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I've been reading my school textbook and found out that a Transistor can act in 3 different modes i.e. active mode, saturation mode and cut-off mode (let's just for now ignore the reverse-active mode). The cut-off mode is the mode when no current flows through the base if I am right (I am talking about the C-E config). The saturation mode is when all the current provided by the external cell flows through the Transistor.

So what actually is the active mode. Also is the value of saturation current different for different values of base current?

JYelton
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radiantshaw
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1 Answers1

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The internal structure of NPN bipolar transistor looks like this:

enter image description here

Cut-off

You're quite right about the cut-off mode - in this mode there is no current flow into base and no current flow into collector. However, usually the cut-off mode is defined in terms of base-to-emitter voltage (for example: \$V_{BE}\le0.7V\$ - cut-off).

The reason why there will be no current in cut-off is because the BE junction is reverse biased and does not allow for current to flow through it.

Note that saying that there will be no current flow is an approximation - there will be leakage currents (these low currents are usually neglected).

Active mode

In active mode of operation the BE junction is forward biased (usually stated as \$V_{BE} \ge 0.7V\$), while the CB junction is reverse biased because \$V_{CE}>V_{BE}\$ (\$V_{CB}>0V\$).

In active mode any current flowing into the base terminal will result in an amplified current flowing into the collector terminal (by a factor of \$\beta\$). This mode of operation is useful for amplifiers.

Saturation mode

In saturation mode of operation the BE junction is still forward biased, but CB junction also becomes forward biased (\$V_{CB}<0\$).

You can think of saturation mode as being the limiting case of active mode: if too much biasing current is flowing into the base in active mode, the current flowing into collector becomes very large, thus causing the collector's voltage to go down. Why this happens? Because the circuit supplying current to the collector will usually have some resistance, thus the increased current will lead to an increase in voltage drop on this resistance, leading to a reduction in collector's voltage.

At the onset of saturation mode (which is the the offset of active mode) the current gain is still \$\beta\$, but it reduces the deeper the device in saturation. This means that there is current amplification in "early" saturation (collector's current is still controlled by the base current), but the differential amplification factor reduces to 0 at some point, thus causing the collector's current to become insensitive to base current.

Answer

There usually be some dependence of collector's saturation current on base current, but the ratio will be lower than \$\beta\$ (which is known as the "current gain of the transistor in forward active mode").

Vasiliy
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  • > In active mode of operation the CB junction is rev-biased But How?? The n-type(collector) is connected to positive and the p-type(base) is also connected to positive... Then how this CB is rev-biased?? – radiantshaw Jan 31 '14 at 02:15
  • @TheDreamCoder17, take a look at the schematics: if \$V_{CE}>V_{BE}\$ then \$V_{CB}>0\$ which implies that CB junction is reverse biased. You might want to take a look at this question in order to get a deeper understanding of voltage concept: http://electronics.stackexchange.com/a/75169/25469 – Vasiliy Jan 31 '14 at 08:18