Explain the working of a transistor as a switch.

Explanation

A switch is a device that is used to open (or break) and close a ciruit. A bipolar transitor, either npn or pnp, may be used as a switch.
Working : Consider a simple using the npn transistor as shown in the figure. The load resistance `R_(L)` is connected between the collector and the positive terminal of the collector bias supply `(+V_(CC)).`

Applying Krichhoff s voltage law to the output part (colleector-emitter), ` + V_(CC) - I_(C)R_(L) - V_(CE) = 0`
since the ground is considered to be at zero potential.
`therefore V_(CE) = V_(CC) - I_(C)R_(L)" "...(1)`
(1) Open switch conditions : Applying no voltage at the base of the transistor, i.e., `V_(i) = 0`, puts the transisitor in the cutt-off region. Then `I_(B) =0 and `I_(C) = 0`
THus, no current passes through the load resistance `R_(L)` .This is the open switch condition.
(2) Closed switch condition : WHen `V_(i)` is positive and greater than 0.7 for a silicon transistor, there is a base current which puts the transistor into the saturation region. The transistor becomes fully conductive. With `C_("CE") = 0.2 V`, almost the entire collector supply voltage `C_("CC")` appears across the laod . From eq. (1).
`I_(C)R_(L) ~~ V_("CC")`
i.e, the current through the load is limited only by the load resistance. This condition is that of a closed swithch. Thus, when used as a switch, a bipolar transisitor is operated in the cut-off region (open switch) and saturation region (closed switch).