Explain how a transistor works as a switch with a circuit diagram.
Dear Student
NPN Transistor Switching Circuit
the transistor as a switch the
transistor needs to be turned either fully OFF (cutoff) or fully ON (saturated). An
ideal transistor switch would have infinite circuit resistance between the Collector and
Emitter when turned “fully-OFF” resulting in zero current flowing through it when turned “fully-OFF” resulting in zero current flowing through it and zero resistance between the Collector and Emitter when turned “fully-ON”, resulting in
maximum current flow.
1. Cut-off Region
Here the operating conditions of the transistor are zero input base current ( I ), zero
output collector current ( I ) and maximum collector voltage ( V ) which results in a
large depletion layer and no current flowing through the device. Therefore the transistor
is switched “Fully-OFF”.
The input and Base are grounded ( 0v )
• Base-Emitter voltage V < 0.7v
• Base-Emitter junction is reverse biased
• Base-Collector junction is reverse biased
• Transistor is “fully-OFF” ( Cut-off region )
• No Collector current flows ( I = 0 )
• V = V = V = ”1″
• Transistor operates as an “open switch
2. Saturation Region
Here the transistor will be biased so that the maximum amount of base current is applied,
resulting in maximum collector current resulting in the minimum collector emitter
voltage drop which results in the depletion layer being as small as possible and maximum
current flowing through the transistor. Therefore the transistor is switched “Fully-ON”.
• The input and Base are connected to V
• Base-Emitter voltage V > 0.7v
• Base-Emitter junction is forward biased
• Base-Collector junction is forward biased
• Transistor is “fully-ON” ( saturation region )
• Max Collector current flows ( I = Vcc/R )
• V = 0 ( ideal saturation )
• V = V = ”0″
• Transistor operates as a “closed switch”
Then we can define the “saturation region” or “ON mode” when using a bipolar
transistor as a switch as being, both junctions forward biased, V > 0.7v and
I = Maximum. For a PNP transistor, the Emitter potential must be positive with respect
to the Base.
Regards
NPN Transistor Switching Circuit
the transistor as a switch the
transistor needs to be turned either fully OFF (cutoff) or fully ON (saturated). An
ideal transistor switch would have infinite circuit resistance between the Collector and
Emitter when turned “fully-OFF” resulting in zero current flowing through it when turned “fully-OFF” resulting in zero current flowing through it and zero resistance between the Collector and Emitter when turned “fully-ON”, resulting in
maximum current flow.
1. Cut-off Region
Here the operating conditions of the transistor are zero input base current ( I ), zero
output collector current ( I ) and maximum collector voltage ( V ) which results in a
large depletion layer and no current flowing through the device. Therefore the transistor
is switched “Fully-OFF”.
The input and Base are grounded ( 0v )
• Base-Emitter voltage V < 0.7v
• Base-Emitter junction is reverse biased
• Base-Collector junction is reverse biased
• Transistor is “fully-OFF” ( Cut-off region )
• No Collector current flows ( I = 0 )
• V = V = V = ”1″
• Transistor operates as an “open switch
2. Saturation Region
Here the transistor will be biased so that the maximum amount of base current is applied,
resulting in maximum collector current resulting in the minimum collector emitter
voltage drop which results in the depletion layer being as small as possible and maximum
current flowing through the transistor. Therefore the transistor is switched “Fully-ON”.
• The input and Base are connected to V
• Base-Emitter voltage V > 0.7v
• Base-Emitter junction is forward biased
• Base-Collector junction is forward biased
• Transistor is “fully-ON” ( saturation region )
• Max Collector current flows ( I = Vcc/R )
• V = 0 ( ideal saturation )
• V = V = ”0″
• Transistor operates as a “closed switch”
Then we can define the “saturation region” or “ON mode” when using a bipolar
transistor as a switch as being, both junctions forward biased, V > 0.7v and
I = Maximum. For a PNP transistor, the Emitter potential must be positive with respect
to the Base.
Regards