SEPIC converter design

[Peter05]

Hello,

i would like to design a SEPIC converter with input voltage ranging from 3 Volts to 19 Volts . On its output will be connected three DC-DC converters ( 1 boost and 2 buck converters , their image is attached) and the output voltage of the SEPIC should be around 8.4V. Could you help me with the design ?

Thank you.

 

[rjenkinsgb]

There is a pretty good article here:

https://www.ti.com/lit/an/snva168e/snva168e.pdf

 

[Peter05]

I ended up with this design :

But even though the output voltage can reach about ~8.4V the power dissipation on the MOSFET( IRF1404) is huge,e.g. 400W or something like this. This means that something goes wrong. What could be the problem?

 

[Nigel Goodwin]

Presumably this is just a simulation?, particularly as it's full of stupid values - so who knows what might need changing to make the simulation work?. You don't consider an imaginary 400W dissipation is a bit crazy?, particularly from an 8V supply.

However - in real life - it's absolutely crucial to drive the FET correctly, and ensure it switches ON and OFF VERY quickly, or it will overheat and die (perhaps very rapidly).

There's obviously no driver in that diagram, and little information on the signal feeding the FET.

 

[schmitt trigger]

As Nigel mentioned, probably a simulation artifact.
For instance, on C29’s left side it shows 88 volts, but on its right side shows -211.8E-24 volts!
211 nano_femto volts!

 

[Peter05]

Yes it is just a simulation. What do you mean full of stupid values? Of course i consider 400W very strange and crazy value that's why i pointed it.
The circuit is simulated in PSpice and as signal feeding the FET i placed a square pulse generating source . In real life i consider something like a pwm signal from a microprocessor. But for now , i would like to focus on the simulation.

 

[Nigel Goodwin]

Yes it is just a simulation. What do you mean full of stupid values?

You don't consider every single value in that to be stupid? (with the possible exception of the 5 ohm - but even that is more likely to be 4.7 ohm) - where do you expect to find a 166.32uF capacitor for example?.

Why would you imagine a simulated imaginary square wave might be suitable for driving a switching FET? - presumably you need to accurately specify it's requirements, and often have to modify things to make them simulate.

 

[Peter05]

A with that meaning you say silly.. Ok there are not these values available but replacing the values with the closest availabe (e.g 166.32uF with 150 uF makes no difference at all,so i didnt consider it as a problem.)
In regard to the square wave, i used it as i saw some other similar circuits using it to drive the fet. So you wpuld propose to change the signal driving the FET?

 

[Nigel Goodwin]

A with that meaning you say silly.. Ok there are not these values available but replacing the values with the closest availabe (e.g 166.32uF with 150 uF makes no difference at all,so i didnt consider it as a problem.)
In regard to the square wave, i used it as i saw some other similar circuits using it to drive the fet. So you wpuld propose to change the signal driving the FET?

Yes, it's essential to drive the FET correctly, and not just a random unknown squarewave.

 

[ChrisP58]

It looks like you're using a gate drive voltage of 5V. That may be enough for some mosfets, but not the IRF1404. They need a higher gate voltage for reliable operation. I'd change it to 10 Volts.

Although I'm not sure that will make a big difference. Can you show the operating waveform of the current through L24?

 

[Peter05]

It looks like you're using a gate drive voltage of 5V. That may be enough for some mosfets, but not the IRF1404. They need a higher gate voltage for reliable operation. I'd change it to 10 Volts.

Although I'm not sure that will make a big difference. Can you show the operating waveform of the current through L24?

Generally, i think that higher gate voltage will increase the power dissipation on the mosfet. Well, the gate threshold voltage of IRF1404 is 2-4Volts so i think that 5Volts will cover it.
The current through L24 is this :
(i replaced irf1404 with irf1010N, there is no difference at gate voltage just better response,power remains huge)

 

[Peter05]

Yes, it's essential to drive the FET correctly, and not just a random unknown squarewave.

Hm do you think about something specific?

 

[Nigel Goodwin]

Hm do you think about something specific?

As fast as you can, both ON and OFF, with decent current capability - 2A is often mentioned when discussing FET drivers.

Here's what I'm currently using - PL2 is for a push on jumper, so I can add the extra inverting 2N7000 or not.

 

[Peter05]

As fast as you can, both ON and OFF, with decent current capability - 2A is often mentioned when discussing FET drivers.

Here's what I'm currently using - PL2 is for a push on jumper, so I can add the extra inverting 2N7000 or not.

View attachment 133619

Maybe the problem is not the fet driving? Cause when i connect on the output a 10 Ohm load (resistor) instead of the 3 converters , the power dissipation drops to normal values of power.

 

[Nigel Goodwin]

Maybe the problem is not the fet driving? Cause when i connect on the output a 10 Ohm load (resistor) instead of the 3 converters , the power dissipation drops to normal values of power.

Who knows?, you're only running a simulation, and don't seem to have anything like a complete SEPIC converter?

 

[Pommie]

In theory (simulation) it's exactly the same as practice. In practice it's not.

Mike.
Yes, I accept simulators are useful.

 

[rjenkinsgb]

Well, the gate threshold voltage of IRF1404 is 2-4Volts

Which is exaxtly what it says: Threshold, where it starts to conduct (and drain current reaches 250 Microamps, according to the data sheet).

For low ON resistance and full switching capability, you need a minimum of 10V on the gate.

Note that the gate-source capacitance of that devioce is over 5nF (not unusual for power FETs), which is why you need a high-current capable drive circuit to achieve fast turn-on and turn-off, for good efficiency.

 

[Nigel Goodwin]

For low ON resistance and full switching capability, you need a minimum of 10A gate voltage.

Seeing as someone picked at my typo earlier!

 

[rjenkinsgb]

minimum of 10A gate voltage.

Doh! Corrected.

 

[ChrisP58]

Generally, i think that higher gate voltage will increase the power dissipation on the mosfet. Well, the gate threshold voltage of IRF1404 is 2-4Volts so i think that 5Volts will cover it.
The current through L24 is this :
(i replaced irf1404 with irf1010N, there is no difference at gate voltage just better response,power remains huge)
View attachment 133617

Yes, the Threshold Voltage of the IRF1404 is between 2 and 4 Volts. But that's the point where the mosfet just begins to turn on. And like a water tap, you'll only get a trickle through it at that point. But if you want it to be fully turned on, you need a higher gate-source voltage. Note: the datasheet shows the condition of the 2-4V threshold voltage is 250 microAmps.

And, while that will help some, I don't think that is all of what's wrong.

The waveform you showed looks like the ringing from initial power up. It's timing doesn't match that of the gate drive signal.

What I'd like to see is the waveform after the converter has stabilized.

And while were at it, let's see the waveforms of the gate voltage, the mosfet drain voltage, and the D38 anode voltage.