I'm trying to reverse calculate the resistors used in a slope
compensation network on a UC3843N current mode PWM controller (i need
to change the current limit). Does anyone know of any material that
explains it.
I've read some app notes from unitrode and texas, but they refer to
different topologies to the one i've got. It's being used in a boost
configuration.
Can anyone point me in the right direction? If further information is
needed just let me know.
Slope compensation only applies to DC-DC converters like the booster
and fly-back where the energy goes to the output when the transistor
is off. It is only needed when the current in the inductor is
continuous. I am assuming that both of these are the case in your
topology.
The first step is to understand the DC-DC converter as it is supposed
to work. The duty cycle of the transistor is usually determined by
the ratio of the voltages on the inductor. Work out what the duty
cycle is intended to be.
Now work out what the current in the inductor is doing and what the
ramp on the current sense resistor looks like. It really helps to
make a little sketch of the inductor current and current sense
voltage. The sketch should be several cycles long. It doesn't need
to be at all accurate just contain the idea of the shape.
Consider the point on the current sense ramp where the transistor is
just about to turn off. This point should be in a cycle near the
right hand edge of your sketch. (Point-1)
Go back from this point one full cycle and assume that the transistor
stays on a little extra "dX" time. Figure out how much extra current
ends up in the inductor.
Since things are sort of linear, this current decays with a L/R time
constant while the transistor is off and then a different one while
the transistor is on. You can figure out how much extra current
("AdX") is flowing at the instant of consideration. Often it is
almost right to just say "it doesn't decay at all" and just carry on
with that current. We want the amount of current when you get to
Point-1.
From the amount of current extra ("AdX") you can find out how much
earlier this turn off would be shifted ("BAdX"). The extra current
makes a little extra voltage in the current sense resistor. You
divide by the slope of the sense resistor waveform to get back to
time.
The minimum slope compensation is just enough that the shift on this
cycle ("BAdX") is less than the shift you assumed to cause it ("dX").
A good practical limit is that it should be under half the amount.
