How to define Vsat for MOSFET as a switch?
I was looking at this design:
It uses MC34063 as switching controller. I wanted to adapt this circuit, and I was busy trying to pick R1 value. The datasheet has following formulas:
2 answers
$$\boxed{\text{What maximum value of } I_{OUT} \text{ did you have in mind?}}$$
Using $I_{OUT}$ and $V_{OUT}$ we calculate power uplift. Using power uplift and switching frequency, we calculate how much energy the inductor needs to store (and transfer) in each switching cycle: -
$$\text{Power uplift is:} \hspace{1cm}I_{OUT}\cdot(V_{OUT}-V_{IN})$$
$$\text{Energy per switching cycle} = \dfrac{\text{Power Uplift}}{F_{SW}}$$
Assuming the boost converter will work in DCM, we can determine how much peak current needs to be drawn through the boost converter's inductor because: -
$$\text{Energy per switching cycle} = \dfrac{1}{2}\cdot LI_{PK}^2$$
However, I can't say if the boost converter should operate in DCM or CCM without knowing what $I_{OUT}$ is. So, you need to come clean on this. If it operates in CCM then the calculations are slightly different to DCM (see this Q and A on how boost converters work). $$$$ This then dictates how large in value we set R1 in order to "set" the protective current limiting circuit of the chip.
If you know it operates in DCM, you can run through the above calculations and estimate $I_{PK}$ in the inductor (all based on knowing $I_{OUT}$). Then you will know how much voltage will be dropped across the MOSFET when it is "on". If you then want to calculate R1, the data sheet should have the information but please do leave a comment if you need further assistance on this.
Taking a stab at what information is contained in the question would see it operate in DCM if the load is 3 kohm and the switching frequency is 10 kHz: -
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$I_{PK}$ would be 3.027 amps and, with an RDS of 0.55 ohms, the volt drop would be about 1.7 volts. Personally speaking I think you could find a more suitable MOSFET with slightly lower on resistance (circa 0.1 ohms) that was OK to run up to 300 volts. Given that the output is set to be 180 volts, that shouldn't be a problem.
First, that's an archaic chip that requires a lot of circuitry around it to do anything useful. Why use this particular chip?
Second, the datasheet gives you an example of how to use this part as a step-up converter with external switch. Why not follow that?
Third, blindly following equations in the datasheet is no way to design anything. You have to actually understand what is going on first.
Note that the example step-up circuit with external switch uses a NPN transistor as the external switch, not a FET. This switcher chip is more suited to driving a BJT than a FET.
Page 8 of the datasheet shows this example circuit:
I see no reason not to follow this, considering you really want to use this chip in the first place.
The NPN transistor must be rated for the full output voltage. Since that is apparently 180 V in your case, it should be rated for 200 V at least.
Your main question seems to be about R1 in your schematic, which is Rsc in the schematic above. That is the current sensing resistor that the chip uses to shut off the switch when the current gets above the threshold. Once you have picked an inductor, this limit should be set to the saturation limit of the inductor. The external NPN transistor also needs to be rated for this current.
Once you know the maximum current you ever want to allow thru the inductor and switch, you look at page 4 of the datasheet. It shows the current limit sense voltage being from 250 to 350 mV. Since 350 mV is the case that results in the highest current, you use that. Rsc is then simply 350 mV divided by the maximum current you want to allow.
Again, you need to understand the circuit to pick part values. Once you understand it, how to pick the values will usually be obvious.
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