View Full Version : Transistor info
04-21-2010, 11:33 PM
I am having some trouble understanding the transistor and have a couple of things that I would like to know about transistors.
1. Why is the output of a transistor at the collecter?
2. How does a transistor invert a signal?
3. When will a darlington resistor go on?
04-21-2010, 11:47 PM
Functional model of an NPN transistor
The operation of a transistor is difficult to explain and understand in terms of its internal structure. It is more helpful to use this functional model:
The base-emitter junction behaves like a diode (http://www.kpsec.freeuk.com/components/diode.htm).
A base current IB flows only when the voltage VBE across the base-emitter junction is 0.7V or more.
The small base current IB controls the large collector current Ic.
Ic = hFE × IB (unless the transistor is full on and saturated)
hFE is the current gain (strictly the DC current gain), a typical value for hFE is 100 (it has no units because it is a ratio)
The collector-emitter resistance RCE is controlled by the base current IB:
IB = 0 RCE = infinity transistor off
IB small RCE reduced transistor partly on
IB increased RCE = 0 transistor full on ('saturated')
04-21-2010, 11:48 PM
A transistor inverter (NOT gate)
Inverters (NOT gates) are available on logic ICs but if you only require one inverter it is usually better to use this circuit. The output signal (voltage) is the inverse of the input signal:
When the input is high (+Vs) the output is low (0V).
When the input is low (0V) the output is high (+Vs).
Any general purpose low power NPN transistor can be used. For general use RB = 10khttp://www.kpsec.freeuk.com/images/ohm.gif and RC = 1khttp://www.kpsec.freeuk.com/images/ohm.gif, then the inverter output can be connected to a device with an input impedance (resistance) of at least 10khttp://www.kpsec.freeuk.com/images/ohm.gif such as a logic IC or a 555 timer (trigger and reset inputs). If you are connecting the inverter to a CMOS logic IC input (very high impedance) you can increase RB to 100khttp://www.kpsec.freeuk.com/images/ohm.gif and RC to 10khttp://www.kpsec.freeuk.com/images/ohm.gif, this will reduce the current used by the inverter.
04-21-2010, 11:49 PM
This is two transistors connected together so that the current amplified by the first is amplified further by the second transistor. The overall current gain is equal to the two individual gains multiplied together: Darlington pair current gain, hFE = hFE1 × hFE2
(hFE1 and hFE2 are the gains of the individual transistors)
This gives the Darlington pair a very high current gain, such as 10000, so that only a tiny base current is required to make the pair switch on.
A Darlington pair behaves like a single transistor with a very high current gain. It has three leads (B, C and E) which are equivalent to the leads of a standard individual transistor. To turn on there must be 0.7V across both the base-emitter junctions which are connected in series inside the Darlington pair, therefore it requires 1.4V to turn on.
04-21-2010, 11:49 PM
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