The PSU2.2 circuit was
designed to be scalable to provide output voltages other than 15V. The following section describes the
changes required to make a 65V regulated supply and provides guidelines
for calculating changes for other voltages greater than 15V.
The obvious place to start
engineering is the voltage ratings of the various components. Almost all of the capacitors need to be
changed from the original BOM. A
Bill of Materials is provided as Appendix B and includes Digikey and
Mouser part numbers.
The main filter capacitors
(C4, 5, 27, 28) are 16 mm diameter with 10 mm lead spacing. Since we are shooting for 65V output,
use at least 80V rated capacitors.
The highest value capacitors that I found to fit are 470 µF 100V
Panasonic FC, although other less expensive capacitors that fit the space
allocated could be used. Since the
total capacitance of the main filter is only 940 µF, output current will
be limited to avoid excessive ripple.
When I wanted to build my
test regulator, Digikey was out of stock of the 470 µF Panasonic caps,
with delivery time undetermined.
Have some 1000 µF 100V Nichicons for another project that are 18
mm diameter with 7.5 mm lead spacing.
There is a little interference cap to cap but not with any other
parts. Zip tie them together and
they’ll sit straight. It all
depends on you tolerance for not quite fully seated mounting. The testing that follows was completed
with two of these caps installed.
Similarly, the output
capacitors (C10, 11, 20, 21) need to be upgraded to 80+V types. The
highest value capacitors that I found to fit are 100 µF 100V Panasonic
FC. Again, other types can be
used, although I was unable to find any higher value.
The large film capacitors
also need to be uprated, except for C14 and C17 which will only see
10V. The highest value that fits
the 5 x 7.5 mm space allocated is 0.47 µF, although a 1 µF capacitor with
a 7.5 x 7.5 mm footprint is available.
It fits all positions as long as you are willing to give up the
trim pot. I used the 0.47 µF caps
for my initial tests.
The BC546C and BC556C
transistors specified in the baseline version are rated at 65V. That’s a bit too close for
comfort, so use MPSA42 and MPSA92 instead. These devices have much lower current
gain than the BCs specified for the original circuit, so expect somewhat
lower performance with them. Stick
with the BCs up to 45-50V for best performance.
WHEN USING MPSA42/92 INSERT THEM OPPOSITE THE ORIENTATION
SHOWN ON THE SILK SCREEN
Now we go through the
circuit to verify that all parts are operating within their current and
power limits.
Starting with the voltage
reference current we find that the LM4040-10 is specified for operation
from 120 µA to 15 mA. There is no
data offered on noise performance or regulation vs. current. Choose R4 and R19 to keep the current
within these limits. The 15V
version has 5 mA in the reference.
To calculate the value of R4 and R19 use:
R4= (Vout-Vref)/Iref
Using our LM4040-10 for Vref,
5 mA and 65 V target output, we come up with 11K for R4. The dissipation in R4 is 55V x 5 mA =
0.275W. Therefore you’ll
need to use a 1/2W resistor. I
had some 10K 1/2W resistors handy, so I used them, increasing the
reference current to 5.5mA. They
don’t quite fit in the space provided, so you will have to mount
them higher than the surrounding resistors. This is probably a good idea anyway,
given the dissipation.
Another option is to reduce
the current in the reference. If
we use 22K for R4 and R19, we get a reference current of 2.5 mA and 0.14W
dissipation. This is still plenty
of current for the LM4040, but if you use a zener reference, it may be a
bit noisier than at 5 mA.
Now choose R3 and R18. These should be 10 times the value of
R4 and R19 to maintain the short circuit protection. Since we used 22K for R4, use 220K
here. The current and power
dissipation in these parts is so low that they can be 1/8W.
Next we’ll look at the
dissipation in the differential pair.
The MPSA42 is rated at 0.6W.
R10 and R12 set the current in the differential pair. The voltage across them is Vref -
.6V (the Emitter to Base voltage of the differential transistors),
therefore the current is:
I = (Vref - .6)/R10
With R10=1K, we have 9.4 mA
in the differential. In steady
state operation the voltage across R1 must be 1.2V, since it is the same
as the two Emitter to Base voltages in the pass devices. Therefore the current in the T3 leg is
1.2 mA, leaving 8.2 mA in the T4 leg.
Since I plan to use a 55V
transformer to drive this regulator, the filtered DC should be around
76V. The voltage across T3
is. This makes the dissipation in
T3 .077W, well within its limits.
Turning to T4, the voltage across it is
65V - Vref + .6 =
55.6V. This makes its dissipation
0.46W. This is within its rated
dissipation, but staying under half its rated dissipation will help
ensure longer life, especially when the regulator is inside a hot
amplifier case.
If R10 and R12 are increased
to 2K, the current in the differential pair drops to 4.7 mA. T3 will still pass 1.2 mA, so the current
in T4 is 3.5 mA. Now the T4
dissipation is only 0.192W. Now I
feel more comfortable, so 2K it is.
The feedback network needs
to be changed to get the proper output voltage. You could reduce R11 or turn P1 down,
but the dissipation in R2 would be 3W.
As discussed when determining the value of R4, somewhere in the
neighborhood of 20K will allow the use of 1/4W resistors. To determine the value of R2 use the
following equation:
R2 = (Vout/Vref -1) x R11
If we use R11 = 3.3K then R2
is 18150. The nearest standard
value is 18.2 K which also happens to be in my parts bin. The power dissipation is 0.17W, safe
enough for a 1/4W resistor. You
could increase them, keeping the ratio the same and reduce the
dissipation further. See the spreadsheet to calculate values.
The last parts to consider
are R6 and R17. We need to
increase them to limit the current drawn by the LEDs as well as their
power dissipation. Left at 3.3K
the LEDs would receive 19 mA and R6 would dissipate 1.2W.
Just a few mA are needed to
give a power indication, although most LEDs can handle 20-30 mA. For the sake of argument, I have
decided that 4.2 mA makes the LEDs bright enough. To find the required resistor:
R6 = (Vout - Vled) / Iled
Most green LEDs drop 2V,
reds 1.2V and blues 5. Plugging
green into the above equation, we get 15K for R6. This gives us .26W dissipation, so a
1/2 W resistor is required. The
leads will fit in the board, but the resistor must be mounted slightly
above the board as the resistor length is too long for the lead spacing.
So, how much current can
this regulator deliver? As
explained in the thread, the pass transistor can dissipate about 5W if
the 2” heat sink is used. In my case, using 77 volts
unregulated and 65 volts regulated, I get a power dissipation limited
current of 417 mA. Modeling the
supply with PSU Designer from Duncan’s
Amps shows that we should expect ~2.8V of ripple at 400 mA output and
that the unregulated voltage stays high enough allow proper regulator
operation. However, testing is
required to see how much current the regulator can deliver with
acceptable performance.