What is the purpose of the resistor in a RC circuit?

Question

I understand that a capacitor can pass high frequencies and block low frequencies, because of its impedance properties.

Now, why do you need to put in series a resistor? Doesn't it already filter frequencies on its own?

For instance, when we use a decoupling capacitor, no resistor is needed.

Can somebody please clarify this?

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

Figure 1. (a) All pass. (b) High pass.

I understand that a capacitor can pass high frequencies and block low frequencies, because of its impedance properties.

Alright, and so you're aware that the impedance decreases with increasing frequency.

Now, why do you need to put in series a resistor? Doesn't it already filter frequencies on its own?

If you want a high-pass filter then you need to create a frequency-sensitive voltage divider.

• In Figure 1a C1 is feeding a buffer with very high input impedance. As a result there is no load on the capacitor and the right side will track the left side even to very low frequencies.
• In Figure 1b C2 and R1 form a voltage (potential divider). C2 has low impedance at high frequencies so the divider effect will be very small. Let's say that R1 is 10 kΩ and C2 is 1 kΩ at the frequency of interest. The output voltage will be \$ \frac {10}{1+10} = 91\% \$ of the input voltage.
• If we drop the frequency so that C2 now has an impedance of 10 kΩ then the output voltage will be \$ \frac {10}{10+10} = 50\% \$ of the input voltage.

If the input resistance of the buffer is suitable then you could omit R1 and still get the frequency sensitive voltage divider - a high-pass filter.

Answer 2

If you are referring to for example a low pass filter, there needs to be a series resistor or inductor to form a "voltage divider". An ideal voltage source would otherwise charge the cap with high frequencies too, even if it's almost a short and the current required would be high.

In bypass caps the track to the device can be considered a series inductance. High frequency currents will flow the short distance from the capacitor to the device.

Answer 3

There is no such thing as an ideal capacitor. There are a variety of models for a "real" capacitor, and the simplest of those is just an ideal capacitor in series with a small resistor.

So, there's at least a small resistor always present. The size of that small resistor is usually given on the data sheet for the family of capacitors you're using. Often, though, you don't care what that value is, because you're going to use your own resistor, which should be a few orders of magnitude bigger than the "effective series resistance" associated with the capacitor, and it's OK to ignore the ESR in many situations (warning: in some situations, the ESR can be critical).

Answer 4

I understand that a capacitor can pass high frequencies and block low frequencies, because of its impedance properties. Now, why do you need to put in series a resistor? Doesn't it already filter frequencies on its own?

The key is to understand the difference between an ideal voltage source in circuit theory and a real power supply in the real world.

Ideal Circuit

An ideal voltage source always keeps its output voltage the same by magic. No matter what is being connected across the ideal voltage source - resistors, capacitors, inductors, they have absolutely no effect on its output voltage. It cannot be "loaded down", even if it means the voltage source has to deliver millions of amperes to achieve this result. It's also why an ideal voltage source in circuit analysis cannot be short-circuited, or be connected to a different voltage source - it's essentially division by zero, an illegal move.

In ideal circuit theory, connecting a capacitor across an AC voltage source does absolutely nothing to its voltage, regardless of whether this voltage is signal or noise - an ideal source is an ideal source.

^{simulate this circuit – Schematic created using CircuitLab}

Practical Circuit

A real circuit, like a battery or a power supply, always has a non-zero output resistance or reactance - it's the "hidden" R in RC. It's why, in a practical circuit, a capacitor can be connected directly across the power supply and works as a filter on its own. Because the existence of a resistor after the ideal voltage source, now its voltage changes depending on the load. The capacitor "loads down" the voltage from the power supply, thus creating a filter.

^{simulate this circuit}

In other words, the large capacitor on the circuit board is only half of the filter, another hidden half is the internal resistance of the power supply itself. When they're combined, a voltage divider is created. In fact, a filter's performance can change dramatically under different source and load impedances.

Forgetting this fact can lead to practical problems - for example, many engineers learned it the hard way that a powerline EMI filter's noise attenuation won't be as good as its datasheet suggests, since it's measured with 50 Ω source and load in standard RF test instruments - a condition you will almost never find in a power supply!

So the bottom-line is, at the very least, R is here to enforce physical reality and to stop the math from breaking down.

To be even more realistic, the power supply is modeled as a voltage source followed by output resistance and wiring inductance. The capacitor is modeled as a ideal capacitor, with parasitic inductance and series resistance.

^{simulate this circuit}

It is important to understand that decoupling is not the process of placing a capacitor adjacent to an IC to supply the transient switching current [...] rather it is the process of placing an L–C network adjacent to an IC to supply the transient switching current [...] All decoupling capacitors have inductance in series with them. Therefore, the decoupling network is a series resonant circuit.

• Henry W. Ott - Electromagnetic Compatibility Engineering

Answer 5

Unlike the other "frequency" explanations, I will use the more intuitive "time" explanations.

V-to-I converter

The purpose of the resistor in an RC circuit is to "convert" the input voltage into output current; so it acts as a voltage-to-current converter (actually, the resistor does not convert but only determines the current according to Ohm's law).

I-to-V integrator

The single capacitor converts the constant input current into linearly changing output voltage; so it acts as a current-to-voltage integrator.

V-to-V integrator

The combination of the two cascaded converters converts the constant input voltage into exponentially changing output voltage; so it acts as a bad voltage-to-voltage integrator. It is interesting to see how we can make it perfect again...

Generalization

Besides in an RC circuit, we can see the same resistor in an R1R2 voltage divider, RD logarithmic converter, RLED voltage indicator, RBE transistor switch, RL differentiator, RA voltmeter made through an ammeter, etc. The resistor converts all these current-input devices into voltage-input devices.

You can follow the RC circuit evolution in detail in my other two answers:

Charging of capacitor in RC circuit

What is an integrator topology?