# Decoupling capacitors

Discussion in 'General Electronics Discussion' started by Madmicra, Feb 14, 2017.

1. ### Madmicra

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0
Feb 14, 2017
Hi, I'm new at electronics, more than a newbie actually, but pretty interested to know more in the field.
I'm currently struggling to understand the proper function of a decoupling or bypass capacitor.
I know how it is suppose to work: you have, say, an IC that is powered by 5V and you want a clean no
ripple power source. Capacitors tend to divert high frequencies away, as they oppose very little resistance (reactance)
to such signals.

But my question is this: in terms of Kirchhoff voltage law (KVL), where in one close branch the sum of voltages drops
are equal to voltage gains, and knowing that in this case Vc should eliminate the ripple making Vc (voltage across capacitor) almost equal to the DC component (5V), where can you find the ripple signal if there is no other component in this circuit?

I'm using superposition theorem to analyse the input signal as the sum of a pure DC component and another signal, the ripple. The DC component analysis is straight forward...
it appears in capacitor leads... but regarding the ripple I do not know were it goes and how to process it under KVL.

Besides that, how come Vout is different from Vin if they are connected to the same node, assuming wires have no resistance?

I'm sure I'm making some kind of an huge confusion here... I already know how to use and the purpose of decoupling capacitors but I really
want to comprehend it in terms of more formal circuit analysis, like the observation of Kirchhoff laws.

Can you help me here?

Thanks!!

Regards

Jorge

2. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,481
2,830
Jan 21, 2010
In theory, theory is no different to practice. In practice it is.

KVL assumes that connections to nodes have zero resistance or inductance. In practice there are stray resistances, inductance and capacitances which have a real effect.

If the IC in question demands a sudden high current, the resistance and inductance between the IC and the power source will cause a voltage drop that is not predicted by a simplistic KVL model. To compound this, your filter capacitor may have significant series resistance.

A small ceramic cap placed close to the IC will have far less resistance and inductance between it and the chip, and it's internal series resistance (ESR) will be much lower. This will allow the brief spike in current to be supplied without such a large voltage drop.

BobK likes this.
3. ### AnalogKid

2,467
712
Jun 10, 2015
You are correct that the capacitor is acting as the shunt leg of a series-shunt voltage divider. The series leg is the resistance and (inductive) impedance of the wiring or traces leading up to the capacitor from the voltage source. This is why it is critical that a decoupling capacitor be placed physically as close as possible to the power pin it is decoupling; the greater the distance between the cap and the pin, the greater the additional series inductance.

At the frequencies of the power rail noise and the ripple (those are two separate things), you can calculate the impedances of the wiring and the capacitor and determine the attenuations as a starting point. A decoupling cap's additional value kicks in with transient currents, such as the small current spikes on the leading and trailing edges of a CPU's clock. Lotsa gates change state with each clock edge, producing short transient current spikes whose fundamental frequency is much higher than the clock frequency, and way higher than the switching noise coming from the power supply. At these high frequencies the capacitive reactance is much lower, making an effective lowpass filter and very-short-term energy store.

ak

Last edited: Feb 14, 2017
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4. ### Madmicra

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Feb 14, 2017
ok I see... thanks for your feedback

I'll assume, in order to clarify things up in my mind, that at high frequencies, like those operating several ICs where decoupling is really necessary, the wires have measurable impedances... even skin effect is observable and can multiply those wires resistance... I guess the ripple in a power supply can be assumed to be an high frequency signal, even if the IC is operating at low freq.
The internal resistance of the power supply, the impedance of the wires, all summed up, can be though as a series resistance where HF ripple sign is drop.

Regards

Jorge

5. ### AnalogKid

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712
Jun 10, 2015
Correct. Also, decoupling capacitors are really necessary for ALL ICs regardless of the signal frequencies. Some tolerate no decoupling better than others, but do not be fooled. For example, the bipolar 555 has a design quirk, where both output transistors are on briefly every time the output changes state. So even at a very low oscillation frequency and with nothing connected to the output, it tries to short out its power source twice every cycle. The current spike can be over 300 mA and can cause problems for everything around it.

Separate from digital and quasi-digital parts, a fundamental assumption of the operation of any opamp is that the effective power source impedance is zero ohms all the way down to DC. Again, some parts tolerate non-zero power source and ground impedances better than others, but decoupling is critical to successful circuit design.

ak

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