Q&A
Voltage, on the other hand, is
equal to V = current/power. In
logarithmic language, that’s a 20
ratio, not 10. The formula is Volts
(dB) = 10(dB/20) and can be calculated
on the PC by using the “x^y” function
of the Windows calculator. First,
divide the dB value by 20 and save it
to memory (MS). Clear the calculator
using the Esc key, type in 10, click on
x^y, read the memory (MR), and hit
Enter. That is your voltage ratio.
The decibel system has another
advantage in that the ratios can be
added and subtracted without going
through a lot of math. For example,
let’s say that you want the voltage
ratio between 25 and 40 dB or
between 40 dB and 34 dB. Just
subtract the values — negative
numbers are permitted and represent
an attenuation rather than amplification.
Run your answer through the calculator
(or look it up in Table 1) and you will
have your power and voltage ratios. In
the first example, 40 minus 25 is 15
dB for a voltage gain of 5. 62. In
the second, - 6 dB equals a voltage
attenuation of exactly half with
one-fourth the power.
I have rewired the rotors
so that the resistance through
each of the 26 possible paths
in each rotor is less than 1 Ω.
That was, comparatively,
easy. The problem is that the
intermediate stators are built
out of contacts and springs
( 98 parts per stator!) and the
resistance through each of
the 26 paths in the stators is not
uniform — varying from about 1 Ω to
about 4 Ω, depending on the particular
stator and path.
Can you show me how to build a
power supply that has the following
characteristics?
LM317
T
IN OU
COM
Figure 1
18V - 35V
0.1
R1
3. 9
Constant-Current
Regulator
320mA Out
• If it sees an open circuit (say, more
than 100 Ω), it doesn’t panic and
produces no output.
voltage is needed to push 320 mA
through a 100 Ω load. Using Ohm’s
Law, E = IR = 0.32 x 100 = 32 volts.
Because the LM317 needs at least 3
volts to operate, the minimum input
voltage is 35 volts DC. If you can live
with an upper limit of 50 Ω, the
minimum input voltage is 18 volts,
which is easily obtained from a cheap
wall-wart. Be sure to heatsink the IC
because it can run hot — up to 10
watts with a shorted output.
• If it sees a resistance of 10 Ω, it
cranks out 3 volts at about 0.3 amps.
High-Power Variable
Voltage Regulator
• If it sees a resistance between 10
and 100 Ω, it puts out whatever voltage
is required to develop 0.3 amps.
Peter Ingerman
via Internet
Wide-Range Current
Regulator
Q. I’m refurbishing a pair of old
Hebern cryptographic machines
and I think I need something rather
peculiar in the way of a power supply.
The machine (from an electrical
point of view) looks like a voltage
source that is connected to the “left
stator.” Then there is an alternating
sequence of rotor-stator-rotor-stator-rotor-stator-rotor-stator-rotor, ending
with the “right stator.” There are 26
possible paths through each rotor
and each stator. The left rotor is
connected to a keyboard; depressing
a key “wets” one of the 26 contacts
on the left rotor. The power is then
transmitted through the rotors and
stators to the right rotor, which
illuminates one of 26 light bulbs to
indicate the enciphered letter. The
orientation of the rotors to the stators
changes on a letter-by-letter basis.
JULY 2004
A. What you need is a constant
current power supply of 0.3
amps or 300 mA. This is easily
accomplished using an LM317
adjustable voltage regulator (Figure
1). Resistor R1 determines the output
current of the LM317 by setting the
current of the internal reference
voltage, which is 1.25 volts. If R1 is
1.25 Ω, then the output current will be
1 amp; a 3. 9 Ω, 1/2 watt resistor
limits the current to 320 mA.
Now, let’s determine how much
Q. Could you suggest a schematic
for a variable power supply to
replace the electronics inside my old
HeathKit (BE- 5) battery eliminator? I
don’t need the 6 volt option, but
something with better 12 volt
regulation would be nice. I’m thinking
of replacing the variable transformer
with a pot and a fixed transformer —
as long as it fits in the original
cabinet, which I have grown to love.
Jim
via Internet
A. This is a simple request that
requires little more than an
LM338 (Figure 2). Basically, the circuit
is a full-wave rectifier, followed by a
Figure 2
100V, 6A
LM338
T
0.1
IN OU
COM
1.2V - 15V
OUT
120
115VAC
18VAC
5A
1N4148
+
10,000uF 5k
25V
+
10uF
+
1uF
15-volt, 6-amp Variable Power Supply
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