When working with microcontrollers it is recommended to place filter/decoupling capacitors between a supply pin and ground. I understand the purpose of this implementation, namely that voltage across a capacitor cannot instantaneously change, but what are the outstanding differences between a singular capacitor and a low-pass filter?

These are not calculated values; I inserted this just as an illustration.


simulate this circuit – Schematic created using CircuitLab

For example, if I were wanting to provide my ADC reference voltage a clean supply to compare input voltages against, I could either realize a low-pass filter to reject the high-frequency fluctuations or simply insert a adequately sized capacitor.

My immediate thought is that the initial current demand of a singular capacitor could momentarily exceed the maximum rating of the MCU, but with a resistor this current would be limited. Would it not be that with a LPF (with a resistor) that one could potentially design such that the output impedance of the filter is infinite as to not load the ADC? Similarly, a capacitor alone would provide sufficient voltage filtering but would it not result in a low output impedance?

What are the pros and cons of each realization of filtering, and when should a designer use one or the other?

Any other thoughts ?


5 Answers 5


The cap near the power pin is not to protect the part from noise, but to keep the part from generating noise as the logic switching causes rapid changes in supply current. Ideally the cap would supply instantaneous demands for more current without increasing current all the way back to the power source.

The sum of the impedances on the PSU side of the circuit - the internal impedance of the PSU plus the inductance, resistance, and capacitance of the traces or planes - is enough to give some low pass filtering on the input side of the cap. I think of the cap as a tiny a power supply that can respond to demands with a bandwidth in the multi-MHz range. The larger regulators that supply a full circuit react far too slowly and the cap is a temporary source of power that replaces or bypasses (or decouples) the PSU. Placing the cap close to the power pin on a chip minimizes resistance and inductance that would slow the response.

CMOS parts consume most of their power while switching state. For microprocessors this means on clock edges and the current draw is in little fast spikes. The size of the spikes varies as fast as the clock as every instruction uses different combinations of internal circuits. Imagine the circuitry used in checking a register for zero versus fetching data from RAM. The power needed fluctuates at the clock rate. The greater the current changes, the bigger the cap. Calculating the right size is a matter of estimating for most of us and the 0.1uF ceramic cap is so common that it is very low cost. Capacitor construction is also a concern as well as change with temperature. Some can respond quicker than others and some vary by 80% over the commercial temperature range.

They are also called bypass caps because: 1) They can "bypass" (short) high frequency PSU noise to ground. 2) They can "bypass" the PSU and respond to high frequency demands for power.

Also called "decoupling caps", a more accurate term for high frequencies as they "decouple" the power demand between the part and the PSU.

  • \$\begingroup\$ I understand the current sourcing capabilities of a capacitor, but I thought that was more of a lower priority consequence of including them in the design from VCC to GND. I thought the main purpose of including the capacitor was to filter out the high-frequency noise that the wire might pick up from the power supply/battery to the VCC pin of the MCU. The idea being that a charged capacitor will maintain a smooth output voltage and short to ground any of the HF fluctuations from the supply. What are the capacitors "decoupling" and how are the capacitors bypassing the PSU? \$\endgroup\$
    – sherrellbc
    Commented Dec 31, 2013 at 14:43
  • \$\begingroup\$ By "sum of impedances" are you referring to the impedance of the wire length to from the cap to the power supply? Also, under what conditions does the MCU demand more current? Of course there would be a difference, but would it be measurably different for a situation when the MCU is idly doing something vs a doing bit heavier calculations? \$\endgroup\$
    – sherrellbc
    Commented Dec 31, 2013 at 14:50
  • \$\begingroup\$ @sherrellbc By "sum of impedances" I mean the internal impedance of the PSU (It does have internal impedance. If you could exactly match it to the impedance of the circuit being powered, you would get the highest efficiency) plus the resistance and inductance of the PCB traces or power planes and I'm sure there are more. I'll edit the answer. \$\endgroup\$ Commented Dec 31, 2013 at 18:59

The short answer:

A capacitor alone is good for delivering power when the MCU power draw changes fast. The RC filter is used to block unwanted high frequency signals.

The looong answer:

The two different circuits are used for different purposes. As you have stated, the voltage across a capacitor cannot change instantly.

I'm sure you know that

  1. An MCU requires a minimum voltage to operate
  2. An MCU requires a varying amount of power during operation

Since power equals voltage * current (P = VI) and voltage must be constant then any change in power manifests itself as a change in current.

For a hypothetical design with a voltage regulator and an MCU:


simulate this circuit – Schematic created using CircuitLab

Say we remove C2:


simulate this circuit

(Sorry for the varying schematics I haven't set up an account for that schematic site and I need to keep redrawing it)

If the voltage regulator that is supplying the power to the MCU was perfect and there were no parasitic inductance or trace resistance the MCU would draw varying amount of current and the regulator would not lower or raise it's voltage. Unfortunately in the real world a circuit board looks more like this:


simulate this circuit

(Quick note: in this context an inductor can be thought of a resistor at high frequency)

Due to parasitic inductance from the board, trace resistance, and the fact that regulators cannot respond to current draw changes instantly the voltage will drop and raise as the MCU draws more or less current respectively.

As a reference here is a graph from an LM7805 Datasheet

ST 7805

Transient Response of an LM7805

This shows the finite response time of the LM7805 regulated output voltage (the triangle dips and humps in the bottom line) as the load increases and decreases. If the regulator was perfect then the 'Voltage Deviation' would not go up or down when there is a relatively fast increase or decrease in current.

I understand inductors can be a little confusing to use at first so for the sake of simplicity you can replace the inductor in the above schematic with a resistor and add the two resistors together and you have a resistor in between your regulator and MCU. This is bad because V=IR and the more current the MCU draws the more of a voltage drop will be seen across the resistor. (I'll explain more about what this resistors does below when I talk about RC Filters.

Back to the original design. The bypass capacitor is put in as close as possible to the MCU so that all the inductances and resistances found on a circuit board and the fact that a regulator cannot respond instantly will not effect the voltage level on the MCU.

For your second (RC) circuit


simulate this circuit

The reason a resistor should not be added to bypass an MCU is because the voltage across a resistor is relative to the current being drawn across it. This is important because if an MCU operates at 5V and draws 10mA quiescent (operating without doing anything) then there is a voltage drop across that resistor of:

R * 10mA = Vdrop

So if you had a resistor of 50 ohms you would drop .5V, this might reset your MCU.

A lowpass filter such as the RC filter you drew up there is not good for supplying power but is useful for filtering out high frequency components of a signal.

This is great for signals that are being read with an ADC because an ADC can only sample at a specific rate, so if a signal is changing at a rate greater then the high frequency signals (really 1/2 the rate due to the Nyquist theorem) will show up as random noise so it's good to remove it with an RC filter.

As an example say you have an ADC that samples at a rate of of 10Khz

and you want to read an analog sensor that only changes at a rate of 1KHz then you can set up your RC filter to filter out signals greater than 5Khz (you probably don't want to start filtering at 1Khz because an RC filter has a small amount of attenuation below the frequency at which it is designed to filter.

So to design an RC filter to achieve this you can use a resistor of:

330 Ohms and a capacitance of .1uF

Here is a great calculator if you need to solve this for any other frequencies:

Awesome RC Calculator

I hope I stayed on topic enough to answer your question.

  • \$\begingroup\$ Where does the parasitic inductance come from? I thought that inductance was a product only of coils in wires. The PCB traces are linear I thought. \$\endgroup\$
    – sherrellbc
    Commented Dec 31, 2013 at 16:01
  • \$\begingroup\$ What would be the effect on the effectiveness of the capacitor if it were placed further away from the MCU? \$\endgroup\$
    – sherrellbc
    Commented Dec 31, 2013 at 16:02
  • \$\begingroup\$ This is bad because V=IR and the more current the MCU draws the more of a voltage drop will be seen across the resistor. Wouldn't this supplement current be sourced from C2 (assuming it was present)? And how would you approximate the amount of current or draw time available from C2 such that the voltage across the capacitor does not change? Also, what is the purpose of C1? \$\endgroup\$
    – sherrellbc
    Commented Dec 31, 2013 at 16:11
  • 1
    \$\begingroup\$ @sherrellbc - If the capacitor were placed further away from the MCU, then you may see more switching noise generated by the MCU. Basically, you are inserting a very, very small inductor and resistor in series in between the filter cap and the MCU, which prevents the filter cap from doing its job as effectively. \$\endgroup\$ Commented Dec 31, 2013 at 16:52
  • 1
    \$\begingroup\$ To answer your question 'Where does the parasitic inductance come from?' All traces, component and vias have a parasitic inductance. Inductance is the property of a conductor by which a change in current in the conductor creates a voltage in both the conductor itself and in any nearby conductors Wikipedia inductors. This definition doesn't say anything about a coil. A trace is a conductor therefore an inductor, an inductor with coils is just a special case where the value of the inductance is more closely controlled. \$\endgroup\$
    – Dave
    Commented Dec 31, 2013 at 18:51

The difference is that placing only the capacitor depends on both the power supply impedance and the chip supply impedance to make up the rest of the low-pass filter. That is, both instances create a LPF, the explicit resistor is simply for tuning it.

  • \$\begingroup\$ I see. I did not considering the pins. How would the capacitor interface with the I/O pin models and how can the output (would it be referred to as output or input?) impedance of the pins be calculated to fine-tine the LPF? \$\endgroup\$
    – sherrellbc
    Commented Dec 31, 2013 at 4:21
  • \$\begingroup\$ You can get a general idea of the load by looking at the supply current requirements of your circuit. This will vary as transistors switch, but in general you want the roll-off frequency to not go too much above a few dozen hertz or so. For a small circuit just a capacitor may do, but for larger circuits you'll want usually an inductor (rather than a resistor) to maintain filtering. \$\endgroup\$ Commented Dec 31, 2013 at 4:35
  • \$\begingroup\$ What is meant by roll-off frequency in this context? And the inductor would be used to filter the current as the capacitor filters voltage fluctuations? \$\endgroup\$
    – sherrellbc
    Commented Dec 31, 2013 at 4:44
  • \$\begingroup\$ Roll-off frequency is generally the -6dB point. An LC circuit can act as a LPF, with each device responsible for holding its respective component stable. \$\endgroup\$ Commented Dec 31, 2013 at 4:54

You are right. This is decoupling technique, and we have to follow manufacturers suggestions. Typical decoupling consists of:

--> A large electrolytic capacitor (10~100μF) not more than 5cm away from chip. The aim of this capacitor is to supply “locally” the instantaneous current requirements, avoiding taking this power from the main power trail and their impedance. o This is a low ESR capacitor. --> A smaller capacitor (0.01 μF – 0.1 μF) closer to the power pins of the IC as possible, to drive HF components out of IC. Both capacitors should be connected to a large area ground on PCB for minimum inductance. --> A ferrite bed on series with Vcc pin of IC, to reduce EMI to and from this IC.

As you can judge, the above are general techniques for Linear and Digital IC. But the RC filter you are draw, it is dedicated for digital IC decoupling. The changes in digital gates state, cause a PS voltage to fluctuate because of the traces impedance. The high frequency noise can be minimized using RC or LC topologies. In the LC filter, noise appears across the coil rather than in chip or passing in power supply circuit. It provides a very efficient filtering but it has a resonant frequency that may radiate EMI. A ferrite bed can be use instead of inductor. enter image description here

The RC filter you are mention, converting noise to heat and as such is dissipated. Con is that the resistor introduce a voltage drop in the supplied voltage. On the other hand RC filter is less expensive. Some times you can find a wire wound resistance instead of inductor

The above are recommented by Silicon Labs and Analog Devices

  • \$\begingroup\$ By radiate are you referring to the possibility that the LC filter might emit EMR and cause interference on some victim component nearby? \$\endgroup\$
    – sherrellbc
    Commented Jan 1, 2014 at 17:27
  • \$\begingroup\$ Right. For example a decoupling net with 100μF and 1μH resonates at 16kHz. If this frequency appears in power line this will cause a problem. The way to fix this is to add a small resistance (i.e. 10Ω close to IC ) in series with coil to lower the Q, or a ferrite bed for frequencies above 100kHZ. Decoupling method it is also depends of the power supply type are in use (linear or switching), and has a major effect in spectrum (SNR issues) and on waveforms shape (ringing etc) in devices with clock above 1.5GHz and in 14-bit ADC \$\endgroup\$
    – GR Tech
    Commented Jan 1, 2014 at 18:46

low pass filter is used for blocking high frequency and noice signals above particular frequency. Resonance occurs at that particular frequency. All signals above resonance frequency will be grounded and about singulasr capacitor you described same.

The R-C filter is used instead of L-C filter for economical purpose.


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