LT8364 Don't start during load

Hello,

Thanks for your query.

I observe that you are using an AC voltage source, with a diode bridge rectifier and capacitor filters at its output (C2 and C6).  For successful start-up of the converter, the input voltage to the SEPIC converter (the voltage across C2, for example) must be above the minimum voltage required by the LT8364.  I believe that during power up, the input voltage to the converter may be dropping below the minimum required voltage due to: (1) input impedance of the AC source, (2) voltage drop on the rectifier diodes and common mode choke, (3) low capacitance at the output of the rectifier, and (4) starting off by charging the 14mF as well.  Please share the input voltage waveform (across C6) during a power up transition.  When you removed the 14mF (or the 100 Ohm load) successful start-up of the converter indicates that a combination of the above reasons is preventing successful start-up with the 14mF in place.

We can help you with calculations and suggestions to size a required C2 for successful start-up.  We need the following inputs:

1. What is the voltage and frequency of the input AC source (RMS, Hz)?  Please indicate the minimum and maximum input voltage and frequency range expected.

2. What are the components chosen for the SEPIC converter?  We can help review of the schematic if you can share the bill of material with component choices (manufacturer part numbers).

3. I observe that the SEPIC inductor has two different values (3.3uH and 2.4uH) on the two sides of the coupled inductor.  Please confirm if this is chosen purposefully, and what the reasons are.  I find that Wurth does not have such a coupled inductor in their published catalogue.

4. Apart from charging the 14mF through the 100 Ohm resistor, what is the load current expected during start-up operation?  Note that this load current adds to the 14mF charging current.  What is the load current expected after the start-up period, after the 14mF is charged to 25V?

- Giridharan.

Thank you very much for your kind reply.

The voltage source are intended to be DC and the rectifier bridge is just placed there because some  technical installers may connect the polarity the wrong way (found out of experience)
The tests I have done have been with a dc supply with a current limit of 4 amp.

The circuit needs to be able to handle input voltages from 9 to 50 volts.
I will make a complete BOM list for you to help investigating the issue.
Regarding the inductor there is a miss on the schematics I sent and the one I used is the one recomended in the PDF, WURTH 744873003.

The expected peak load current will be 350 mA.

Well, one test I have done is a bit strange.
If I have the 14m capacitor disconnected and start the sepic without load it will, as I mentioned, start and give the expected output voltage but if I then connect the 100 ohm resistor at the output it will directly lower the output to some few volts.

I will get a pic of the voltage at C6 as soon as possible and send to you together with the BOM.

I'm must happy for all help I can get because the customer wish to have this problem solved and we are planning for some 1000;s a year.

Thank you and best regards.

Gunde

Thank you very much for your kind reply.

The voltage source are intended to be DC and the rectifier bridge is just placed there because some  technical installers may connect the polarity the wrong way (found out of experience)
The tests I have done have been with a dc supply with a current limit of 4 amp.

The circuit needs to be able to handle input voltages from 9 to 50 volts.
I will make a complete BOM list for you to help investigating the issue.
Regarding the inductor there is a miss on the schematics I sent and the one I used is the one recomended in the PDF, WURTH 744873003.

The expected peak load current will be 350 mA.

Well, one test I have done is a bit strange.
If I have the 14m capacitor disconnected and start the sepic without load it will, as I mentioned, start and give the expected output voltage but if I then connect the 100 ohm resistor at the output it will directly lower the output to some few volts.

I will get a pic of the voltage at C6 as soon as possible and send to you together with the BOM.

I'm must happy for all help I can get because the customer wish to have this problem solved and we are planning for some 1000;s a year.

Thank you and best regards.

Gunde