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