This
PSU has been especially designed for current-hungry ham
radio transceivers. It delivers safely around 20Amps at
13.8V. For lower currents, a separate current limiting
output, capable of 15ma up to a total of 20A has been
added. Let us see what we have got here. The power
transformer should be capable to deliver at least 25A at
17.5 to 20V. The lower the voltage, the lower power
dissipation. The rectified current will be “ironed” by the
C1, whose capacity should not be less than 40.000uF, (a
golden rule of around 2000uF/A), but we recommend up to
50.000uF. This capacity can be built up by
several smaller capacitors in parallel. The base of this
design is a simple 12V regulator (7812). The output voltage
can be brought to desired value (here 13.8V) by two
external resistors (R5 and R6) using this formula:

U=
12(1+R5/R6)

The
low currents (here 15mA) will keep the 7812 in its regular
function. As soon as the current rises over 15ma, the
voltage drop on R4 will “open” the Q3, actually handling
the high output current. This is a PNP transistor (Ic>25)
and current amplification factor of at least 20. The one
that has been tested and proven here is the 2N5683. The
current limiting resistance RL, for the maximum output of
20 Amps should be 0.03 Ohms, rated at

least
15W. You can use the resistance wire or switch several
resistors in parallel, totaling the resistance/power
values. Values for other currents can be calculated by the
rule:

RL=0.7/Imax

The
RL and Q2 (3A PNP such as BD330) form a short circuit
“automatic fuse”. As soon as the maximum current reaches
20Amps, the voltage drop over the resistor RL will open Q2,
and thus limit the B-E Current of Q3. Parallel to Q2 is Q1,
which lights the LED 1 whenever the current limiting
circuit is active. When the “fuse” is active, the Q2
bridges the R3, so the full current would flow through the
IC1, and damage it. Therefore the R4 is inserted, as to
limit the IC1 current to 15mA. This makes it possible to
run the IC1 without any cooling aid. The LED 2 will light
up every time the PSU is switched on.

There
is an adjustable current limiter in parallel to the fixed
output, thus providing adjustable current source for
smaller currents.

This
circuit is very simple too. You will notice that there is
no current sensing resistor. But it is really there, in a
form of the Rds-on resistance of the N-channel FET, which
actually handles the load cutoff from the source. The
function of the FET is shown in the diagram 2. When the
current Id is rising, the tension Uds over the resistance
Rds rises very slowly in the beginning, but very fast after
the knick. This means, that before the knick the FET
behaves as a resistor but after it, works as constant
current source.

The
D2, R3 and B-E connection of the Q4 will sense the Uds
voltage of the FET1. When the voltage rises enough, the Q4
will shortcut the FET1 gate to mass, and cut the current
flow through the FET 1 off. However, to enable the FET1 to
open, there is certain gate voltage necessary, which in
this case is brought up by the voltage divider consisting
of R8, Z1, P1 and R9. So the maximum Gate voltage will be
the one of the Z1, and the minimal will be around 3V6. The
Z1 voltage (Uz1) will thus determine the max current
flowing through the FET 1.

The
diagram 2 will show that for 5 Amps the Uz1 should be 5V6,
and for 20Amps around 9V6. The Capacitor C4 will determine
the “velocity” or the reaction time of the limiter. 100 uF
will make the reaction time to be around 100ms, and 1n will
make it 1us.

Within the designed limits, the P1 will limit the current
output in the range of 15mA to 20A. You can use both output
simultaneously, but the total output current will be
limited by the value of the RL. This PSU can be built also
for higher outputs, as long as the transformer will handle
the current requirements, and you provide sufficient
cooling for the Q3.

If
somebody will be interested, there is a PCB design ready.

Have
fun.

Bob

REV1.

I
have received several requests for some modifications, and
the one I find useful is the addition of an amp meter.
Therefore the slightly modified diagram is included in this
revision. All elements within the dotted border are now
placed on the pcb. There is also elements placement design
included. Should one have an 25Amp instrument on hand,
there is nothing easier. Just mount it in line and there it
goes. However, a ham would probably find an instrument
somewhere in his “junk-box”, but the scale would be
something completely different, let say an “S” or
Voltmeter. No problem. We already have a “shunt” for the
amp-meter, and it is there as the Current limiting resistor
“RL”. As
already known from before, there is a voltage drop of 0V7
over the resistor at current flow of 20A. What we now have
to do, is to simply measure the voltage drop over the
resistor, and co-relate it with the current. Let us say
that our instrument has an internal resistance of 13R, and
has a full scale reading of 60mV. The voltage drop over the
RL is 0V7 for 20Amps. Therefore, we need another resistance
in line with the instrument, that would bring the 0V7 to
60mV, or an voltage drop of 640mV.

The
formula is simple:

**
U1:R1=U2:R2**

60:13=640:X

X=13x640/60

X=138.66

Therefore, the resistance that has to be inserted in line
is around 140R. I suggest to insert a trimmer (VR1) of
around 200R, to fine trim the reading when calibrating the
instrument. Using your favorite drawing software, design
your scale to your likings, (at least 20A for the full
scale) and insert it in the instrument you have. Due to the
many requests for the PCB Layout, I have included the
design here. The exact dimensions of
the
pcb are 160x100mm. Please remember, that the pcb has to be
printed as a mirror image, to obtain higher quality when
transferring it to the copper side of the board.

Wish
you a good time.

Bob