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
WHAT’S UP:
Join us as we delve into the
basics of electronics as applied
to every day problems, like:
✓Blinking Lamp Schematic
●✓Solenoid Driver
✓Stereo Amp
60 VOLT, 13 AMP
S QUPPLY
I need a power supply that
puts out (max.) 60V at
13 amps. Ripple is not too
important, as long as it is
no more than 0.5V or 1V. I don’t
want to use expensive transformers,
though. Is there a switching mode
that can run off house current?
— Craig Kendrick Sellen
ABoth these requests are
similar enough to use the
same basic circuit. The
power output at 20 volts,
40 amps is 800 watts and the power
at 60 volts, 13 amps is 780 watts. The
power input will be higher due to
losses, so plan on 1,000 watts input
( 8. 3 amps at 120 VAC). Since the big
deal for me is the transformer design,
I will do that first. I am using the
Magnetics, Inc., manual for ferrite
cores. For high power, E core is
recommended and a formula is
provided for estimating the size:
WaAc = (P 8 o*C*10 )/4eBfK where
Po = Power output
C = Current capacity in sq.
cm per amp (.00507 for square
waves)
e = Transformer efficiency
(90% or 80% for an inverter)
B = Flux density in gauss
f = Frequency in Hz
K = Winding factor (0.3 for
24 December 2008
primary side; 0.2 for toroids)
Using a flux density of 1,000 and
frequency of 100 kHz, I get WaAc =
4. 23. Checking the catalog, I find
core 45021-EC has WaAc = 4; that is
close enough because my calculation
was conservative. Checking the back
inside cover, I see that this core is a
standard; available in R material (if it
is not a standard, then a minimum
order is required). The catalog gives
the bobbin window area = 1.78 sq
cm = 0.276 sq in. Now see if the
windings will fit: The mH/1000 turns
(Al) for the core is given = 4600.
I want the no load primary
current to be low compared to the
full load current of 8. 3 amps, so try
0.4 amps. The rectified line voltage
will be 120*1.41 = 170 VDC, so the
inductive reactance of the primary
should be: Xl = 170/0.4 = 423 ohms.
The inductance at 100 kHz is
therefore = Xl/2πf = 673 μH. Now,
using the formula N = (L*106/Al)½, I
get less than one turn; use 15 turns
so there is more than one turn on
the secondary. The catalog gives this
formula for the primary turns: Np =
Vp*108/4BAf. I don’t know what A is
but if I use the minimum area given
for the core, Np = 170*108/4/1000/
2.13/105 = 20, that is a good
number; go with it and see if it fits.
For the eight amp primary
current, I think #16 wire will be okay,
but winding #16 solid wire will be
difficult. Use four parallel #22 wires
for a total of 80 primary turns. Each
wire is .02535 inches in diameter.
If I square that times 80 turns, the
primary wire area is .051 sq in.
For the 40 amp secondary, the
number of turns is: 20V*20T/170V =
2.4. Can’t have partial turns, so use
three turns and that might be enough
to compensate for losses. For 40
amps, you should use #10 wire but
that would be impossible to bend
around the small bobbin. Use 16 #22
wires in parallel for a total of 48
turns. The secondary wire area in
this case is .031 sq in. Adding the
primary and secondary, the total is
.082 sq in and will easily fit in the
available 0.276 window.
For the 13 amp secondary, the
number of turns is: 60V*20T/170V =
7.06. Use nine turns to allow for
losses and regulation. For 13 amps,
you should use #14 wire but again,
that is too difficult to bend. Use eight
#22 wires in parallel. With 72 turns
#22 wire, the area requirement is:
.02535*.02535* 72 = .046 sq in. This
plus the primary will also fit in the
window.
You can wind the multiple wires
individually and parallel them when
you are done. If you wind them all
together, only fasten them at the start
end and parallel them when you are
done winding. You will need to
compute the length of wire needed
and allow plenty of extra because it
will take more than you think. I
recommend using a compression
splice to connect the parallel wires
and convert to a single solid wire for
