explain this. There are also Tony Kuphaldt's -free- books "Lessons in Electric Circuits".

I cannot stress enough on the importance of the basics: If you skip, you'll get a knowledge that's like swiss cheese, with gaping holes and you'll struggle later. You'll build on shaky foundations and inevitably fail to wrap your head around relatively more complex things..

Answer 14

Theoretically, the resistor isn't neccesary. If you draw both the HPF and LPF circuits with only the capacitor, you get the filtering effect. The reason filters add the resistor is to have control over the cutoff frequency: $$ f_{3dB} = \frac{1}{2\pi RC} $$ As an example, many times when designing a circuit capacitors are added between the power supply and ground without a resistor to create a LPF that discharges AC.