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## Q. 5.5

Light Timer

RC circuits can be used to build simple timers. Figure 5.17 shows a circuit that turns off a light after a given period of time. The transistor Q1 and the relay ensure that the light goes from fully ON to fully OFF. Transistor circuits are discussed in detail in Chapter 8.

When the momentary contact switch SW1 is pushed, the capacitor charges and the light turns on. When the voltage between the base and the emitter of the transistor, $v_{BE} (t)$, is below 0.6 V, the transistor, and as a result the relay, turns off. This turns off the light. Given the simplified circuit shown in Figure 5.18, how long after pressing SW1 will the light turn off?  ## Verified Solution

In Figure 5.18, the switch has been closed for a long time and capacitor is fully charged to 9 V. Using KCL, the equation for $v_B (t)$, that is, the voltage across the capacitor after the switch SW1 is turned off, corresponds to:

$0=\frac{v_{B}(t)}{2200}+1000 \mu F \frac{dv_{B}(t)}{dt}$

Thus:

$0=\frac{v_{B}(t)}{2.2}+\frac{dv_{B}(t)}{dt} \rightarrow v_{B}(t)=K e^{-t / 2.2}$

Because the initial charge of the capacitor is 9 V, in the initial conditions, $v_{B}\left(0^{+}\right)=9 V$, accordingly:

$9=K e^0 \rightarrow K=9$

Thus:

$v_{B}(t)=9 e^{-t / 2.2}$

Now, using a simple voltage division rule, the voltage across the base and the emitter, $v_{BE}(t)$, and the voltage across the capacitor are related by:

$v_{BE}(t)=\frac{200}{2000+200} \times v_{B}(t)=0.09 v_{B}(t)$

Therefore:

$v_{BE}(t)=0.818 e^{-t / 2.2}$

Solving for t:

$0.6=0.818 e^{-t / 2.2} \rightarrow \ln \left(\frac{0.6}{0.81}\right)=\frac{-t}{2.2} \rightarrow t=0.682 s$

Thus, the light turns off after 0.682 s. It is clear that the delay time can be increased if the 2 kΩ increases. For example, increasing it to 20 kΩ, the delay will increase to about 6 s.