The switch is closed Almost 10 V drop across bub. Its +10V The base resistor is

Question

The switch is closed Almost 10 V drop across bub. Its +10V The base resistor is chosen small enough so that the base current dkives the transistor nto saturaticn the collector current. +10 V In this example the mecharical switch is used to produce the 1 K base curent to close the transistor switch to show the The switch is open There is no current to the base, so the transistor is in the cut off condition with no colector curent. AI the voltage drop is accoss the transistor =0.05 to 0.2V 1 KS2 principles. In practice, an .4 mA voltage on the base sufficient -0.6 v to drive the transistor to saturation will close the switch and light the bulb. Closs 1.1 What is the math expression for the total power drawn by the circuit? a. (10Xc) c. (10 Volc) 1.2 What is the math expression for the power dissipated by the bulb? a. (10Ic) b. (l0)(lc +IB) c. (10-Vclc 1.3 If -9.4mA, Ic 1Aand 0.1V, compute the efficiency the switching circuit when it's turned on. Efficiency (Power by bulb)LTotal power consumed) 1.4 If the transistor is not fully turned on, its voltage and currents may settle to I 4mA, 1.* 0.4A &?26 V. Compute the efficiency the switching circuit when it's in this state. Efficiency- (Power by bulb) (Total power consumed)

Explanation / Answer

Solution:

1.1) the power drawn by the circuit is the multiplication of supply voltage and the total current drawn by the circuit fro the battery

Pdrawn = Vcc * Itot

Itot = IC+IB

where IC is collector current and IB is base current.

Pdrawn = 10(IC+IB) watts

S option is (b)

1.2) Exression for the power dissipated by the bulb

Pdiss = V * I

where V is voltage across bulb and I is current flowing through the bulb.

Trnasistor is in saturation . let Vc is collector to emitter voltage.

Voltage across the bulb = (10-Vc)

Power dissipated by bulb = (10-Vc) * IC

So option is (C)

1.3) Efficiency = (Powerdrawn by bub/total power)

Efficiency = (10 - Vc)*IC / 10 (IC + IB)

= 9.8*1/10(1.0094) = 97.08%

1.4) Efficiency = (Powerdrawn by bub/total power)

Efficiency = (10 - Vc)*IC / 10 (IC + IB)

= 4*0.4/10(0.4004) = 40%

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