Editing Emitter-coupled logic (section)

== Operation ==

The ECL circuit operation is considered below with assumption that the input voltage is applied to T1 base, while T2 input is unused or a logical "0" is applied.

[[:Image:ECL transition 1000.jpg|During the transition]], the core of the circuit – the emitter-coupled pair (T1 and T3) – acts as a differential amplifier with single-ended input. The long-tail current source (R<sub>E</sub>) sets the total current flowing through the two legs of the pair. The input voltage controls the current flowing through the transistors by sharing it between the two legs, steering it all to one side when not near the switching point. The gain is higher than at the end states (see below) and the circuit switches quickly.

[[:Image:ECL logical0 1000.jpg|At low input voltage]] (logical "0") or [[:Image:ECL logical1 1000.jpg|at high input voltage]] (logical "1") the differential amplifier is overdriven. The transistor (T1 or T3) is cutoff and the other (T3 or T1) is in active linear region acting as a [[Common emitter#Emitter degeneration|common-emitter stage with emitter degeneration]] that takes all the current, starving the other cutoff transistor.<br>The active transistor is loaded with the relatively high emitter resistance ''R''<sub>E</sub> that introduces a significant negative feedback (emitter degeneration). To prevent saturation of the active transistor so that the diffusion time that slows the recovery from saturation will not be involved in the logic delay,<ref name = "unitd04" /> the emitter and collector resistances are chosen such that at maximum input voltage some voltage is left across the transistor. The residual gain is low (''K''&nbsp;=&nbsp;''R''<sub>C</sub>/''R''<sub>E</sub>&nbsp;&lt;&nbsp;1). The circuit is insensitive to the input voltage variations and the transistor stays firmly in active linear region. The input resistance is high because of the series negative feedback.<br> The cutoff transistor breaks the connection between its input and output. As a result, its input voltage does not affect the output voltage. The input resistance is high again since the base-emitter junction is cutoff.