Math Keeps Me Busy

@aud098 Sorry, I assumed you saw the edit I made to my answer. I made an LTSpice model and checked it at both 5 and 24 volts input. Don't see any problems with the component values in that model. Posted it in the answer.

I'm still working on a PSPICE model. Talk to you tomorrow.

Give me 20 minutes or so.

This get's worse as your input voltage goes up. Ugh. As your input voltage goes up, your duty cycle needs to go down. Then your peak switching current needs to go up. So, limiting your switching current is going to limit your max input voltage. Ugh.

I take back what I said about the 2.2 Ohm. That won't work. I was thinking about buck converters. Since you are using an inverting buck-boost, a 5 V input requires a 50% duty cycle, not a 100%. Therefore, your input current is only available half the time, and needs to be at least twice your output current.

The switch turns off after a small delay after the voltage across Rsense reaches 330 mV.

You could set Rsense to 2.2 Ohms, which will limit your current to about 150 mA, and then reduce your inductor to 62.5 uH.

Are you trying to fit this in a small space?

You could set the Rsense to 1 Ohm. that will limit your current to about 330 mA (+ overshoot). You can then decrease the inductor to about 125 uH. Is that too much?

Unfortunately, in order to handle 24 V input (or approximately 30 between Vin and Vout) and a maximum t_on of 20 uSec, you would need a minimum inductance of about 500 uH. I calculated this based on a high duty cycle during startup. This can be avoided by utilizing the current limiting feature, but that feature is not quick. The actual current will overshoot what you set in Rsense, especially if your inductance is low. How much current do you need?

For some reason, in Fig 13, they use an unnecessarily large time capacitor, resulting in an unnecessarily low frequency. That, in turn, will lead to an unnecessarily large inductor. I would change C_T to 470 pF. This will give a normal switching frequency between 48 kHz and 84 kHz.

OK, I see what they did. They didn't connect the gnd pin to ground, but to the -12V output in fig 13. R1 should be 1.2k, R2 should be 3.6k. I have to check about the inductor value.

@aud098 I'm looking at the top schematic of fig 13, and it appears wrong to me, i.e. the voltage at the feedback pin is negative. I'm getting older and making lots of mistakes lately, but it looks wrong. Let me investigate for a bit.

@aud098 Yes. It should. If you have a sufficient load for the converter to operate in continuous current mode, the average duty cycle will be about 18% when going from +24V to -5V. If you do not have sufficient load, the converter will do pulse skipping, lowering the switching frequency. This sometimes leads to unwanted noise from the inductor. See electronics.stackexchange.com/a/706254/268467

@ChristianB. The switch is always turned off when the oscillator reaches its peak, so one cannot get 100% duty cycle. The max duty cycle without current limiting is about 83%. So, at 5.1 V input, even without switching voltage drop, the output is limited to about 4.2 V. With an external PNP switch (fig. 11b) and 0.2 V saturation voltage, your are down to 4.0 V.

you ar welcome

The assumptions are made to simplify problems so that they are mathematically solvable. Again, with a field solver, you can get approximate results where those assumptions are not made.

"But the magnetic field though not uniform along the loop would be uniform along a surface cross section" That is also an assumption we make, that doesn't correspond to reality. If you ever see the results of a field solver, you will see that B is not uniform across a cross-section. This has a variety of causes, such as the sharp corners, turns, and the different lengths of loops in the core. Near the core window(s) the loop length is smaller than near the outer rim of the core.

The theory of electromagnetics, which includes maxwell's equations, but also an understanding of how electromagnetism works in matter has been extensively tested by virtue of this theory being used to accurate predict electromagnetic phenomena over a very wide range of cases.

You could also have uniformity over a surface, but I specifically talk about uniformity over the loop, because it is that uniformity that allows us to solve the integral, even though that uniformity is really a fiction.

Yes, B is uniform along a loop, if it has the same value at any point on the loop.

yes. B is different in different parts of the core

It is like water leaving the river, entering the ditch, and then returning to the river at another point.

Leakage flux, in this context, is flux that leaves the core. In the area surrounded by the coil, that flux is in the core. But as you move around the core, the leakage flux leaves the core, then returns to the core at a different point.

Yes, exactly.

Does that make sense?

the speed of the river is equivalent to the flux density in a magnetic circuit.

Now, if the cross sectional area of the river is the same everywhere, because there is less flow in one section of the river than in another section of the river, the speed at which the river flows cannot be the same everywhere.

yes

Imagine there is a circular river. Don't ask me how it works, just imagine there is one. Now imagine that a small ditch is dug that allows some of the water to flow from one point in the river to another point in the river. This diversion of some of the flow will mean less flow in that part of the river away from which the water was diverted. Do you understand that?

We assume uniformity so that we can calculate an integral. But the assumption does not match reality. The assumption entails that there is no leakage flux. We know that is incorrect. There IS leakage flux.

Comments are not the place for extended discussion. But to answer your question, No, the flux density will not be uniform along a loop through the core if some of the flux is leaking out of the core.

"when we calculate the amperian loop for the core we assume it is uniform". Yes, we assume something that is not strictly true. "Will it not be uniform INSIDE the core if the core has leakage as well?" No, B will vary from point to point along the loop.

In some very specialized cases, B has a uniform value in every position within a loop. For example, if the loop is circle centered on an infinitely long current carrying wire. But in general, B may (and in reality will) vary from one point to another along a loop.

@LEXORAI if some of the flux leaks out of the core, then part of the core will have less flux. If the core has the same cross sectional area both where there is the full flux, and where some has leaked out, then the flux density must be lower at the cross section with less flux.

The primary factor that determines collector current is the number of free electrons in the base reaching the base-collector junction, which in turn primarily depends upon the number of free electrons injected from the emitter into the base.

In an NPN transistor, in either active mode or in saturation mode, the bulk of the current in the transistor is from free electrons. Holes contribute much less. This is intentional, and is achieved by making the base doping much lighter than the emitter doping.

@LvW There is physically no current that flows directly between C and E without flowing though the base region.

The electrons do move from B to C. (NPN, saturation) They are not attracted b the less positive potential, they are in fact repelled by the less positive potential. They move from B to C because the base is teaming with free electrons, and so they diffuse. My statements a) repeat the fact that electrons move from B to C, regardless of intuition, and b) explain this as the result of diffusion.

That is what is so counter-intuitive about charge flow in a saturated (NPN transistor).

@LvW think about the situation again. In an NPN transistor in saturation, conventional current is flowing from the collector into the base region. Physically electrons are flowing from base region into the collector. But the base-collector junction is forward biased, That means the base is more positive than the collector. So electrons are flowing from the more positive base region into the more negative collector region.

@LvW think about the situation again. In an NPN transistor in saturation, conventional current is flowing from collector to base. Physically electrons are flowing from base to collector. But the base-collector junction is forward biased, That means the base is more positive than the collector. So electrons are flowing from the more positive base to the more negative collector. That is what is so counter-intuitive about charge flow in a saturated (NPN transistor).

@EdinFifić I would like to point out that your theory of the "channel" between the collector and emitter obscures the fact that in a saturated NPN transistor, electrons are flowing from the N region of the emitter into the P region of the base, and then from the more positive P region of the base to the less positive N region of the collector. They are flowing against the gradient of the electric potential. Asserting the existence of this "channel" between collector and emitter just side-steps addressing how it is that the electrons are flowing against the gradient of the electric potential.

"The "+" & "-" markings are not congruent with the polarity of the bridge diodes." Yes, that is a problem.

You did a good job!

My error. I copied the number from your table incorrectly. @LoicDaigle

You would need an "8" for 25V, or a "7" for 16V.

The "6" in the code CC0402MRX5R6BB104 refers to 10V.

from page 2 of yageo.com/upload/media/product/app/datasheet/mlcc/…