Is the knee voltage in a BJT transistor is the same as the saturation voltage?
That is, in the cutoff region, is the collector emitter voltage drop the same as the knee voltage, the voltage above which any further increase induces no increase in the collector current?
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2\$\begingroup\$ Question: Does a definition for such a quantity Vce(knee) exist? Where did you find such a term? \$\endgroup\$– LvWCommented Dec 20, 2023 at 8:55
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The short answer is no.
Knee voltages are those across a collector-emitter junction when the forward current spikes (think of the graph of an approximate diode model). This is commonly considered 0.7 V, like the voltage drop across the base-emitter junction, for practicality's sake. Actual values vary a bit.
Saturation occurs across the collector-emitter junction at the point that coincides with the top of the load line, which is a graphical means of easily finding a transistor's Q or quiescent voltage/current point. It has collector current on the y-axis and collector-emitter voltage on the x-axis.
The true saturation point is a little below the upper limit of the load line, as the collector-emitter voltage is close to but not exactly zero. The collector-emitter voltage at saturation should inform you of the max collector current. Upon saturation, a transistor experiencing increases to its base current will experience no more increase to its collector current.
Load lines are implemented in this way because they make handy visual references of the main points of a given transistor's behaviour. This allows you to decide how you want to use any given model, whether that's as a switch or an amplifier.
