Discussion on a booster circuit for solar cell power supply

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Drive problem of switching device under input voltage

Start-up problem of the boost circuit under input voltage

Maximum duty cycle problem with input voltage

solution:

The DC/DC boost circuit is powered directly from the input or from the output.

Add a start-up circuit to the original DC/DC

Portable products are generally powered by batteries, and because of cost and volume considerations, there is a tendency to reduce the number and size of batteries used. If the number of batteries is reduced, the power supply voltage will be higher than the operating voltage required by the equipment. Low, then you need to use DC / DC boost circuit. In addition, due to global energy issues, various types of battery use have received much attention, including solar cells.

Generally, the minimum voltage of a single-cell solar cell is between 0.4 and 0.7V. Under such a low input voltage, the following three problems are encountered:

Switching device drive problem

Today's DC/DC boost circuits generally have two modes of power supply, one is to supply power directly from the input, and the other is to supply power from the output. If the power is supplied from the input, the high level of the driving NMOS is normally equal to the input voltage. When the input voltage is low, the NMOS with a low turn-on voltage is required, and if the output power is selected, the EXT driver can be obtained. It is equal to the higher driving voltage of the boosted voltage, which not only makes the NMOS easier to turn on, but also achieves a lower Rdson to improve efficiency. Of course, these are all under the premise of IC startup work, but when the power supply voltage is lower than the startup voltage of the whole IC, the latter will be more difficult to start because it will pass a diode, which brings a problem, how to start This IC?

Boost circuit startup problem

The operating voltage of the conventional DC/DC is generally above 1.0V, and if the input voltage drops below 0.6V, the internal circuit of the DCDC does not work properly. At this time, we need to consider adding a startup circuit to the original DC/DC. This circuit needs to include the following main parts: an oscillator that can still operate as low as 0.3V, a charge pump voltage doubler circuit, and a voltage sense comparator.

The basic working conditions are as follows: first, 0.3V is connected, the oscillator works, and then the charging pump starts to double the voltage. When the required IC driving voltage is obtained, the VDD is supplied to the IC. When the IC enters normal operation, the output is supplied by the output. Instead of powering up the startup circuit, the startup circuit goes to sleep at this time. After the booster circuit is working properly, people will care about the maximum voltage they can get. This leads to another question, what is the maximum duty cycle?

Maximum duty cycle problem

For ultra-low input boost circuits, in order to achieve high output voltage, a large duty cycle must be supported.

In continuous current mode, the duty cycle (Duty) is calculated as Duty=1-Vin/Vout. According to this formula, if the input is 0.5V and the output wants to get a 5V boost circuit, the maximum duty cycle is 90%, and the general boost circuit duty cycle is 80% to 90%, which is not complete. Meet the requirements.

In theory, the larger the duty cycle, the stronger the maximum boosting capability (of course, the duty cycle cannot reach 100% like the buck), but due to non-ideal factors: inductance parasitic resistance, drive tube internal resistance, Schottky forward voltage drop Wait, after the duty cycle is large enough, its boosting capacity will worsen. Therefore, we need to improve the above non-ideal factors and get a suitable high enough duty cycle to meet our boost demand.

In short, solving the above three problems can solve the problem of the design of the booster circuit that requires the ultra-low voltage input to start with the solar single-segment power supply.

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