I have read about capacitive loading, but what does it actually mean intuitively?

I just asked this question so as to get an intuitive answer and also know what should I understand when one says that a load is capacitive.

Question 1:

When we have a capacitive load, I know that current leads voltage in phase and therefore we will get the current on the capacitor before the voltage reaches its maximum value.

What does it mean when one says,"It is a capacitive load" and what should I understand (intuitively) when one says so?

Question 2:

It would be great if one can also add on "why output impedance should be low for capacitive loads."

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    \$\begingroup\$ Do you understand what it means when one says "It is a resistive load"? Do you fully understand that phrase, why they are important and where they are used? \$\endgroup\$
    – Andy aka
    Aug 12, 2020 at 16:55
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    \$\begingroup\$ Yes, resistive loads are easy to understand. But I am not able to understand the capacitive load. Understood its plain theoretical meaning. But require some intuitive understanding \$\endgroup\$
    – user220456
    Aug 12, 2020 at 17:08
  • \$\begingroup\$ An antenna can be a capacitive load. At different frequencies it can be an inductive load. Piezo is a highly capacitive load period. Crystals can be capacitive, resistive or inductive over the space of a few hertz. Loudspeakers are mainly resistive but with some inductance in series. Cable can be fairly capacitive at low frequencies but become resistive at high frequencies. \$\endgroup\$
    – Andy aka
    Aug 12, 2020 at 17:22
  • \$\begingroup\$ The first thing to understand is what does a voltage square wave (with series resistance) look like when it drives a capacitive load, a resistive load, and an inductive load? This will give you some intuition about the effect. \$\endgroup\$
    – user57037
    Aug 12, 2020 at 17:50
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    \$\begingroup\$ What is your background? Have you studied complex impedance? \$\endgroup\$
    – J...
    Aug 13, 2020 at 13:35

6 Answers 6


Your question is so fundamental that it can be answered (if you want and if you have the courage to accept it) fully intuitively without any math or knowledge of electricity. The capacitive load is simply a capacitor; so you have to have a good intuitive notion about the capacitor. But you already have it from life - a tank that can store something (air, water, money, data, charge, energy, reputations:-), and your task is to fill or empty it... fast or slow... Fish tank and communicating vessels are the most popular fluid analogies of the capacitance phenomenon. So you have the answer to the question when these tanks (capacitors) will charge quickly and why...

Now let's put the analogies aside and discuss something "more serious". Like anything in this world, capacitive load can be both useful and harmful:

A useful capacitive load is, for example, the capacitor in an RC integrating circuit. In this case, its slow charging is something we want, because it allows us to get an idea of ​​the time through the voltage (hence the resistor in series to the capacitor). In this way, we can make timers (555), ramp generators and more.

In other cases (e.g., dynamic RAM, sample&hold circuits, peak detectors), the capacitive load is also useful since we used it to memorize (store) digital or analog data... but now we want to quickly charge the capacitance. In these cases, the charging voltage source has to be powerful enough so as not to "dip up" (as you guessed).

Strictly speaking, this arrangement is incorrect because the capacitor short-circuits the voltage source... but it is still used. The capacitor differentiator (without a resistor) is another well-known example.

An undesired capacitive load is the parasitic (stray) capacitance of elements and wires that causes them to behave to some extent as capacitors. The undesired capacitance appears "in parallel" to the useful property of resistance or inductance... or an open circuit (no load connected). In these cases, the slow charging is undesired. And we know how to solve this problem from our routine (the analogies above) - by charging with high current through low resistance (hence the absence of resistance).

Capacitive loads may cause problems at the complementary outputs of digital circuits. At high output voltage ("1") they are charged through the upper transistor (the charging current exits the output). At low output voltage ("0") they are discharged through the lower transistor (the discharging current enters the output).

Finally, about the phase shift... it is a great challenge to explain it in a fully intuitive way...

Imagine you slowly started to open the tap to fill a bucket with water (the water flow is the "current" and the water level is the "voltage"). As a result, the "voltage" increases when the "current" increases.

Then begin slowly closing the tap (to empty the bucket, as you probably think:-). Supprisingly, the "voltage" continues increasing although the "current" decreases. This is an example of the so-called "phase shift"...

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    \$\begingroup\$ Thank you for the wonderful answer. I am just not able to get the understanding of this one line - "In other cases (e.g., dynamic RAM, sample&hold circuits, peak detectors), the capacitive load is also useful since we used it to memorize (store) digital or analog data... but now we want to quickly charge the capacitance." - Why do you want to charge the capacitance quickly? And my last question - When you say low impedance - It is actually kind of 0ohms of resistance. So, during to low resistance, high current will flow and the voltage will actually dip atleast for some amount right? \$\endgroup\$
    – user220456
    Aug 13, 2020 at 6:43
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    \$\begingroup\$ You are welcome. I can answer later because I am on my way... \$\endgroup\$ Aug 13, 2020 at 9:29
  • \$\begingroup\$ @Newbie, I think that, first of all, you need to develop a very good intuitive idea of such an electrical phenomenon as ​​"capacitance". Generally speaking, this is a property of accumulating something ... that takes time... and cannot happen instantly. You ask me, "Why do you want to charge the capacitance quickly?" Let's consider an example from photography. In an electronic flash, we charge a capacitor to high voltage and then discharge it through a flashtube to produce a flash. We want to be ready to produce the next flash right after that. So, we need "to charge the capacitor quickly". \$\endgroup\$ Aug 13, 2020 at 19:18
  • \$\begingroup\$ Another example: You have an electric car powered by a supercapacitor. Unfortunately, it is discharged... but you need to travel urgently. So, you have "to charge the supercapacitor quickly"... \$\endgroup\$ Aug 13, 2020 at 19:19
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    \$\begingroup\$ thank you for the wonderful answer which helped me to understand the concept intuitively. Accepted the answer. Thank you \$\endgroup\$
    – user220456
    Aug 14, 2020 at 3:00

But what does it mean when one says,"It is a capacitive load"

It means the load behaves like a capacitor. You have to deliver charge to it before the voltage across it will change.

why output impedance should be low for capacitive loads?

Because if you want to change the voltage across this load, you'll have to deliver a significant current. You need a low-impedance load to deliver a current without the voltage sagging.

  • 1
    \$\begingroup\$ Thank you for the immediate answer. So, if I have a 5V Regulator output connected to a capacitive load, the output impedance of the regulator should be low so that it can give the current to the capacitive load without any dip in its output 5V. Am I right? One doubt : Why are capacitive loads important or where is it used widely? \$\endgroup\$
    – user220456
    Aug 12, 2020 at 16:25
  • \$\begingroup\$ In the case of a regulator, you should check the specs to be sure it can power a capacitive load without loss of stability. \$\endgroup\$
    – The Photon
    Aug 12, 2020 at 16:40
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    \$\begingroup\$ Some regulators, if the load is capacitive, will oscillate, producing big voltage variations on their output. This generally isn't good for the operation of those loads. Having a low output impedance is one part of avoiding this problem, but there's more to it than just that. \$\endgroup\$
    – The Photon
    Aug 12, 2020 at 16:51
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    \$\begingroup\$ I am trying to understand that. But I am finding difficulty in understanding this - Low impedance along with capacitive load. When one says low impedance - It is actually kind of 0ohms of resistance. So, during low resistance, high current will flow and the voltage will actually dip atleast for some amount right? So, how can you provide huge current (to a capacitive load in this case) without the voltage dipping even for a small value? Could you please explain @Justin \$\endgroup\$
    – user220456
    Aug 13, 2020 at 16:26
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    \$\begingroup\$ The lower the source impedance, the less the output voltage dips for a given current flow. No source is perfect, the voltage will always dip. The question is how to minimize the dip (knowing you can't eliminate it altogether). \$\endgroup\$
    – The Photon
    Aug 13, 2020 at 16:27

This is just a conceptual model, and won't hold up to much scrutiny, but might help get your mind going in the right direction.

Imagine your circuit is a piece of tubing. The power source is you blowing into one end. DC could be you blowing with a steady pressure, while AC could be you blowing in and out.

Now, cap the other end and put a tiny hole in it...a restriction if you will. This is a resistive load. Constant pressure = constant flow.

For the capacitive load, imagine replacing the small hole with a balloon instead. Now constant pressure does not equal constant flow. The flow slows down as the balloon/capacitance "fills up".

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    \$\begingroup\$ Nice air (pneumatic) analogy... For drivers, you could tell it like pumping a tire (and using a valve to explain the diode behavior). \$\endgroup\$ Aug 12, 2020 at 20:11

Simple answer

Big Capacitors store charges like a tiny battery. They draw more current with the rate of change of voltage, just as shock absorbers react with more force with pot-hole step size or rate of change of position = speed. Current is a good analogy to force as it is proportional to torque in motors. So Capacitive loading is like choosing bigger shocks for a heavier vehicle to create a greater force to slow down the axle over a bump. A washboard road creates an AC force on the axel and the shocks must be selected to match the spring-mass resonance for best damping.

Ticky tacky technical answer

All Capacitors have some Effective Series Resistance, ESR, therefore creating the minimum possible resistance at some breakpoint frequency. Then they become inductive above this.

Power supplies using negative feedback have limited gain-bandwidth products so the output impedance is reduced by this gain. Thus as load frequency increases this output impedance is the opposite of a low ESR shunt capacitor.

Often the bandwidth of acceptable low impedance of a cap spans only a few decades for low impedance, and ESR also changes with bulk size and chemistry of the cap. Therefore in critical SMPS applications it may be common to see a range of output Caps from Ceramic, Film to Electrolytic.

The impedance of a cap is defined as \$\dfrac {1}{\omega C}+ESR\$ up to the breakpoint frequency where these are equal.

Thus when paired with an active LDO or SMPS that a rising low impedance defined from the load regulation error and up, the goal is to achieve a near flat output impedance, yet often never achieved due to complexity and tolerance errors.

Load regulation error for some load impedance is like any impedance divider as a ratio of load to total (Load+ source impedance) often spec’d at 98 to 99% or 1 to 2% load regulation error, where the load can be static DC or AC.

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    \$\begingroup\$ I agree but I'm not sure hydaulics are taught in kindergarden. \$\endgroup\$ Aug 12, 2020 at 19:22
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    \$\begingroup\$ Tony, as far as I can remember, I was a few years old (before kindergarten) when I filled empty cans with water, drilled small holes in them and watched with interest how the water flowed slowly... and I also didn't know what hydraulics was... but i figured out what an integrator was:-) Your "shock absorber" analogy is wonderful! \$\endgroup\$ Aug 12, 2020 at 20:13

Our professor has used a very simple fact to make the behaviour clear:

at an ideal capacitance voltage can not jump, but current can.

When a discharged capacitor is connected to some source (ideal capacitance!), it's initial voltage is 0V. But the initial current which flows into the capacitance is infinite. Then the Voltage will slowly rise with an exponential function, at the same time current will start to go down with an exponential function until it is 0A again.

So if you connect a capacitive load it represents a huge load for a certain amount of time which migth blew fuses, or damage other stuff. In case of large capacitance often a inrush current limiter is used to control the large currents.

"why output impedance should be low for capacitive loads" Its because, when turned on, huge currents want to flow (and will if the power source is strong enough) - low impedance will guarantee that the capacitor gets filles fast and nothing gets too hot

If you turn off the power source, an ideal capacitance will hold the stored energy forever, voltage will never change. If any load is sill connected parallel to the ideal capacitance, its voltage will initially stay constant and start to go down with an exponential function until it is 0V again. Also initially the current will "jump" to whatever current the load needs.

the same thing can be defined for an ideal inductance.

here the current can not jump but the voltage can.

so an inductance connected to a source will initially not change its current. If it was e.g. 0A the inductance will work against the source by producing a inverted voltage... Well, it will work against the power source in any case where the current which was flowing prior to connecting the power source is any different to the current which physicaly wants to flow with the source connected (hope thats a good explaination...) as soon as the inductance has no more stored energy, it will lose the fight with the power source.

The same happens when you turn off the power source. now the inductance will again try to keep the current steady by acting as a power source itself (now we can assume it has stored energy) - with infinite voltage if resistance is low enough. The result can be arcing in switches or relays for example. Thus when turning off a inductance you must provide a path for the energy to compensate or turn the inductance off when the current is 0A

capacitive load - be carefull when turned on (e.g. phase control: use a device which turns on at zero crossing - but you can turn of at any time)

inductive load - be carefull when turned off (e.g. phase control: use a device which turns off at zero crossing, but you can turn on at any time )

  • \$\begingroup\$ Thank you for the insightful answer. \$\endgroup\$
    – user220456
    Aug 28, 2020 at 3:47

Question 1: What you should recognize is that since you have a capacitive load, what ever is supplying current directly into the capacitor weather it be an opamp or power supply, That device/ circuit will need to be able to handle and/or supply the current needed to charge that capacitor.

Question 2: If you have an opamp for example that has an output impedance of 1kohm, and lets say the capacitor is 10uF. If you try to charge the capacitor to 5v from the output of the opamp, it will take around 50ms to charge because that 1kohm is limiting how fast the capacitor can charge. If you change the output to a lower impedance of 100ohm, it will take around 5ms to charge.

  • \$\begingroup\$ Thank you for your answer. From your answer to the question 2 - In both cases, say 1kohm and 100ohm, during charging, the output 5V of the Op-amp should not dip right? That is the main thing? And why are capacitive loads important and where are they used? \$\endgroup\$
    – user220456
    Aug 12, 2020 at 16:48

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