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2N3904 is a NPN transistor hence the
collector and emitter will be left open (Reverse biased) when the base pin is
held at ground and will be closed (Forward biased) when a signal is provided to
base pin. 2N3904 has a gain value of 300; this value determines the
amplification capacity of the transistor. The maximum amount of current that
could flow through the Collector pin is 200mA, hence we cannot connect loads
that consume more than 200mA using this transistor. To bias a transistor we
have to supply current to base pin, this current (IB) should be limited to 5mA.
When this transistor is fully biassed then it
can allow a maximum of 200mA to flow across the collector and emitter. This
stage is called Saturation Region and the typical voltage allowed across the
Collector-Emitter (VCE) or Collector-Base (VCB) could be 40V and 60V
respectively. When base current is removed the transistor becomes fully off,
this stage is called as the Cut-off Region and the Base Emitter voltage could
be around 600 mV.
When a transistor is used as a switch it is
operated in the Saturation and Cut-Off Region as explained above. As discussed
a transistor will act as an Open switch during Forward Bias and as a closed
switch during Reverse Bias, this biasing can be achieved by supplying the
required amount of current to the base pin. As mentioned the biasing current
should maximum of 5mA. Anything more than 5mA will kill the Transistor; hence a
resistor is always added in series with base pin. The value of this resistor
(RB) can be calculated using below formulae. RB
= VBE / IB
Where the value of VBE should be 5V for
2N3904 and the Base current (IB depends on the Collector current (IC). The
value of IB should not exceed mA.
A Transistors acts as an Amplifier when
operating in Active Region. It can amplify power, voltage and current at
different configurations.
| Product Details | |
| Product Name | 2N3904 NPN Transistor |
| Price | Rs. 10/- |
