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
Electronic
Surplus
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.
sales1@electronicsurplus com
Toll free: 1-800-642-1123
8755 Munson Rd
Cleveland, OH 44060
TIMER
electronicsurplus.com/specials.cstm
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
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12VDC 0.24A
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18VDC 3500RPM
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Motor
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Binding post
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July 2010 25