Circuit Response
every model has an edge, and the figure carries it
Concept

Positive feedback — where it appears

Returning part of an output to add to the input, so that a change reinforces itself. With a loop gain above one it gives a circuit more than one stable state or an oscillation, and it is what makes latches, comparators and thermal runaway.

Named by 2 essays across one field — each of them below, with the objects they name alongside it.

Three bias points below a coupling of 56.6 kΩ, one above — and the middle one is where Newton from nought lands. computed by solving, not by drawing. Two transistors, each with 1 kΩ to a 5 V supply and each biased from the other's collector through a coupling resistor into a 10 kΩ base resistor, solved by Newton on the netlist from six starting guesses and de-duplicated. Below a coupling of 56.6 kΩ there are three operating points: two latched, with one collector at 0.061 V and the other at 4.615 V at 10 kΩ, and a symmetric one at 1.702 V whose Jacobian determinant is negative, so the circuit cannot rest there. Above it the three merge into one symmetric point, and the circuit is an amplifier rather than a memory.

The circuit with three bias points

Two transistors, each biased from the other's collector, have three operating points: one collector at 0.061 V and the other at 4.615, the same the other way round, and both at 1.702. All three solve the circuit to a part in 10⁹, and Newton started from nought — where a simulator starts — finds the symmetric one, which is the one the circuit can never sit at: its Jacobian determinant is negative, so any positive capacitance at the nodes makes it run. Weaken the coupling past 56.6 kΩ and the three merge into one, the loop gain around the symmetric point having fallen through one. Which answer a solver returns depends on where it begins, and from 1,681 starting guesses at 10 kΩ, 437 return the one that is not a state.

semiconductors · Operating point At θ = 4 K/W the transistor holds at 311.7 K; the second crossing at 333 K runs away. computed by solving, not by drawing. The power a transistor dissipates against its junction temperature (solid), each point an electrical operating point solved at that temperature with the saturation current and thermal voltage it has there, for a transistor biased from a stiff 1 Ω Thévenin source to carry 100 mA from 15 V at 300 K with no emitter resistor; and the heat a sink of 4 K/W removes (dashed). Operating points are the crossings: 311.7 K with a thermal loop gain of 0.64 and 332.5 K with a thermal loop gain of 1.43. Above θ* = 4.43 K/W the two meet and vanish.

The bias point that heats itself

A transistor driven from a stiff base voltage to carry 100 mA from 15 V dissipates 1.5 W, and every kelvin that heats it raises the current it carries. On a 4 K/W heat sink the heat balance has two solutions — 311.7 K, which holds, and 332.5 K, which runs away — and above 4.43 K/W it has none. A tenth of an ohm in the emitter moves that edge to 7.69 K/W, and from a quarter of an ohm there is no edge below 520 K at all. The designer's estimate from the loop gain at room temperature puts the edge at 11.04 K/W, two and a half times too high, because the loop gain grows as the part warms.

semiconductors · Operating point

Named alongside it

The objects these essays reach for when they reach for this one.

Loop gainNewton raphsonOperating pointBistabilityConvergence orderFixed pointJunction temperatureSaturationThermal resistanceThermal runawayVerification

All concepts