Flyback slope compensation?

[Hammy]

I've been reading different opinions on when to apply slope
compensation on a current mode SMPS in CCM. If for example my flybacks
worst case maximum duty is 41% and it's not deep in CCM is slope
compensation needed?

I've read papers where they recommend slope compensation no matter the
duty cycle some say only if the duty exceeds 50%.

I would like to be able to get rid of the 5 components cluttered
around my controller if they are not necessary. The NCP1203 requires
external components to generate the ramp for slope comp. I've been
trying to get the 1217 or 1216 which have internal slope compensation
but nobody I buy off stocks them.

 

[legg]

I've been reading different opinions on when to apply slope
compensation on a current mode SMPS in CCM. If for example my flybacks
worst case maximum duty is 41% and it's not deep in CCM is slope
compensation needed?

I've read papers where they recommend slope compensation no matter the
duty cycle some say only if the duty exceeds 50%.

Is it needed?

Test your circuit at rated minimum load (including reasonable error
margins) and under rated load transient conditions, including
start-up, at the environmental temperature extremes.

Then decide.

RL

 

[MooseFET]

I've been reading different opinions on when to apply slope
compensation on a current mode SMPS in CCM. If for example my flybacks
worst case maximum duty is 41% and it's not deep in CCM is slope
compensation needed?

I've read papers where they recommend slope compensation no matter the
duty cycle some say only if the duty exceeds 50%.

I would like to be able to get rid of the 5 components cluttered
around my controller if they are not necessary. The NCP1203 requires
external components to generate the ramp for slope comp. I've been
trying to get the 1217 or 1216 which have internal slope compensation
but nobody I buy off stocks them.

You can do a little math.

(1)
Assume that the thing is making exactly the average current and
perfectly stable.

(2)
Displace the turn off time of one cycle by some small dX

(3)
Work out how much the amount of extra current the dX adds to the
inductor current.

(4)
During the off time, this extra current causes extra voltage in the
various resistances. This reduces the amount of current difference
that will be brought into the next on time. The off time will be
shorter but the current will decrease faster.

(5)
During the on time, the extra current causes extra voltage in a
different set of resistances.

(6)
When you get to the next point of turn off, see how much time you have
to move along the ramp to get the current difference.

When you get to step (6) you want the answer to be much less than the
dX you started with. Under 1/2 is a good boundary to use.