■ WITH TJ BYERS
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
✓UV photometer, a clock heater, and
a vacuum tube filament supply.
✓Solutions for the deceased MM5369
clock chip and a new 15-minute timer.
✓Plus, a power transformer tutorial
and how to prevent PA feedback.
VDC = 0.9 VAC
IDC = 0.28 IAC
VDC = 0.45 VAC
IDC = 1.0 IAC
VDC = 0.9 VAC
IDC = 0.62 IAC
■ FIGURE 1
into saturation and ceases to
function properly. This is
especially important when
the transformer has more
than one secondary winding.
The resistance of the primary
and secondary windings also
influences the amount of
current a transformer will
handle because increased resistance and higher current
makes heat — heat that must
be dissipated. Your particular transformer has a rating
of 240VA which means it will
deliver 10 amps at 24 volts.
But I don’t think this is
the answer you’re looking
for, because I assume you
want to turn this AC into DC. The DC
current output is determined by the
configuration of the rectifier circuit —
of which there are three:
QI have a power transformer
with the following specifications: 24VCT (12 - 0 - 12) 10A.
If the center-tap terminal is
not used — and we used it as a 24-volt
transformer — then will its rating remain
at 10A or will it be half of that value (5A)?
1. Half-wave (single diode)
2. Full-wave center-tapped (two diodes)
3. Full-wave bridge (four diodes)
ondary at a time — which results in an
output voltage that is one-half (0.45, to
be exact) the full voltage across the
secondary, but takes advantage of the
full 10 amps your transformer has to
offer. The full-wave bridge rectifier outputs a voltage that is 0.9 percent that
of the transformer voltage, but can only
use 0.62 percent of the transformer’s
current. To achieve the desired DC load
current, the transformer current should
be 1.6 times higher. For example, to get
10 amps of DC, you need to have 16
amps available to the rectifiers.
Why? Because of the surge current required by the filtering capacitor. Transformers are not ideal and
have an internal impedance or “
regulation” characteristic. As the load
increases, the output voltage decreases. Consequently, the transformer
current must be sufficient enough to
overcome the extra current imposed
by the charging capacitor. A way to
eliminate this requirement is to insert
a filter choke in series with the line to
the filter cap. Doing this will let you
use the full potential of the transformer’s current rating — at the cost
of an extra inductor and added space.
ATransformers are rated in volt
amps (VA). It defines the
limit of the magnetizing field
inside the transformer. If this
limit is exceeded, the transformer goes
12 January 2006
Refer to Figure 1 for the following
discussion. The advantage of half-wave
rectification is in its simplicity — one
diode and a capacitor. They are generally viable only for power supplies of
one-half watt or less, and require more
filtering than full-wave rectification.
The full-wave center-tapped rectifier uses only half the transformer sec-
QCould you suggest a circuit for
measuring the UV output of a
mercury vapor fluorescent type
bulb? This type of bulb is used