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Part 1 · Simple: schematic circuits

1 · Voltage divider

Status: ✅ Runs today

Goal: run a simulation and find your answer in the output.

Build​

01-divider.circuit.tsx
import { analog } from "tscircuit"

export default () => (
  <board routingDisabled>
    <voltagesource name="V1" voltage="5V" />
    <resistor name="R1" resistance="10k" footprint="0402" />
    <resistor name="R2" resistance="4.7k" footprint="0402" />
    <trace from=".V1 > .pin1" to=".R1 > .pin1" />
    <trace from=".R1 > .pin2" to=".R2 > .pin1" />
    <trace from=".R2 > .pin2" to=".V1 > .pin2" />
    <voltageprobe name="VOUT" connectsTo=".R2 > .pin1" referenceTo=".V1 > .pin2" />
    <analog.dcoperatingpointsimulation name="DIVIDER_OP" spiceEngine="ngspice" />
  </board>
)

This is what each line does:

  • <voltagesource> is an ideal 5 V source. pin2 is its negative terminal, and we use it as ground.
  • <trace from to> is a wire. ".R1 > .pin2" reads as "pin 2 of R1".
  • <voltageprobe> marks the node you care about, and the node it's measured against.
  • <analog.dcoperatingpointsimulation> is the analysis, the same as .op in SPICE.
  • routingDisabled skips PCB autorouting, which a schematic simulation doesn't need.

Run​

bunx tsci simulate 01-divider.circuit.tsx

What you see​

Index  v(n1)        v(n2)        i(vsimulation_voltage_source_0)
0      5.000000e+0  1.598639e+0  -3.401361e-4
  • v(n1) = 5 V is the source node.
  • v(n2) = 1.5986 V is the divider output, your VOUT.
  • The last column is the source current: −340 µA, flowing out of the + terminal.

Check by hand​

  • Vout = 5 × 4.7k / (10k + 4.7k) = 1.5986 V ✓
  • I = 5 / 14.7k = 340.1 µA ✓

Try this​

Change R2 to 10k. Before you run it, predict both numbers.