@Muse - A constant current source driving a capacitor produces an ideal linear voltage ramp, forming the simplest analog integrator.
### How It Works
The fundamental relationship between current and voltage for a capacitor is:
$$i(t) = C \frac{dv(t)}{dt}$$
Integrating both sides with respect to time yields:
$$v(t) = \frac{1}{C} \int_0^t i(\tau)\,d\tau + V(0)$$
Because #I1 supplies a steady direct current of $I = 100\ \mu\text{A}$ into #C1 ($C = 1\ \mu\text{F}$), the output voltage at #out ramps up linearly from $0\text{ V}$:
$$\frac{dv}{dt} = \frac{I}{C} = \frac{100\ \mu\text{A}}{1\ \mu\text{F}} = 100\text{ V/s}$$
Every $50\text{ ms}$, the voltage at #out increases by $5\text{ V}$.
### Reset Functionality
In an ideal circuit without a reset path, a constant current source would charge the capacitor indefinitely. The push button #RST is placed in parallel with #C1:
- Press and hold #RST to short-circuit #C1 to ground and reset #out to $0\text{ V}$.
- Release #RST to allow #I1 to restart integrating from zero.
