# Characteristic of a common emitter transistor [closed]

How is Base emitter junction and collector emitter junction biased? How do we determine the value of potential difference between emitter and collector required to be maintained in order to determine input characteristic of transistor? Explanations covering nitty gritty are most welcome.

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## closed as off topic by David Z♦Dec 23 '12 at 3:01

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This question is rather broad. I'd suggest trying one of the many treatments available: e.g. wikipedia or eecs.berkeley.edu/~hu/Chenming-Hu_ch8.pdf . For specific questions on using transistors, you might want to try the Electrical Engineering stack exchange site, where you will find many knowledgeable folks. –  Art Brown Dec 22 '12 at 18:22
electronics.stackexchange.com would probably be a better home for this question. –  Qmechanic Dec 22 '12 at 19:59

Well, the biasing of the Base-Emitter (BE) and Collector-Emitter (CE) junctions is determined by the operation mode. Based on the "common emitter" in the title and your goal of determining input characteristics I am guessing that you are interested in the operation of a BJT transistor as a common-emitter small-signal amplifier. For this case the transistor will be operating in the active mode. I'll use the following figure as reference

A short answer to your question is: In the active mode the BE junction is forward biased and Collector-Base (CB) junction is reversed biased.

Here is the explanation and some quantitative estimates for the biasing of these two junctions:

Base-Emitter Junction

For values of current that are typical in these types of circuits $V_{BE} \approx 0.7\,V$ (at room temperature). The transistor is in the cut-off mode if $V_{BE} < 0.7\,V$. In other words, the transistor "turns on" at $0.7\,V$. In the active mode $V_{BE}$ deviates very little from its typical value of $0.7\,V$. As a result, in analytical calculations (done by hand) $V_{BE}$ is commonly approximated as a constant fixed voltage (in the active mode of course). If you're interested, the $V_{BE}$ can be exactly determined from the relation $$I_C = I_S \left(\exp\left(\frac{eV_{BE}}{k_B T}\right)-1\right)$$ where $I_C$, $I_S$, $e$, $k_B$, $T$ are the collector current, saturation current, electron charge, Boltzmann constant, and temperature respectively.

Collector-Base Junction

The collector voltage $(V_C \:\text{or}\: V_{\text{out}})$ will drop as $I_C$ increases $(V_C = V_+ - I_C R_C)$. If $V_C$ drops to a point such that $V_{CB} < -0.4$ (while $V_{BE}$ still forward biased) then the transistor enters the saturation mode. In other words, the CB junction becomes forward biased. The transition between the saturation and active modes can be clearly seen in the figure

The transition occurs for $V_{BC} = -V_{CB} = 0.4\text{-}0.5$. This is analogous to the $0.7\,V$ of the BE junction. The channel from base to collector "turns on" as well in the saturation mode.

So to summarize, for a common emitter amplifier you need to be in the active mode. To be in the active mode $V_{BE}$ and $V_{CB}$ must at least be $0.7\,V$ and $-0.4\,V$ respectively. From these constraints we can infer that $V_{CE}$ must at least be $0.7\,V+(-0.4\,V) = 0.3\,V$

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