I chose values gleaned from previous posts:

Supply voltage input:
Vin min=24
Vin nominal= 27
Vin max=30

Required output voltage:
Vout max=30
Vout nominal=17.5
Vout min=5

Required output current:
Iout max=5A

See below (new values circled in green, modified parts circled in red)

 

I chose values gleaned from previous posts:

Supply voltage input:
Vin min=24
Vin nominal= 27
Vin max=30

Required output voltage:
Vout max=30
Vout nominal=17.5
Vout min=5

Required output current:
Iout max=5A

See below (new values circled in green, modified parts circled in red)

View attachment 277021

Post #137 gave the inductor calculation for an inverting buck-boost converter, and that is not what you have. I'm not even sure what this would actually be called.

 

Post #137 gave the inductor calculation for an inverting buck-boost converter, and that is not what you have. I'm not even sure what this would actually be called.

its a non-inverting septic converter

 

its a non-inverting septic converter

It's SEPIC (Single Ended Primary Inductance Converter)
Apparently, the usual design assumption for a SEPIC converter is that both the inductors and capacitors are large. e.g. 100μH and 100μF
Simulation shows a typical open loop circuit.

As the duty cycle goes over 50% the control of the output voltage gets very coarse and sensitive making precise closed loop control increasingly difficult. Notice also the increase in current and voltage ripple at high duty cycles. There are also losses to account for in the switching time and the diode.

 

It's SEPIC (Single Ended Primary Inductance Converter)
Apparently, the usual design assumption for a SEPIC converter is that both the inductors and capacitors are large. e.g. 100μH and 100μF
Simulation shows a typical open loop circuit.
View attachment 277033
As the duty cycle goes over 50% the control of the output voltage gets very coarse and sensitive making precise closed loop control increasingly difficult. Notice also the increase in current and voltage ripple at high duty cycles. There are also losses to account for in the switching time and the diode.

Trying to work with the inductors the TS has on hand (hand-made, 30uh x 2).
Duty cycle is 50%@Vin=30v and 56%@Vin=24v for the input range and 30v out. Actual L=35uHx2 but see how it goes with 30uh. The coupling cap doesn't need to be large (its 2x4.7u || ) since its not really carrying the load. So based on those numbers, output capacitor is 280uf minimum.

See below.

 

I could not find any explicit calculation for required inductor values, but you can still calculate average output current and ripple currents via the usual methods.
The average source current in L1 is:

\( I_{L1}\;=\;\cfrac{V_{o}^2}{V_{s}R} \)

The L1 ripple current is

\( \Delta i_{L1}\;=\;\cfrac{V_{s}D}{L_{1}f} \)

The average L2 current is the same as the output current determined by Vo and R.
The L2 ripple current is:

\( \Delta i_{L2}\;=\;\cfrac{V_{s}D}{L_{2}f} \)

The output ripple voltage is given by

\( \Delta V_{o}\;=\;\cfrac{V_{o}D}{RC_{2}f} \)

So if we are given an output ripple voltage specification, we can rearrange the previous expression to give:

\( C_{2}\;=\;\cfrac{D}{R(\Delta V_{o}/V_{o})f} \)

Similarly for C1 we have

\( C_{1}\;=\;\cfrac{D}{R(\Delta V_{C1}/V_{o})f} \)

 

The coupling capacitors tend to be chosen for ripple current rating (or low ESR), not for capacitance. It usually takes an oversized capacitor to get sufficient ripple current rating.

 

It is not possible to build or test your design if you don't have a design specification to work from.
It indicates you have no idea what you are building. So how can anyone help if you don't know what your building?

Thank you for your explanation
I know exactly what I want
I just don't understand your terms.
The input of my circuit is 27 volts and 15 amps.
I don't know how to tell you Vin max or Vin min
I expect the circuit to give me at least 15 amps when the output voltage is equal to or less than the input voltage.
But I don't need more than 8 amps in output higher than 27 volts.

 

Thank you for your explanation
I know exactly what I want
I just don't understand your terms.
The input of my circuit is 27 volts and 15 amps.
I don't know how to tell you Vin max or Vin min
I expect the circuit to give me at least 15 amps when the output voltage is equal to or less than the input voltage.
But I don't need more than 8 amps in output higher than 27 volts.

You should remember that in any DC-DC conversion scheme the one immutable rule. The power out will always be less than the power in. In some cases, it will be much less. Your stated goal may not be achievable in practice. For example, 27V @ 15 A = 405 watts of input power. Assume the conversion is 85% efficient (just to be charitable), and the output voltage is 26 VDC; then 405 watts /26 VDC = 13.24 A. In short, I don't think your chances of success look very promising. Lacking the ability to articulate achievable results you should probably get used to disappointment.

 

Also worth noting:
1) The output is discontinuous, so for 15A output, there will be pulses of at least 30A through the diode, maybe up to 60A if the inductance is small.
2) the current through the MOSFET is input current + output current =45A
3) the voltage across the MOSFET is input voltage + output voltage,
so at least 60V

 

Also worth noting:
1) The output is discontinuous, so for 15A output, there will be pulses of at least 30A through the diode, maybe up to 60A if the inductance is small.
2) the current through the MOSFET is input current + output current =45A
3) the voltage across the MOSFET is input voltage + output voltage,
so at least 60V

Ok, thank you very much
This was a very good tip.
I don't want to make the impossible possible.
I want the best real values
I was able to improve the voltage drop by changing the value of some parts.
Also, the heating of the MOSFET is slightly reduced.
Can I ask for 25V 10A from this circuit?
Can I ask for example 40V 5A?
Guys, I may not be able to say what I mean well with little information, but I would like some guidance to get the maximum "real" power available from this circuit.
Thanks

 

2) the current through the MOSFET is input current + output current =45A
3) the voltage across the MOSFET is input voltage + output voltage,
so at least 60V

According to the MOSFET datasheet, it supports this much.

 

Ok, thank you very much
This was a very good tip.
I don't want to make the impossible possible.
I want the best real values
I was able to improve the voltage drop by changing the value of some parts.
Also, the heating of the MOSFET is slightly reduced.
Can I ask for 25V 10A from this circuit?
Can I ask for example 40V 5A?
Guys, I may not be able to say what I mean well with little information, but I would like some guidance to get the maximum "real" power available from this circuit.
Thanks

We have two different things to work with on the table. The original design with the IC controller chip and the fundamental SEPIC circuit detailed in post #144. We can outline possible outcomes very quickly by simulating various combinations of input voltage, and duty cycle which will yield various results in terms of output voltage, ripple current and ripple voltage. We know from the formulas in post #146 that things we do not want (current and voltage ripple) are linear functions of duty cycle. We can automate the data collection and display of this data so you can get an idea of what is possible.

I strongly recommend you download a copy of the e-book mentioned in post #137. the SEPIC converter is described in §6.8

 

one question:
How is "power dissipation" calculated in MOSFET?

 

Rds(on)*I^2 + 0.25(Vpeak.Ipeak.tslew) is not a bad approximation.
t slew is the time taken for the waveform to slew from the on voltage to the off voltage.
Don't forget that the value of 300W only applies if the device is attached to an infinite heatsink, and Rds(on) is higher when it is warmer.

If your dissipation is approaching 300W, you might as well have made a linear power supply.

The moral of the story so far: High power switched-mode supplies appear on paper to be the same as low-power switched-mode supplies.
Don't let that fool you.
Make a low-power version first and learn from it.

 

one question:
How is "power dissipation" calculated in MOSFET?

The instantaneous power dissipated in a MOSFET is the instantaneous drain current, Ids, times the instantaneous drain to source voltage, Vds, integrated over a single switching period. That is:

\[ P_{D}\;=\;\int_{0}^{T}I_{DS}(t)V_{DS}(t)dt \]

where T is the period of the switching frequency. Drawing the current and voltage waveforms can be illuminating especially during the interval where the MOSFET driver is charging and discharging the Miller capacitance. This is a major source of losses if not done properly.

 

The instantaneous power dissipated in a MOSFET is the instantaneous drain current, Ids, times the instantaneous drain to source voltage, Vds, integrated over a single switching period. That is:

\[ P_{D}\;=\;\int_{0}^{T}I_{DS}(t)V_{DS}(t)dt \]

where T is the period of the switching frequency.

To be pedantic we're both wrong as we have both given the energy not the power.
The average power is the energy per cycle multiplied by the frequency.
If the slewing waveform is trapeziodical and the current and voltage waveforms are in phase, your integral simplifies to my approximation.

 

Maybe your inductor connection is not correct. Make sure Drain connected to winding toward v+ and anode to winding toward gnd.

 

Maybe your inductor connection is not correct. Make sure Drain connected to winding toward v+ and anode to winding toward gnd.

thank you for your guidance
I have checked the connections many times
All connections are correct

 

I also currently building sepic converter for input voltage 8.5v(minimum uc3843 supply) to 24v and output to 2.5v to 24v. Mosfet FDP036N10A (100 V, 214 A, 3.6 mΩ). I'm expecting to get 150 watt output and 10A max current. At 12v input, I'm testing 12v 6A load by halogen lamp, Input current show 8.5A (efficiency about 70%). But inductor and R sense get hot. I have trying to rewinding different inductor with different core color (yellow-white, green-blue and blue), size and turns. But all have same issue. So i think that sepic converter is not suitable for such high power application.
I have noticed about 4 switch buck boost non inverting converter for higher power, with single inductor. But circuit is seem so complex and i hate that.
I also meet same issue like yours, when output get drop. Try to increasing R sense > 0.05ohm, increase input bulk capacitor above 2200uf or adding one npn transistor at pin 4 to pin 3 as loop compensation.

For reference, see https://danyk.cz/univ_m_en.html
Maybe it will help.