Transistor darlington pair question

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The discussion centers on the operation of a Darlington pair, specifically how TR1 and TR2 interact to control a lamp. When TR1's base-emitter junction is forward-biased, it allows current to flow, which subsequently turns on TR2 and brightens the lamp. The collector of TR1 is connected to the lamp instead of Vcc to maintain saturation and allow for effective current flow through the load. Concerns are raised about the potential wattage dissipation in TR1 when connected to a higher voltage, highlighting the importance of understanding transistor switching mechanisms. Overall, the conversation emphasizes the need for clarity on how Darlington pairs function in saturation mode.
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A bit confused about how it works.
Link - http://www.technologystudent.com/elec1/transis2.htm (3rd fig from top)

The way I understand, transistor TR1's Base-emitter conducts, which fwd biases B-E junction of TR2. This in turn will turn on the lamp, but it won't be really bright.
Once its on, the collector of TR1 is at +Vcc minus the drop across the lamp. This will increase emitter current of TR1, which in turn increases the base current going to TR2, which is amplified and makes the lamp glow even more.
Correct?

Also, why is TR1's collector connected to lamp instead of Vcc?
 
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please observe that the darlington pair is being operated in saturation mode/cutoff mode in quiescent point. The gain of the transistor pair is a fixed value : hfe^2.
when base of TR1 is driven high VBE is > 0.7V hence TR1 enters sat mode, gushing out emitter current. which inturn drives TR2.
The lamp is the load here, it is turned on when current flows through it.
this is possible only if the Darllington pair in saturation to provide a path to ground.
Hope this explanation helps.
 
likephysics said:
Also, why is TR1's collector connected to lamp instead of Vcc?

I'm puzzled by this too. If TR1's collector were connected to Vcc, then TR2's collector could be brought to about a diode's drop away from ground. As it is, TR1's collector can't go closer than about two diode's drops from ground without taking the transistors out of saturation.
 
Chandra214 said:
please observe that the darlington pair is being operated in saturation mode/cutoff mode in quiescent point. The gain of the transistor pair is a fixed value : hfe^2.
when base of TR1 is driven high VBE is > 0.7V hence TR1 enters sat mode, gushing out emitter current. which inturn drives TR2.
The lamp is the load here, it is turned on when current flows through it.
this is possible only if the Darllington pair in saturation to provide a path to ground.
Hope this explanation helps.

Initially when TR1 has no base current, its collector is open. Just base current is enough to drive it into saturation? No need for BC junction to be reverse biased?
Now, I can't figure out how a transistor switch works. How can collector current flow when there is no potential at Collector terminal.
 
Figure the wattage dissipated by TR1 when its collector is tied to +9V. It will unnecessarily be dissipating well over a watt.
 
Averagesupernova said:
Figure the wattage dissipated by TR1 when its collector is tied to +9V. It will unnecessarily be dissipating well over a watt.

Good point.
 
likephysics said:
Initially when TR1 has no base current, its collector is open. Just base current is enough to drive it into saturation? No need for BC junction to be reverse biased?
Now, I can't figure out how a transistor switch works. How can collector current flow when there is no potential at Collector terminal.

likephysics , I would suggest you to go through transistor operation @ tpub.com
I have found the explanations simple and accurate.
 
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