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Transistor Switching Circuit

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expelleior

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Hi,

I dont understand the theory behind the two transistor switching circuit . This is because I have used the parallel port to switch on a BC338 witch lit up a lamp from a secondary power source.

In the picture, if it helps could you describe the operation of the circuit please.
 

The two transistors are connected as a darlington transistor that has a high current gain.
When the input voltage to the 10k resistor goes positive more than about 2.0V then the two transistors turn on and the output device gets the supply voltage minus 0.7V.
When the input voltage to the 10k resistor goes down to 1.0V or less then the two transistors turn off.

The diode arrests the inductive voltage spike when the output device turns off.
 

thanks but I still dont get why you dont just use 1 transistor ?
 

If the load (in your case a light or LED) can be driven by a single transistor (usually Ic is less then 100mA) you don’t need to add another one, but if you want (or need) to increase the output current and in the same time maintain the input current at very low level you just add another transistor (darlington configuration) or use a relay driven by a single transistor ..

BC639 Ic(max)=1A
BC548 Ic(max)=0.1A

See this website for detailed explanation on how to choose a suitable switching transistor ..
http://www.kpsec.freeuk.com/trancirc.htm

Rgds,
IanP
 

Since the schematic includes an inductive kickback diode I will assume that it is intended to sink high current. The darlington configuration has a current gain of the first transistor beta multiplied by the second transistor beta, i.e. it's high.
 

Two transistors are used here as Darlington Pair, in which the current gain is β1xβ2.

And, by increasing the current gain, one can drive a Relay, which can be used in place of "Output Device". (And that's why there is a Diode here.)

If, you don't want those 2 transistors, then go for ULN2003 IC, which is an array of Darlington Pair Transistors.
 

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