Electronics > Beginners
MC34063 high voltage dc-dc boost converter
magic:
The flat part is called "Miller plateau". It occurs due to so-called Miller capacitance, which is capacitance from gate to drain. These parts are close to each other and they form a capacitor hidden inside the FET. It's also called "reverse transfer capacitance" and as such you will find it on the datasheet.
As drain voltage raises, Cgd charges up and sends current into the gate. The rate of drain voltage raise is limited by the ability of your 200Ω resistor to pull that current out of the gate. In fact, you will see a perfectly linear ramp when you look at drain voltage over time.
dazz:
So if I got it right, it's some sort of stray capacitance opposing voltage change, hence slowing down the falling edge of the signal. I've read the pcb design is critical in switching regulators and power supplies, I guess stray capacitance is one of the main concerns for these reasons?
dazz:
--- Quote from: T3sl4co1l on January 01, 2020, 05:18:43 pm ---Regarding the sim, the inductance looks awfully large. Consider how much dI is obtained in the, what, 4us or so expected on-time, at 24V and L. dI = V dt / L
Tim
--- End quote ---
Something's amiss here. I used the calculator for a flyback converter that you suggested yesterday: http://schmidt-walter-schaltnetzteile.de/smps_e/spw_smps_e.html
With 24V input, 300V and 200mA output, the calculator suggests 120uH in the primary and a turn ratio of 80... but as far as I know that would mean 80^2 = 6400 times more inductance in the secondary. That would be 7.68H in the secondary... can't be right. I must be making one of my stupid mistakes
T3sl4co1l:
The suggested values are for, I think, ripple fraction 10 or 20%. You can always set it lower, and in this case you need to, with a peak current mode controller which will not behave well with a high inductance.
You misread the scientific notation: it's 79.81e-3 or 0.07981. It's also labeled N1/N2 so your misread would imply the secondary is 80 times shorter! ;)
Tim
dazz:
--- Quote from: T3sl4co1l on January 02, 2020, 02:16:24 am ---The suggested values are for, I think, ripple fraction 10 or 20%. You can always set it lower, and in this case you need to, with a peak current mode controller which will not behave well with a high inductance
--- End quote ---
Ripple fraction.. just another term that goes down on my list of things I need to research. Transformers are a world of their own apparently. Jeez, why does this stuff have to be so complicated?
--- Quote from: T3sl4co1l on January 02, 2020, 02:16:24 am ---You misread the scientific notation: it's 79.81e-3 or 0.07981. It's also labeled N1/N2 so your misread would imply the secondary is 80 times shorter! ;)
Tim
--- End quote ---
Damn it! That was a clear case of confirmation bias. I was expecting a ratio > 1 so I missed the exponential term |O
But I'm not sure I understand what you mean. If N1/N2 = Nprimary/Nsecondary = 79.81e-3, then the N2/N1 = 0.07981^-1=12.53
So if the suggested primary inductance is 120uH, the secondary would be (12.53^2)= 157 times that, hence 120uH x 157 = 18.8mH... assuming I'm not having another brain fart, that is
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