🔴 TRANSISTOR - Part 3 || Transistor as a SWITCH || Semiconductor - 19 || for Class 12 in HINDI

In this Physics video in Hindi for class 12 we explained how a Bipolar Junction Transistor (BJT) is used as a Switch. In this video we considered n-p-n bipolar junction transistor in common Emitter (CE) configuration as reference.
When transistor is used as a switch it operates in the cut off region or in the saturation region.
Click here to visit the playlist on this chapter
• Physics 12th : Semicon...
Other Related Video:
🔴 TRANSISTOR - Part 1 | Construction and Working | Bipolar Junction Transistor (BJT)
• 🔴 TRANSISTOR - Part 1 ...

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Case 1: Vi < 0.7 V
When the input voltage (Vi) is less than the threshold voltage of 0.7 V, the base-emitter junction is reverse-biased, and the transistor is essentially turned off. In this case, both Ib and Ic are approximately zero, and the transistor acts like an open switch. The voltage drop across the collector-emitter junction (Vce) is equal to the supply voltage (Vcc).
Now, suppose we gradually increase Vi. As Vi increases, the base-emitter junction becomes increasingly forward-biased, and the transistor begins to turn on. The base current (Ib) starts to flow, which in turn causes the collector current (Ic) to flow. As Ic increases, the voltage drop across the collector-emitter junction (Vce) decreases, according to the formula Vce = Vcc - Ic*Rc.
Eventually, at some Vi value above 0.7 V, the transistor enters the active region, where it is able to amplify the input signal. In this region, the collector current is proportional to the base current, and the voltage drop across the collector-emitter junction is much smaller than Vcc.
Case 2: Vi > 0.7 V
When the input voltage (Vi) is greater than the threshold voltage of 0.7 V, the base-emitter junction is forward-biased, and the transistor is turned on. In this case, both Ib and Ic are nonzero, and the transistor acts like a current amplifier. The voltage drop across the collector-emitter junction (Vce) is given by the formula Vce = Vcc - Ic*Rc.
Now, suppose we gradually decrease Vi. As Vi decreases, the base-emitter junction becomes less forward-biased, and the transistor starts to turn off. The base current (Ib) decreases, which in turn causes the collector current (Ic) to decrease. As Ic decreases, the voltage drop across the collector-emitter junction (Vce) increases, according to the formula Vce = Vcc - Ic*Rc.
Eventually, at some Vi value below 0.7 V, the transistor enters the cutoff region, where both Ib and Ic are negligible. In this region, the voltage drop across the collector-emitter junction is equal to Vcc, and the transistor acts like an open switch.

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what are the on/off terminals of a transistor when it is operated as a switch?

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17:00 After being forward baised (CB diode), why Ic couldn’t increase by the same rate of Beta times of Ib???

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