There are so many things wrong with this circuit that I scarcely know where
to start.
The basic operation of an up convertor is that a certain amount of power
from the low voltage supply is stored as magnetic energy in the coil, and
then released as high voltage when the field collapses. Hhopefully most of
it goes into the output capacitor. You cannot of course get more power out
than goes in, and every milliWatt of it has to be stored and then released
by the magnetic core on the way.
Let's start with C8, the cap directly on the coil. At the moment of turn
on, this capacitor supplies essentially all of the coil current until it
discharges - thus completely defeating the current limiting function of the
34063, which senses only the current flowing through R3-R6. The coil could
saturate and the chip still think it has no current at all.
The coil is 220uH and there is some series resistance, 0.5R in the FET,
0.25R in R3-R6 and some unknown series resistance in the PSU and elsewhere.
Let's guess it's 1 ohm total. The time constant of the coil = L/R, or in
this case 220us. Note that the bigger the series resistance, the smaller is
the time constant - counter-intuitive, but true. The time constant is the
time it takes the inductor current to rise to 63% of its final value, which
in this case would be about 12V/1R or 12 amps. Unless you run it from a car
battery the PSU probably can't source that much current, but that doesn't
matter in the short term. 63% of 12A is 7.56A. If we assume the current
rises in a linear fashion, which it does approximately, the straight line
from 0 to 7.56 takes 220 microseconds. If you were to plot that on a graph
you would see that the current passes the coil saturation point of 1.25A
after 36 microseconds - actually it will be a bit less, because the current
rises faster at the beginning. (If the current limit was working properly
it would limit the current to about 1.2A.) The switching frequency is set
by C3 and with 1nF, the frequency is about 33kHz. The maximum on period
is therefore 30 microseconds, which is kind of close to the saturation
point. You must not, therefore, make the timing capacitor C3 any larger
than 1nF unless you also increase the size of the coil.
Now, power-in versus power-out. If you want 180V at 35mA, that's 6.3W at
the output. Allowing for 85% efficiency (the 34063 spec) that's about 7.5W
at the input. From a 12V supply, that requires a mean forward current of
625mA in the coil. The current waveform is roughly a triangle and the
maximum permissible peak current is less than twice the required mean -
therefore, you can only just barely put enough energy into the inductor to
get close but still short of the output you want, running flat out with no
losses. To get more, you need to use a physically bigger inductor and
reduce the current limit resistors, to input and store more energy.
Let's look next at the losses. Only one thing really matters in a flyback
convertor and that's how fast you can turn off the switching transistor,
because every nanosecond the drain voltage takes to rise, it's dissipating
magnetic energy that isn't going into the output. The dual emitter follower
configuration Q1-Q3 in the schematic doesn't do a good job here. For a
start, the NPN Q1 is not only unnecessary, it's harmful. We don't care how
slowly the FET turns ON, only how fast it turns OFF. We want to wham that
FET gate down to ground so hard and so fast it doesn't know what it it, and
all Q1 does is fight it. Q1 could be replaced with a 4.7k resistor (and R12
removed). Then there's the configuration of Q3. Emitter followers are fast,
but they have a very high input impedance (rie * hfe) - for the 2N3906 in
this configuration, it would be in the region of 20-50k. Since the 34063 in
this configuration can only source current and not sink it, any stray
capacitance on the base side has to discharge through Q3 and R11, with R11
being dominant. If the stray capacitance is 25pF, the time constant is 125
nanoseconds, which is the time for the voltage to fall by 63%. At (100-63)%
of 12V the base of Q3 is still at 4.5V. and its emitter is Vbe above that,
ie at 5V - the FET is therefore still turned hard on. Then it gets worse,
because the discharge curve flattens after 1 time constant - it takes about
5 time constants to lose 95% of the charge, so over the next microsecond
the FET gate voltage falls slowly from 5V to 1V, and the FET channel
resistance rises equally slowly. During that whole time, the magnetic
energy in the coil has an easier path to ground than through the rectifier
to the output. It's all wasted by making the FET hot.
So without totally ripping up the circuit, here's my suggestion.
- Remove C8, the cap across the coil, and replace C2.
- Remove Q1 and replace it with a 4.7k resistor. Remove R12.
- Replace R11 with a 680 ohm 1/4 watt to shorten the rise time.
If you really want to rip up the circuit,
