Nuts & Volts

■ WITH TJ BYERS
Q & A
In this column, I answer questions about all
aspects of electronics, including computer
hardware, software, circuits, electronic theory,
troubleshooting, and anything else of interest
to the hobbyist.
Feel free to participate with your questions,
comments, or suggestions.
You can reach me at: TJBYERS@aol.com
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WHAT’S UP:
APPLICATIONS
January’s power transformer
advice was a hit, so here’s
the sequel — plus the usual
scattering of circuits.
● Digital wind vane.
● High-power T-Bird lights.
● USB power booster.
● Vacuum tube power amps.
TRANSFORMER
RATINGS, AGAIN
QI have the same 24VCT transformer
that you talked about
in the January ‘06 issue. I am
using the transformer for my
three stepping motors (2A/stepper) and
would like to eventually add a fourth
stepper. Using your data, that means the
transformer won’t be able to supply the
extra current — unless I use an inductor.
What value of inductor would you use?
— Michael Elledge
Christmas, FL
AThe drawing in Figure 1
shows the voltage and
current values to expect from
a full-wave rectifier with a
Full-Wave Center-Tapped
Full-Wave Bridge
24 April 2006
+
+
+V
+V
choke input. Notice that the output
current of both configurations has
increased by 33% — and, in the case of
the center-tapped rectifier, it even
exceeds the transformer’s AC current
rating. The choke input can do this
because energy is stored in the magnetic
field which wants to keep current
flowing at a constant rate, and will take
energy from its own magnetic field if
needed to maintain that current flow.
The value of the inductor
depends on many factors. First, there
must be sufficient inductance to
ensure continuous operation of the
rectifiers and provide good regulation.
The absolute minimum inductance
is given by the equation, L min =
(K / f) * R L , where L is expressed
in Henries. For 60-Hz line operation,
the formula becomes L min = R L / 1000.
You may see this
VDC = 0.45 VAC
IDC = 1.5 IAC
VDC = 0.9 VAC
IDC = 0.94 IAC
■ FIGURE 1
formula reduced
to L min = V out / I out .
Let’s say that your
power supply is 12
volts at two amps.
Plugging these
values into the
equation, we get
6 mH (L = 12V /
2A). Remember
though, this is the
absolute minimum
inductance
needed to sustain
filtering — and
not a practical
value. A rule of thumb is to multiply
L min by at least 10. For our example —
Figure 2 — that value is 60 mH.
Next, we have to calculate for the
value of the capacitor. The LC product
must exceed a certain minimum to
ensure a desired ripple factor. For
this, you need to solve the equations
for a first-order low-pass filter. This is
more math than I care to do on a daily
basis, so I use software simulation
instead. Any decent schematic
capture/simulator package will work.
If you don’t have a simulator, I recommend
eSketch from Schematica
(www.schematica.com), which sells
for $69. Using the values shown, the
ripple is a low 44 mV.
For an eight-amp output, you
need to do the math all over again,
starting with the inductor. If you do,
calculate for the lowest anticipated
current. That is, if you don’t run all
four steppers all the time, go with the
lowest total current because it’s the
most critical inductor factor.
In actuality, the output voltage
will be 10.8 volts, and that assumes
perfect devices. You need to take into
account resistance loss in the inductor
and other wiring, which lowers the
voltage even more. Inductors like this
are hard to come by — and expensive
when you can find them. The alternative
would be to wind your own. On
the upside, it does reduce the size of
the filter cap considerably. Me? I’d
spend my money on a bigger power