How To Control The Circuit Of Solar Lawn Lighting?

How to control the circuit of solar lawn lighting?

1> During the conduction period of the switch tube Vr: During this period, the switch tube Vr is turned on, and the voltage on the capacitor C2 makes the diode D2 reverse bias and cut off. At this time, the input current iL2 makes L1 store energy; the discharge current iL2 of C2 makes L2 store energy, And supply power to the load, as shown in the equivalent circuit diagram (a) of the continuous operation mode of the Cuk converter.

2> During the period when the switch tube Vr is turned off, the switch tube Vr is turned off, the diode D2 is forward biased and turned on, the energy release current iL1 of the power supply and L1 is charged to C2, and the energy release current iL2 of L2 is maintained to maintain the load, such as the Cuk converter is continuous. The equivalent circuit diagram of the working mode is shown in (b). Therefore, during the off period of Vr, C2 is charged, and during the on period of Vr, C2 is discharged to the load, and C2 plays the role of energy transfer. The circuit principle of solar lawn light is relatively simple. Its controller is implemented using a boost circuit.Component selection: BT1 uses 3.8V/80mA solar panel, monocrystalline silicon is better, polycrystalline silicon is the second; BT2 uses two 1.2V/600mA Ni-Cd batteries, if you need to increase the luminosity or prolong the time, you can increase accordingly Solar panel and battery power.The β of VQ2, VQ3, and VQ5 is about 200, and VQ4 needs a transistor with a large β value. VD1 is selected as low as possible, such as germanium tube or Schottky diode. LED can choose white, blue, green ultra-high brightness astigmatism or spotlight. When using low voltage drop LEDs such as red, yellow and orange, the circuit needs to be reset.It is recommended to use 1% precision resistors for R3 and R5; for R4, use a photoresistor with a bright resistance of 10kΩ～20kΩ and a dark resistance of more than 1MΩ. Other resistors can choose ordinary carbon film (1/4)W, (1/8)W resistors. L1 uses (1/4) W color inductor, and the DC impedance should be small.The working principle of the circuit: when there is sunlight during the day, BT1 converts light energy into electrical energy, and VD1 charges BT2. Due to the light, the photoresistor is low resistance, and VQ4 b is extremely low and cut off.When there is no light at night, the photoresistor has a high resistance, VQ4 is turned on, and VQ2 b is also turned on at a very low level. The DC boost circuit composed of VQ3, VQ5, C2, R6, and L1 works, and the LED is energized to emit light.The core of the DC boost circuit is a complementary tube oscillator circuit. Its working process is as follows: when VQ2 is turned on, the power supply charges C2 through L1, R6, and VQ4. Since the voltage across C2 cannot be abruptly changed, VQ3 b is extremely high, and VQ3 does not On, as C2 is charged, its voltage drop is getting higher and higher, VQ3 b pole potential is getting lower and lower, when it is low to VQ3 turn-on voltage, VQ3 is turned on, VQ5 is turned on successively, C2 passes through VQ5 ce junction, power supply, VQ3 eb junction (because VQ2 is turned on, we assume that its ec junction is short-circuited, and the VQ3 e pole is directly connected to the positive pole of the power supply) to discharge.After discharging, VQ3 is turned off, VQ5 is turned off, and the power supply is charged to C2 again, then VQ3 is turned on, VQ5 is turned on, and C2 is discharged. The ce junction is connected to the ground, and the current is stored by L1. When VQ5 is turned off, L1 generates an induced electromotive force, which is superimposed with the power supply to drive the LED, and the LED emits light. It is possible to increase the battery voltage to directly drive the LED to improve the efficiency, but the battery voltage is increased, and the corresponding solar cell price is also greatly increased. As long as the circuit components are properly set, the efficiency is still acceptable. When the charging is not enough during the day (such as rainy days, etc.), BT2 may be over-discharged, which will damage the battery. For this reason, R5 is added to form over-discharge protection: when the battery voltage drops to 2V, due to the partial pressure of R5, VQ4 The base potential is not enough to turn on VQ4, thus protecting the battery. Increasing R5 affects the conduction depth of VQ4.