Tutorial: an RC filter from scratch

Draw a low-pass filter, solve it three ways, read its corner frequency off the plot, then put it on a slide with a slider that moves the corner while you talk. About fifteen minutes, no prior Circuit Studio.

We are building the simplest filter there is: a resistor in series, a capacitor to ground, the output taken between them. With R = 1.59 kΩ and C = 100 nF the corner lands at almost exactly 1 kHz, which makes every reading easy to check.

Evaluate the idea first#

The public Signal Conditioning sample lets you change a filter’s resistance and inspect its cutoff readout without an account or Python. Use it to try the relationship before building the circuit below. Creating and saving your own circuit requires an invited beta account; the one-slide workflow explains how to try it alongside an existing presentation.

1 · Open a blank sheet#

In the dashboard sidebar, open Studio, choose Circuit Studio, press New circuit, and pick Blank circuit. You get a grid, an empty library count, and a status bar whose left-hand readout shows the grid coordinate under your pointer. That readout is how you place things exactly; the numbers below are the ones this tutorial was drawn at.

2 · Place four parts#

Press R once before dropping a part to turn it upright, and Esc when you are done placing.

1. Press 4 to arm the DC voltage source, press R to stand it upright, and click at 0, 4. It lands as V1, with its + pin at the top.
2. Press 1 for the Resistor and click at 7, 0 — that is R1, lying flat.
3. Press 2 for the Capacitor, press R to stand it upright, and click at 14, 4. That is C1.
4. Press 6 for Ground and click at 7, 8.
5. Press Esc to stop placing.

The banner over the canvas tells you what is armed the whole time: "Placing Capacitor · click to drop · R rotates · Esc stops".

3 · Wire it up#

Press W for the Wire tool. Wires run orthogonally and turn a corner for you; clicking a pin ends the run there.

1. Click V1's top pin at 0, 2, then click R1's left pin at 5, 0.
2. Click R1's right pin at 9, 0, then C1's top pin at 14, 2.
3. Click V1's bottom pin at 0, 6, then click 14, 8 on the way, then C1's bottom pin at 14, 6.

The ground symbol sits on that bottom run, so it joins it — a dot appears where three arms meet. Connection happens at vertices: a pin, a wire end, or a bend that lands on another wire's segment. Two wires that merely cross do not connect.

Press V to go back to the Select tool.

4 · Give the parts their values#

Click R1. In the Inspector, put 1.59k in its value field and press Enter — it reads back as 1.59 kΩ. Click C1 and type 100n. Click V1 and set its value to 5.

Then, still on V1, set AC amplitude to 1. That is the small-signal stimulus, and it is deliberately a separate number from the 5 V bias: the operating point and the signal are different questions.

Finally, click the wire between R1 and C1 and set its Name to OUT. Naming a wire names its whole net, so the measurement tables, the response plot and any waveform on a slide will all call it OUT from here.

5 · Read the operating point#

Open Measurements at the bottom. The circuit solves as you draw, so it is already there:

• Node voltages — OUT reads 5.00 V, and so does the node between V1 and R1. No current flows, so nothing drops across R1.
• Components — V1, R1 and C1 all read 0 A. C1 holds 5.00 V across it; R1 holds none.

Open Simulation and it says why in one line: "· steady state — capacitors are open at DC; the transient needs time". A capacitor is an open at DC whatever its size, so this circuit has nothing to say until we either move a signal through it or run it over time.

6 · Sweep it — the corner frequency#

Open Response. Set from to 10, to to 1000000, leave per decade at 20, and check that plot is showing V(OUT) — it defaults to the net you named, since a net the stimulus sits on can only ever plot as a flat line.

You get a Bode plot: magnitude in decibels on the left axis, unwrapped phase on the right, one shared logarithmic frequency axis. Decibels are taken against the stimulus, so 0 dB is unity.

Now click the plot at about 1 kHz to pin a frequency. The readout under it should say:

1.00 kHz · -3.01 dB · -45.0°

That is the corner, and both halves agree on it — which is the check. The magnitude is down 3 dB, and the phase is at exactly half of the 90° a single pole eventually gives. It is where you expect it:

fc = 1 / (2πRC) = 1 / (2π × 1590 Ω × 100 nF) ≈ 1.00 kHz

A decade further out, at 10 kHz, the magnitude should read about −20 dB: one pole, 20 dB per decade. Pinning also opens a table of every phasor at that frequency beside the plot, each as a magnitude and an angle, so you can see the capacitor's current leading its voltage by 90° rather than take it on faith.

Click again to unpin.

7 · Watch it charge#

Open Transient. Set run for to 0.001 and step to 0.00001, and leave the trace alone — with nothing chosen it traces the nets a store is attached to, which here is OUT.

The trace climbs from 0 V toward 5 V. Click it at about 160 µs and the cursor readout should show OUT at roughly 3.1 V — 63% of the way, which is the definition of the time constant:

τ = RC = 1590 Ω × 100 nF = 159 µs

Try coarsening the step to 0.0001 and the run refuses instead of drawing a smooth lie:

> The step is too coarse to see C1's time constant (159 µs). Use 15.9 µs or finer, or the waveform will look smooth and be wrong.

Put the step back to 0.00001. The panel also states its method every run — backward Euler, stable and slightly damped by construction, so a ring you can see is a ring the circuit has. That damping is also why the cursor reads a touch under the textbook 3.16 V.

8 · Put it on a slide#

Press Add to deck. Choose Interactive schematic, pick your deck, choose New slide, and leave Link to this circuit checked so the slide follows the circuit as you keep working on it. Press Add to deck to confirm.

Open that deck. The circuit is on the slide, solved, with its readout underneath. Because it is driven only by a signal, the readout says where the action is: · 1.00 V AC in.

9 · Give the corner a slider#

Press P to add a Parameter panel to the same slide, and set its first variable up in the inspector:

• name — R
• min 1000, max 5000, step 10, val 1590
• unit — Ω

Then, in that variable's Drives row, open Choose a value… and pick Circuit · R1 · resistance (Ω). That is the slot the circuit offers for R1's value; slots are keyed by the component, so renaming R1 later will not break the wire you just made.

Drag the slider. The value on the schematic follows it, and the whole circuit re-solves as you drag.

To see it bite, select the circuit object and switch its View to Waveform, then pick V(OUT) under Trace. Now the slider is moving a real curve: at 1 kΩ the trace snaps up in about 100 µs, at 5 kΩ it is still climbing when the run ends. Same circuit, same slide, no screenshots anywhere in the chain.

10 · Present it#

Press Present. The slider stays live in front of the room, and so does everything it drives — drag it while you explain why a bigger resistor means a slower charge and a lower corner. Esc ends the talk.

What to try next#

• Set C1's Initial V to 5 and run the transient again — the same circuit, discharging.
• Add a switch in series and give it a Flips at time; the waveform turns exactly where you said it would.
• Open the Solver tab, pick R1 and ask for its voltage: the hand calculation, with a chip confirming it matches the simulation.

Next#

• Analyses — what each solver models, and every refusal it can give.
• Circuits in your deck — the four views, linking, and parameter slots in full.
• Reactive documents — the parameter system the slider comes from.