Q&A
■ WITH RUSSELL KINCAID
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.
Send all questions and comments to:
Q&A@nutsvolts.com
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FREQUENCY
DIVIDER
QI need a circuit to produce
3250 Hz and 65 Hz at the
same time. I don’t know if
a CD4040, CD4060, or a
PIC will do it; 50% duty cycle would
be nice.
— Craig Sellen
AIf you want crystal control,
the frequency of the oscillator should be above 1 MHz
for least cost. You need
two dividers to get two arbitrary
frequencies, so I will use the CD4040
twice. The oscillator frequency has
to be divisible by both numbers, so
multiplying 65*3250* 6 = 1.267500
MHz. Multiplying by 5 also is above
1 MHz but requires more gates to
decode. You could order a custom
crystal but I will design using an R-C
oscillator. A divide by 2 is needed to
get 50% duty cycle, so I will use a
CD4013 dual D flip-flop.
In Figure 1, the first divider
output is 6,500 Hz. The decoder
taps on the divider are found by
determining the division ratio (195)
and subtracting the largest power of
2 that will fit until the remainder is
1 (195-128-64-2). The taps are at
the first, sixth, and seventh flip-flop
in the divider. A divide by 2 provides
the 3,250 Hz signal which is then
divided down to 130 Hz in the
24 January 2008
CD4040 IC. The division ratio is 25,
so 25-16-8=1 and the taps are at the
third and fourth flip-flops of the
CD4040. A divide by 2 gives the 65
Hz output. A CD4024 seven stage
counter could have been used instead
of the CD4040.
TRANSFORMER
CURRENT
QSeveral months prior to his
untimely demise, TJ Byers
stated in the Q & A column,
in answer to a reader’s
question about power supplies, that
current in a transformer is not a sine
wave. Mr. Byers did not elaborate on
that subject as he was answering a
somewhat different question, but it
raised questions in my mind as to what
is the current in a transformer if not
a sine wave? I have been patiently
waiting for some other reader to re-raise
the question to see if you could tell us
what current in a transformer truly is, if
not a sine wave.
I am as perplexed now as I
was then: If current in a transformer
is not a sine wave, how does one
go about calculating how much
current is available when the AC is
rectified? Since voltage is very
definitely a sine wave and the DC
value is markedly different from the
AC value, it stands to reason the DC
value of the current would follow
the same rules.
To me, that makes sense as
current and voltage are traded
back and forth in a transformer and
we calculate that trade-off via the
60 Hz sine wave. How is it that
the current is not a sine wave? I
don’t quite get that. Could you please
elucidate?
— Charles R. Rhines
AI believe Mr. Byers was
speaking of a transformer
driving a rectifier in a power
supply. Without getting into
nit-picking details of transformer
nonlinearity and effects of nonlinear
loads on the source voltage, the current
in the power supply transformer pulses
and is not sinusoidal.
Consider the case of a full wave
circuit with an output capacitor and
DC load as in Figure 2. The upper
(blue) line is the output voltage; the
lower line is the primary current (scale
on the right). The capacitor is slowly
discharged by the load; the rectifier
replenishes the voltage only at the
peak of the waveform. It was common
in the old days to put an inductor in
series between the rectifiers and the
capacitor to make the pulses wider
(see Figure 3).
Note that the output voltage is
smoother, takes longer to reach a steady
state, and the primary current pulses
are wider. In fact, it is possible to make
the inductor large enough that the
current is sinusoidal, but I have never
seen it done.
