This little circuit prevents a buck converter from putting a panel into a death spiral in low light. Uses little known technique to control panel voltage by faking the feedback pin. Normally the FET conducts and pulls down the voltage of the resistor below the threshhold of the switcher. That voltage is 1.25V + LED or diode forward voltage. A diode canbe used in place of the LED. In low light, the pullup resistor flows additional current to the switchers voltage divider. This causes the switcher to think the output is over voltage and thus shutting it down momentarily. Simple circuit uses the FET gate threshold voltage as the reference voltage. When panel voltage drops, the voltage on the pull up resistor feeds through the diode or LED to the feedback pin (#2). This keeps the panel at power point voltage for highest energy transfer. A voltage divider feeds the gate of the FET. Typically FET turn on voltage is between 3 and 4 volts. The buck converter will still work when the panel voltage drops and it will act as a linear current booster. The LED, if used in place of a diode, will light when the panel voltage starts to drop below the set point. Not brightly as the currents are small. Any power FET can be used as these are very common even though only a few milliamps passes through it. Set buck supply first to the desired voltage. The pot pin you connect to will have a voltage of about 1.25V when operating.
As an example, the board was set to 5.25V. A load was placed on it till the voltage dropped to 1.9 volts. The board provided 1.5A at that voltage while maintaining a panel voltage of 17 volts.
Similar method can be used on almost any other popular buck converters if higher voltage and power is needed.
