Solar battery power supply up voltage circuit -Lithium - Ion Battery Equipment

Out of the treatment method of solar cell power supply power supply -Lithium - Ion Battery Equipment

Use a pen to power the power supply or power supply from the DC/DC boost circuit

Add a starting circuit based on the original DC/DC

Portable products generally use battery power supply, and because of the consideration of cost and volume, there is a tendency to reduce the number of batteries and volume in design. If the number of batteries is reduced, it will cause the power supply voltage to be lower than the working voltage required by the equipment. At this time, the DC/DC boost circuit is used. In addition, due to global energy issues, various types of batteries have attracted much attention, including solar cells.(Lithium - Ion Battery Equipment)

Generally, the minimum solar cell voltage of solar batteries is between 0.4-0.7V. Under such low input voltage, the following three major problems will be encountered:

Driver problem of switching device

There are generally two types of power supply methods of DC/DC boost circuits, one is to power power supply from the pen, and the other is to supply power from output. If the power supply is normal, the high level of the NMOS is the most equal to the input voltage. When the input voltage is very low, select the NMOS with a low opening voltage. It is equivalent to a higher driving voltage after the voltage voltage, which can not only make NMOS easier to turn on, but also get lower RDSON to improve efficiency. Of course, these are under the premise of the startup work of IC, but when the power supply voltage is lower than the starting voltage of the entire IC, the latter will even be more difficult to start than the former because it will pass through a diode, so it will bring a problem. How can it be? Let's start this IC?

Starting problem of boost circuit

The working voltage of traditional DC/DC is generally above 1.0V, and if the input voltage drops below 0.6V, the internal circuit of DCDC cannot work properly. At this time, we need to consider adding a startup circuit based on the original DC/DC. This circuit includes the following important parts: an oscillator that can still work as low as 0.3V, the charging pump voltage circuit, and the voltage test comparator.

The basic working conditions are as follows: First of all, 0.3V access, the oscillator works, and then the charging pump begins to voltage. When the IC driving voltage is obtained, power to the IC VDD. When the IC enters the normal work, then the power supply is replaced by the output power instead instead The power supply of the launch circuit, at this time the starting circuit enters the sleep. After the boost circuit work normally, people will care about the maximum voltage they can get. This leads to another problem. How much can the maximum duty ratio be?

The issue of the maximum duty ratio

For ultra -low input boost circuits, in order to obtain high output voltage, there must be a large occupation ratio.

In continuing the current mode, the calculation formula of the duty ratio is duTy = 1-VIN/VOUT. Calculated according to this formula, if the output wants to get a 5V boost circuit when the input is 0.5V, the maximum duty cycle is 90%, and the general increase of the voltage circuit is 80%to 90%. Meet the requirements.

The larger the theoretical occupation ratio, the stronger the maximum voltage capacity (of course, the duty ratio cannot reach 100%as the voltage voltage is reduced), but due to non -ideal factors: inductance parasitic resistance, driving tube internal resistance, Shawitki positive voltage drop drop drop Wait, after the duty ratio is as large as a certain degree, its voltage ability will become worse. Therefore, it is necessary to improve the above non -ideal factors, and to obtain a suitable high enough duty cycle to meet the needs of our boost.

In short, the above three problems can be dealt with to make solar energy supply power supply. In this way, the design of the booster circuit to enter the boost circuit to enter the boost circuit can be solved.



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