This is a lab question. Could someone please explain what would happen in this c

Question

This is a lab question. Could someone please explain what would happen in this case and the reasoning as to why? Consider a blinking LED circuit, where the LED is equivalent to a 100 ohms resistor and SCS is a silicon controlled switch (3&1 legs of PN2222A). When the voltage Vscs is increasing but less than 9 v, it is an open circuit (see the bottom-left figure). Once the voltage reaches 9 v, it functions like a short circuit and the capacitor discharges that turns on the LED. This behavior will continue as long as Vscs remains above 3 V (see the bottom-right figure). At this point, the SCS becomes an open circuit again. Vscs 15 V LED R1-100 SCS C 330 uF 100 R 10015V LED LED 15 V R1-100 R1 =100 C 330 ulF C: 330 uF The above figure shows the actual voltage across the capacitor within the period kT to (k+1)T for k=0,1,2, The voltage is between 3V to 9V. The time constants of the charging part (kT to kT+T) and the discharging part (kT+Ti to (k+1)T or (k+1)T-T2 to (k+1)T) are respectively, RC and (RI 200)c 200R C 200C, if R>> 200 200+R

Explanation / Answer

The resistor R basically determines the charging time for the capacitor.

When the resistance is 1.4 Kohms, the time constant of the RC circuit is small. Hence, the capacitor gets charged faster and the frequency of blinking is more.

When the resistance is raised to 10kOhms, the RC time constant increases too. This causes the charging time to increase and we get a slower blinking rate.

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