🔋 Capacitors + RC Time Constant

A capacitor is the first component in this course that makes time part of the circuit. Resistors relate voltage and current at an instant. Capacitors remember where voltage was a moment ago, and that memory changes the shape of signals, power rails and button presses. Once you understand the RC time constant, you can look at a circuit and ask a better question: how fast is this node allowed to move?

By the end, you can

1. Explain capacitance as charge per volt, using $Q = C V$
2. Predict why a capacitor's voltage cannot change instantly unless an enormous current is available
3. Compute and interpret the RC time constant $\tau = R C$
4. Use the 1τ, 3τ and 5τ landmarks to estimate charging and discharging without a calculator
5. Choose a capacitor family for a rough job: ceramic, electrolytic, film, tantalum or supercapacitor
6. Recognize RC timing in debounce, reset, filtering, coupling and power-rail hold-up circuits

The part that stores voltage history

A capacitor is two conductors separated by an insulator called a dielectric. Push charge onto one conductor and an equal opposite charge is pulled onto the other. The insulator prevents DC current from crossing the gap, but the electric field through that gap stores energy and creates a voltage between the plates.

The defining equation is short:

$$Q = C V$$

$Q$ is charge in coulombs, $V$ is voltage in volts, and $C$ is capacitance in farads. Capacitance means "how much charge must move to change the voltage by one volt." A larger capacitor needs more charge to move its voltage, so it changes more slowly for a given current.

The second equation is the one that makes capacitors feel alive:

$$I = C \frac{dV}{dt}$$

Read it in plain language: current is how fast charge is moving, and a capacitor's voltage changes only when charge moves. If you want a capacitor's voltage to jump instantly, $dV/dt$ becomes enormous, so the current would have to become enormous too. Real circuits cannot provide infinite current. Therefore a capacitor's voltage moves with a slope, not a jump.

This is the most useful sentence in the lesson:

a capacitor resists sudden voltage change.

Not because it is stubborn, and not because it is a battery in the chemical sense. It resists sudden voltage change because changing its voltage requires moving charge, and the rest of the circuit can move charge only so fast.

Capacitance stores energy, not free power

Because a capacitor stores an electric field, it stores energy:

$$E = \frac{1}{2} C V^2$$

That square on voltage matters. Doubling voltage stores four times the energy in the same capacitance, which is useful in power supplies and dangerous in high-voltage gear. It also explains why voltage rating is not a decorative line in a datasheet. A capacitor rated for 6.3 V is not a safe replacement on a 12 V rail.

Small capacitors do not store much energy, but they can deliver it quickly. Big capacitors store more energy, but their construction often makes them slower and less ideal. That tradeoff is why real boards carry several types instead of one magic value.

Surface-mount ceramic capacitors are the default small reservoir on digital boards. They are not polarized, they come in tiny packages, and their low parasitics make them useful at high frequency. They also have an annoying real-world behavior: many high-K MLCCs lose a large fraction of their capacitance when DC voltage is applied. A "10 µF" 0603 part may be much less than 10 µF on a 3.3 V rail. The label is a beginning, not a guarantee.

The resistor sets the speed

A capacitor by itself says, "my voltage changes when charge moves." A resistor says, "for a given voltage difference, current is limited." Put them together and the result is a circuit with a natural clock.

Imagine a 3.3 V source, a resistor and an empty capacitor to ground. At the first instant the capacitor is at 0 V, so the full 3.3 V appears across the resistor. Ohm's law allows a large initial current:

$$I_0 = \frac{3.3\ \text{V}}{R}$$

That current pushes charge into the capacitor, raising its voltage. As the capacitor voltage rises, less voltage remains across the resistor, so the current falls. The charging slows as it approaches the final voltage. That is why the curve is not a straight ramp. It is an exponential.

The time constant is

$$\tau = R C$$

Use ohms and farads and the answer is seconds. If $R = 10\ \text{k}\Omega$ and $C = 100\ \mu\text{F}$:

$$\tau = 10{,}000 \cdot 100\times10^{-6} = 1\ \text{s}$$

The charging equation is

$$V_C(t) = V_{\text{final}}\left(1 - e^{-t/\tau}\right)$$

You do not need to calculate exponentials every time. Memorize the landmarks:

| Time | Charging capacitor is about | Why it matters | | --- | ---: | --- | | $1\tau$ | 63% of final voltage | First useful timing marker | | $2\tau$ | 86% of final voltage | Most of the way there | | $3\tau$ | 95% of final voltage | Often "good enough" | | $5\tau$ | 99% of final voltage | Practically settled |

Discharging follows the matching decay:

$$V_C(t) = V_0 e^{-t/\tau}$$

After one time constant a discharging capacitor has about 37 percent left. After five, it is practically empty for most circuit work.

?A 10 kΩ resistor charges a 100 µF capacitor. What is the RC time constant?

• A 1 second
• B 1 millisecond
• C 100 seconds
• D 10 microseconds

Drive the charging curve

Move the resistance and capacitance sliders. The curve always has the same shape, but the scale of the time axis changes. A larger resistor limits current more tightly. A larger capacitor needs more charge per volt. Either one makes the node move more slowly.

Try these moves:

1. Set capacitance to 100 µF and resistance to 10 kΩ. The time constant is 1 s and 5τ is about 5 s.
2. Double the resistance. The curve shape is the same, but every time landmark doubles.
3. Return to 10 kΩ, then change capacitance from 100 µF to 10 µF. The timing shrinks by a factor of ten.
4. Ask what a GPIO input would see if its logic threshold were around half the supply. It switches before 1τ, not at the final voltage.

What real capacitors are good at

The schematic symbol hides construction. Real capacitors are built from different dielectrics, electrode materials and packages, and those choices decide capacitance, voltage rating, leakage, tolerance, frequency behavior, polarity and failure mode.

Aluminum electrolytic capacitors are the tall cans you see near power connectors and regulators. They provide tens, hundreds or thousands of microfarads in a cheap package. They are usually polarized, so reversing them is a real failure, not a harmless swap. They also have higher equivalent series resistance than ceramics and their electrolyte ages. That makes them good bulk energy reservoirs and poor high-frequency parts.

Ceramics sit near chips. Electrolytics sit near power entry and regulators. That is not a law of nature, but it is a strong first guess because the two parts live on different time scales. The electrolytic handles slow bulk demand. The ceramic handles fast local events. Lesson 42 turns that into a full power-distribution-network design rule.

Film capacitors use plastic film dielectrics. They are larger for the same capacitance, but they can be stable, low leakage and robust. They show up when the capacitor's value and loss need to behave well: timing networks, analog filters, audio paths, snubbers and mains-rated safety parts.

Tantalum and polymer capacitors are compact bulk parts. They can be useful where space is tight, but old-school tantalums have a reputation for failing short if abused. Do not treat them like generic ceramics. Check polarity, surge current and voltage derating.

Supercapacitors are the other extreme. They can store enough energy for backup power, short ride-through or energy harvesting, but their voltage ratings are low, leakage is significant, and charging current must be controlled. You use them when the job starts to look like "small battery, many cycles, fast charge," not when you need a 100 nF ceramic near an IC pin.

The same RC idea keeps returning

Once you see RC timing, you start seeing it everywhere.

Button debounce. A mechanical switch does not make one perfect transition. It bounces. A resistor and capacitor can slow the input node so the chatter is smoothed before the GPIO threshold is crossed. Software debounce is still common, but hardware RC debounce is a useful first line of defense when the edge is dirty.

Reset delay. Many chips need their reset pin held in a known state until the power rail is stable. A resistor charges a capacitor on the reset node, creating a slow release. The reset input does not care that the capacitor is not fully charged; it cares when the voltage crosses the chip's threshold.

Low-pass filtering. A resistor feeding a capacitor to ground makes a one-pole low-pass filter. Slow signals survive. Fast noise is shunted into the capacitor. The corner frequency is

$$f_c = \frac{1}{2\pi R C}$$

This is a first pass, not a magic noise eraser. If the source impedance, load impedance or ADC sample capacitor is not considered, the "filter" can become part of a new problem.

Coupling. A series capacitor blocks DC once it charges, but changing signals still produce current through it. That is why capacitors appear between amplifier stages and on communication lines that need to remove DC offset.

Power rail hold-up. A capacitor across a rail supplies charge while the source is briefly slow or absent. The rough hold-up estimate comes from the same $Q = CV$ idea:

$$I \Delta t = C \Delta V$$

If a load needs 50 mA for 2 ms and the rail may fall by only 0.2 V, the ideal capacitance is

$$C = \frac{I\Delta t}{\Delta V} = \frac{0.050 \cdot 0.002}{0.2} = 500\ \mu\text{F}$$

Then you add margin, check ESR, check surge current, check leakage and remember that the ideal equation only got you started.

?A capacitor on a reset pin charges through a resistor. The supply jumps to 3.3 V, but the capacitor starts at 0 V. What is true at the first instant?

• A The capacitor voltage is still near 0 V, so most of the supply appears across the resistor
• B The capacitor instantly becomes 3.3 V because it is connected to the supply
• C No current flows because capacitors block all current
• D The time constant is zero until the capacitor is charged

↘ Leakage, ESR, ESL and why the simple model eventually breaks

The ideal capacitor model is one capacitance. Real capacitors are a capacitance with leakage resistance in parallel and parasitic resistance and inductance in series. Leakage slowly drains charge, so long timing intervals and backup circuits need low-leakage parts. Equivalent series resistance, ESR, wastes energy and causes a voltage step when current flows. Equivalent series inductance, ESL, makes the capacitor stop behaving like a capacitor above its self-resonant frequency.

That is why a 100 µF electrolytic and a 100 nF ceramic are not interchangeable even though both are capacitors. The electrolytic stores much more charge at low frequency. The ceramic has much lower parasitics and can handle fast local spikes. In a real power network, the two are partners across time scale. In a precision timing circuit, leakage and tolerance may dominate. In a filter, source impedance and load impedance reshape the pole. In a high-speed board, mounting geometry can matter more than the nominal value.

The first model is still worth learning because every correction is attached to it: start with $Q = CV$, $I = C\,dV/dt$ and $\tau = RC$, then ask which non-ideal term is large enough to matter for this job.

Practice 1 warm-up

A 47 kΩ resistor charges a 10 µF capacitor. Find the time constant and estimate the time to be practically settled.

↘ Show worked solution

Convert the capacitance to farads:

$$10\ \mu\text{F} = 10\times10^{-6}\ \text{F}.$$

Then multiply:

$$\tau = RC = 47{,}000 \cdot 10\times10^{-6} = 0.47\ \text{s}.$$

Practical settling is often estimated as 5τ:

$$5\tau = 2.35\ \text{s}.$$

So the node reaches about 63 percent of final voltage after 0.47 s and about 99 percent after 2.35 s.

Practice 2 core

A reset input releases when the capacitor node crosses 1.6 V on a 3.3 V rail. The capacitor is 1 µF and the resistor is 100 kΩ. Estimate whether release happens before or after one time constant.

↘ Show worked solution

The time constant is

$$\tau = 100{,}000 \cdot 1\times10^{-6} = 0.1\ \text{s}.$$

At one time constant the capacitor is about 63 percent of the final voltage:

$$0.632 \cdot 3.3\ \text{V} \approx 2.1\ \text{V}.$$

The reset threshold is 1.6 V, which is below 2.1 V, so release happens before one time constant. The exact time solves

$$1.6 = 3.3\left(1-e^{-t/0.1}\right)$$

which gives about 65 ms. The rough landmark was enough to know the direction.

Practice 3 stretch

You want to smooth a noisy sensor line with a simple RC low-pass. You choose $R = 10\ \text{k}\Omega$ and $C = 1\ \mu\text{F}$. Estimate the corner frequency. Would this be reasonable for a force sensor read only 10 times per second?

↘ Show worked solution

Use the one-pole corner estimate:

$$f_c = \frac{1}{2\pi RC}.$$

The time constant is

$$RC = 10{,}000 \cdot 1\times10^{-6} = 0.01\ \text{s}.$$

So

$$f_c = \frac{1}{2\pi \cdot 0.01} \approx 15.9\ \text{Hz}.$$

For a force sensor read 10 times per second, a 15.9 Hz analog corner might be reasonable as a first pass: it damps much faster noise while keeping slow force changes. You would still test response time, ADC acquisition behavior and whether the sensor source can drive the 10 kΩ plus 1 µF load cleanly.

Key takeaways

• ▸Capacitance is charge per volt: $Q = C V$.
• ▸A capacitor's voltage cannot jump instantly in a real circuit because $I = C\,dV/dt$.
• ▸An RC circuit has a natural time scale: $\tau = RC$. After 1τ a charging capacitor is about 63 percent charged; after 5τ it is about 99 percent settled.
• ▸Real capacitor families trade density, speed, leakage, polarity, stability, voltage rating and failure mode.
• ▸RC timing appears in debounce, reset delay, low-pass filtering, coupling and rail hold-up.
• ▸The simple model is the spine. ESR, ESL, leakage and layout are the corrections you add when the job demands it.