Nuts and Volts - July 2010

QUESTIONS & ANSWERS

Protection against shorts at the load is provided by Q1 and R4. When the load current exceeds five or six amps, the output is pulled down to 1.2 volts. If the current is still excessive, the output will go lower; a short pulling 10 amps at zero volts would be unusual. Resistor R4 is rated three watts, so it would burn up if 10 amps persisted. In the parts list (Figure 7), axial means that the leads are in line with the body and you have to bend them to mount in a printed circuit board (PCB) or perf board. Radial indicates that the leads come out one side so bending is not required. PC indicates that the part has pins intended for soldering to a PCB.

period and stays there until power is removed or the restart switch is momentarily closed. The capacitor, C2, resets the counter on power-up to insure that it starts from zero.

MYSTERY COMPONENT

QI am working on a Litton microwave oven, model 1508, which contains a transformer which has

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failed on its power supply board. The transformer was made by Keystone Transformer, Co., Pennsburg, PA and is marked “Catalog No. PTL-1, Part No. TG.” The primary is 120 VAC and it has three secondaries with voltages under load of 7. 5 VAC, 7. 5 VAC, and 23 VAC. Dissection of the transformer reveals a two-terminal component, in series with the primary, approx. 1/4” x 1/2” x 1/8” in size. No markings are visible on it.

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TIMER

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QI would like to build a timer to provide a reliable time on period of two hours; I believe I can’t use a 555 for this long a period. Can you steer me in the right direction?

— Bill Woods

120VAC 10A
DPDT
Relay

4-digit resettable 6VDC 3W Counter

AWhat you need is a 555 timer and a count down circuit. Two hours is 120 minutes or 7,200 seconds. My first thought was to decode an eight-bit counter, but eight bits is not enough if the clock is one second ( 2^ 8 = 256). I have some 12-bit counters (CD4040B), but this is a ripple counter and has glitches which make decoding unreliable. If I use the last output (see Figure 8), the division is 2^ 12 = 4,096, but the last output goes high after 2,048 clock pulses and completes the cycle after 4,096 clock pulses, so my time delay will be 2048 clock pulses. Dividing 7,200 by 2,048, I get 3. 52 seconds for the clock period. Using the equation on the datasheet for the LM555 and choosing C = 10 µF, Ra = 470K, then Rb = 18K. I show a CD4012 dual four input NAND because I had started out planning to decode, but any inverter will do. The counter starts as soon as power is applied and stops after two hours. The output goes high at the end of the time