Calculation of the booster circuit to obtain maximum power from the battery
The purpose of this article is to show how it is possible to calculate the operating mode of a DC booster (BPN) DC voltage converter (BPN) so as to extract the maximum possible power from a battery with constant internal resistance.
Often, integrated circuits (ICs) are used in practice, in which the received power is controlled at the hardware level (for example, SPV1040), but the price of such devices is quite high.
There are times when a simple battery-powered device is created, for example, an LED flashlight. I want the battery to be used rationally, but the cost of this should be adequate.
For simple circuits, where the load resistance and internal resistance of the battery are known in advance, there is a method for calculating the operation mode of the voltage boost booster.
Let's see what kind of converter it is.
The general scheme of a typical BPN is shown in the figure:
n batteries with their own booster units are shown here. In the particular case, you can do one.
When the key is closed, the current rushes to “minus”. The current through the inductance increases (but it should not reach the value of the short-circuit current, otherwise the energy will be wasted), the inductance accumulates energy.
Then the key is opened. The current through the inductance cannot change instantly, it creates an additional EMF, giving up the accumulated early energy to the circuit. T.O. The voltage at the load at this moment is greater than the battery gave.
A diode is needed so that the current from the capacitor at the output does not go back. Most often this is a Schottky diode with a small voltage drop.
From the school course in physics, we know that the maximum power of a current source with a non-zero internal resistance is achieved when the input and external resistance are equal.
We write the main currents of the circuit:
Here emf is the EMF of the battery, r is the internal resistance, Rn is the load resistance.
Without going into details of the conclusion, I will give formulas for increasing and decreasing current:
After some transformations we get average values:
Under the following conditions, the maximum power will be achieved:
This system can be solved numerically in mathematical programs, for example: MathCad, MathLab.
1. This method can only be used for a known constant value of Rin and a known load resistance;
2. In parallel with the battery, you need to turn on the capacitor;
3. After this circuit, use the DC-DC module;
4. It is best to put a cheap weak controller with support for PWM generation, for example: ATtiny series, ATmega, you can find something from PICs;
5. Since the circuit works as a current generator; several units can be combined into a common capacitor without coordination.
Often, integrated circuits (ICs) are used in practice, in which the received power is controlled at the hardware level (for example, SPV1040), but the price of such devices is quite high.
There are times when a simple battery-powered device is created, for example, an LED flashlight. I want the battery to be used rationally, but the cost of this should be adequate.
For simple circuits, where the load resistance and internal resistance of the battery are known in advance, there is a method for calculating the operation mode of the voltage boost booster.
Booster voltage converter
Let's see what kind of converter it is.
The general scheme of a typical BPN is shown in the figure:
n batteries with their own booster units are shown here. In the particular case, you can do one.
When the key is closed, the current rushes to “minus”. The current through the inductance increases (but it should not reach the value of the short-circuit current, otherwise the energy will be wasted), the inductance accumulates energy.
Then the key is opened. The current through the inductance cannot change instantly, it creates an additional EMF, giving up the accumulated early energy to the circuit. T.O. The voltage at the load at this moment is greater than the battery gave.
A diode is needed so that the current from the capacitor at the output does not go back. Most often this is a Schottky diode with a small voltage drop.
Maximum power
From the school course in physics, we know that the maximum power of a current source with a non-zero internal resistance is achieved when the input and external resistance are equal.
We write the main currents of the circuit:
Here emf is the EMF of the battery, r is the internal resistance, Rn is the load resistance.
Without going into details of the conclusion, I will give formulas for increasing and decreasing current:
After some transformations we get average values:
Under the following conditions, the maximum power will be achieved:
This system can be solved numerically in mathematical programs, for example: MathCad, MathLab.
Recommendations for practical use:
1. This method can only be used for a known constant value of Rin and a known load resistance;
2. In parallel with the battery, you need to turn on the capacitor;
3. After this circuit, use the DC-DC module;
4. It is best to put a cheap weak controller with support for PWM generation, for example: ATtiny series, ATmega, you can find something from PICs;
5. Since the circuit works as a current generator; several units can be combined into a common capacitor without coordination.