I am trying to convert a XL4015-based CC/CV buck converter module to CV with undervoltage lockout. At this point I am hoping someone can double check my schematic for glaring errors before I attempt to implement it.

The overall project is to revamp the electrical system on a 1980 Yamaha QT50. I am motivated to get rid of the 6V lead/acid battery, to use 12V LED lights instead of 6V incandescent, and by curiosity.

I plan to use XL4015-based CC/CV buck converter modules to discharge a 45mF (45kuF) capacitor tied to the <200Hz AC source through a full-wave rectifier.

I have one module set to output 12VDC which I will float to lights and things. I have another module set to output ~25VDC which I plan to tie to a floating ~35W resistive dummy load to keep the generator output voltage from going too high under no-load conditions and damaging the capacitor, which is rated at 63V.

In my research on these systems, I have read that it is unlikely for the system to output more than 18V with what I have to assume is an 8W load. It reads about 45V max no-load on an unreliable cheap DMM. I want the dummy load to turn on around 25V because it will leave some charge on the cap while still protecting it from overvoltage, and protect the XL4015 modules from their ~30V max input.

The problem is that the XL4015 modules have no undervoltage lockout. I only need this on the 25V supply. My plan is to convert the CC portion of the circuit which contains 7805, TL431, and LM358. The CC circuit is already set up to interface with the XL4015's FB pin via LED, which turns XL4015 off at voltages > 3.3V and turns it on at voltages < 0.8V.

Here is the original schematic: original schematic for XL4015 buck converter with CC/CV, sourced from circuitdigest.com

Here is my plan for revisions to the circuit:

XL4015 buck converter with CC circuit modified, hopefully to allow for undervoltage lockout

I changed TL431 from current source to comparator mode, referenced to a V divider of Vin. I delete the TL431 output voltage divider and sent straight to op-amp comparator, referenced to a V divider of 78L05. I tie off the extra op-amp, hopefully in the linear region.

With these values, TL431 ref pin sees 2.5V with Vin of ~23.63V which seems fine. TL431 K node connects to op-amp non-inverting output, TL431 puts out 5v with a ref below 2.5V and 2V with a ref above 2.5V. The inverting input of the op-amp comparator is referenced to ~2.57V. The LED is set to draw 18.5mA. TL431 draws about 7mA on low condition. Since this circuit is dedicated to a shunt resistor, leaving the sense resistor at Vout- doesn't seem to hurt.

It seems to me like this is a feasible solution. Am I doing anything wrong? I think there is a 50% chance I made an error in the logic tables. I realize now that I could just skip the TL431 but I don't see how using it hurts. Am I fundamentally misunderstanding what the shutdown function of these chips is? i.e. Is it not possible for a buck converter to have 0V on the output with V>0V on the input? Did I miss anything?

This is what I have and I would like to make it work. I think it was Voltaire who said something like "metaphysics is a game where you make extremely complicated moves in order to end up where you started" Please let me know what information I have failed to provide Thank you

  • \$\begingroup\$ So I went to look up Yamaha QT50 to see what sort of musical instrument that was, immagine my surprise!. Automotive is a harsh enviroment for electronics \$\endgroup\$ Sep 18 at 5:02
  • \$\begingroup\$ HMM switcvhing power supplies have negative resistance that's going to make regulating the generator output interesting. \$\endgroup\$ Sep 18 at 5:16
  • \$\begingroup\$ Please simulate your circuit to get the best answer to your question. \$\endgroup\$
    – Andy aka
    Sep 18 at 8:22


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