Question

Look at this circuit component by component: led: this is a standard 5 mm red led. This type of led has a voltage drop of 1.8 v and is rated at a maximum current of 20 ma. R1: this 330 resistor limits the current through the led to prevent the led from burning out. You can use ohm's law to calculate the amount of current that the resistor will allow to flow. Because the supply voltage is +6 v, and the led drops 1.8 v, the voltage across r1 will be 4.2 v (6 1.8). Dividing the voltage by the resistance gives you the current in amperes, approximately 0.0127

a. Multiply by 1,000 to get the current in ma: 12.7 ma, well below the 20 ma limit. Q1: this is a common npn transistor. A 2n2222a transistor was used here, but just about any npn transistor will work. R1 and the led are connected to the collector, and the emitter is connected to ground. When the transistor is turned on, current flows through the collector and emitter, thus lighting the led. When the transistor is turned off, the transistor acts as an insulator, and the led doesn't light. R2: this 1 k resistor limits the current flowing into the base of the transistor. You can use ohm's law to calculate the current at the base. Because the base-emitter junction drops about 0.7 v (the same as a diode), the voltage across r2 is 5.3 v. Dividing 5.3 by 1,000 gives the current at 0.0053 a, or 5.3 ma. Thus, the 12.7 ma collector current (ice) is controlled by a 5.3 ma base current (ibe). Sw1: this switch controls whether current is allowed to flow to the base. Closing this switch turns on the transistor, which causes current to flow through the led. Thus, closing this switch turns on the led even though the switch isn't placed directly within the led circuit.

Answer 1

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