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Lesson 12
A flip-flop is a ball on a hill. Its latch is two inverters in a ring with two valleys to rest in, and a hilltop between them where it can balance. Land the data edge at the wrong instant and the ring starts near the top, and takes its own time to roll off. Commit to how long, then fire up the scope. Lesson 11, Ground bounce, is the same chip from its pins: here, what happens inside one of its flops.
Predict · 1 of 2 · the hang
A latch on 1.2 V samples D on the clock's rising edge. There is one instant, Δt*, at which a data edge leaves it perfectly balanced. With the data edge 1 ps past Δt*, Q takes 68.9 ps after the clock edge to settle (the scope). Now the edge lands 100 fs past Δt*, 10 times closer. How long does Q take now? Drag on the scope or use the slider.
Result
Waiting on your call.
Make your mark, then lock it in. The answer, your error and the reason behind it land here, and the scope and every view below fire up.
Engine: the latch at transistor level, two CMOS inverters (SPICE level-1 MOSFETs) and pass devices, solved in the browser by fourth-order Runge–Kutta on a fixed 100 fs step; Δt* by bisection on the decision to 10⁻²⁴ s; τ measured by a least-squares fit over 34 decisions at offsets from 1 ps to 10 zs either side; checked against an independent implicit solve · LTspice and ngspice: its numbers show once you're in the lab (they'd give the answers away) · noise stated, not simulated; one latch for the flop (its slave stage only passes the decision on); pass devices as linear conductances, not transmission gates; the transistors' values illustrative, chosen for τ = 20 ps (an older-process figure: modern flops resolve in a few ps, so the MTBFs here are for this τ); square-law devices with no channel-length modulation, so the inverters' gain at balance is infinite (see How it works).