Sorting a recovered component bag for a 9 V LED indicator
A Pre-Voc student at a Ratnagiri school is given a plastic bag of recovered components for a bench project: a red LED rated 2.0 V at 20 mA, a 9 V battery, a small stripboard, and a mixed bag containing a 1N4007 diode, a BC547 transistor, a disc capacitor printed '104', an electrolytic capacitor whose polarity stripe has faded, and loose resistors. The only resistor clearly readable is a 470 ohm (yellow-violet-brown-gold). The teacher asks for a single LED indicator that draws about 20 mA and is protected against a reversed battery connection. The student also must decide whether the faded electrolytic can be used in the 9 V DC path and how to test the transistor before soldering. The multimeter available has diode-test, continuity and resistance ranges but no capacitance range. A classmate suggests connecting the LED directly to the 9 V battery 'because it is only a small battery'.
1. Calculate whether the 470 ohm resistor is suitable for the LED on a 9 V supply. State the nearest preferred value you would use if a current closer to 20 mA is required, and comment on resistor power rating.
Read the worked answer
Use Kirchhoff's voltage law: the resistor must drop 9 V - 2.0 V = 7.0 V. With 470 ohm, I = 7.0 / 470 = 0.0149 A, about 14.9 mA. That is below the LED's 20 mA rating, so it is electrically safe and only slightly dimmer; it is a good conservative choice for a student project. For exactly 20 mA, R = 7.0 / 0.020 = 350 ohm. The nearest E12 values are 330 ohm (21.2 mA, slightly over the rating) and 390 ohm (17.9 mA, safe). I would choose 390 ohm unless brightness is critical. Resistor power is I squared times R = 0.0149 squared times 470 = 0.10 W, so a 0.25 W resistor is adequate. Connecting the LED directly to 9 V is wrong: the battery's small internal resistance is the only limit, so current will far exceed 20 mA and destroy the LED.
2. Explain how to identify the anode and cathode of the 1N4007 and the LED, and how the diode-test range confirms a healthy 1N4007. Then state how the diode can provide reverse-battery protection and recalculate the LED current with that diode in series.
Read the worked answer
On the 1N4007 the painted band marks the cathode. On a standard LED the longer lead is the anode and the flat side of the body marks the cathode. Set the multimeter to diode test: with the red probe on the anode and black on the cathode, a good 1N4007 reads about 0.5 to 0.7 V; reversing the probes should show overflow or open circuit. A reading near 0 V both ways means shorted; open both ways means failed. For reverse protection, place the 1N4007 in series with the battery so it is forward-biased in normal operation; a reversed battery then blocks current. The diode drops about 0.7 V, so the resistor now drops 9 - 0.7 - 2.0 = 6.3 V and the current is 6.3 / 470 = 13.4 mA. The LED is still safe, but the loss must be considered. A parallel reverse diode is another option, but without a fuse it may not protect the battery or wiring.
3. The electrolytic capacitor's polarity stripe is faded. Explain how to determine polarity safely and whether it should be used across the 9 V supply. Also state what '104' means and whether the disc capacitor is a substitute.
Read the worked answer
On a radial electrolytic, the stripe with minus signs marks the negative terminal. If the stripe and the printed voltage rating cannot be read with confidence, do not use the capacitor: reverse connection can cause heating, venting or explosion, and an unknown rating may be below 9 V. If the marking is legible and rated at least 16 V, it may be connected with the negative stripe to the battery negative. A multimeter without a capacitance range cannot reliably prove its value or condition; a resistance check may show a rising reading as it charges, but this is only a rough check. '104' means 10 followed by four zeros picofarads: 100,000 pF = 100 nF = 0.1 microfarad. The disc is non-polar and useful for decoupling, but it is not a substitute for a large electrolytic reservoir. For a simple LED indicator, neither capacitor is essential.