![]() At most, a total voltage of 12 volts across a 100 kΩ resistance yields a current of only 0.12 mA, or 120 ♚! The connection made by your touching the wire to a positive point in the circuit conducts far less current than 1 mA, yet through the amplifying action of the transistor, is able to control a much greater current through the LED. The transistor 2N2222 is one of the important and very commonly used transistor type which finds numerous switching application in electronic circuts. It takes 20 mA to fully illuminate a standard LED, so this behavior should strike you as interesting, because the 100 kΩ resistor to which the loose wire is attached restricts current through it to a far lesser value than 20 mA. toggle or slide switch 9 or 12 V battery, power supply or 8 AA cells. When the transistor is turned on, current flows through the collector and emitter, thus lighting the LED. Q2, Q3, Q4 BC 548, 2N2222 or any general - purpose NPN silicon transistor. Also, the 2n2222 transistor is getting quite hot. Why am I getting approximately 3.2V at the relay terminal and not a value close to 5V 3.2V are not enough for switching my relay. I calculated that the base resistor should be 270ohms. R1 and the LED are connected to the collector, and the emitter is connected to ground. 1 Hi all, I am using a 2n2222 transistor for switching a 5V 185mA 27ohms relay. If you touch the end of the loose wire to any point in the circuit more positive than it, such as the positive side of the DC power source, the LED should light up. A 2N2222A transistor was used here, but just about any NPN transistor will work. If anyone has any ideas on how to get this to work or if they notice something I've wired backwards (highly embarrassing) please let me know.The red wire shown in the diagram (the one terminating in an arrowhead, connected to one end of the 100 kΩ resistor) is intended to remain loose, so that you may touch it momentarily to other points in the circuit.
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