A night light made of LED

LED has the advantages of high brightness, low power consumption, long life and small size. The night light made of LED is very suitable for home use.

1. Circuit and Principle

This circuit uses a capacitor step-down bridge rectifier circuit (see attached figure). The capacitive reactance generated by capacitor C1 in the AC circuit is used to achieve voltage reduction. Completely different from resistor step-down, the use of capacitor C in the AC circuit to step down the voltage does not consume active power on capacitor C1, which is beneficial to power saving. Another advantage of capacitor step-down is that in the case of low-power applications, a power transformer can be saved, which is beneficial to reducing the size of the electrical device and reducing the production cost. However, capacitor step-down also has its disadvantages. Its input and output cannot be isolated through coil windings like power transformers, so the output end is energized. Especially in the case of no-load or open-load conditions, the output voltage is often relatively high, and it must not be touched casually. Pay attention and take some insulation measures during production. Another disadvantage of capacitor step-down is that the internal resistance of the power supply is very high and the power supply quality is poor. In addition, due to manufacturing technology and cost reasons, there are currently fewer capacitors with large capacity, high withstand voltage, and small overall area. Therefore, the output current of capacitor step-down is limited, and it is not suitable for equipment with large power load changes and high requirements for power output accuracy.

In the figure below, C1 is a step-down capacitor, and R1 is a discharge resistor of Cl, which can be selected according to the size of C1 (generally between 200k and 1MΩ). C2 is a filter capacitor, which is mainly used to smooth the DC and absorb (filter) the instantaneous spike voltage of the power supply to prevent the LED from being broken down and eliminate the light flickering phenomenon. For this reason, C2 should be a small electrolytic capacitor ≥20uF/25V.

D5 is a 10V Zener diode. When the power supply voltage increases, the voltage at both ends of the load LED will also increase slightly. Because LED is a device with poor overload capacity, in order to make the night light really "long-lived", it must be "stabilized" to 10V with a Zener diode.

2. Component selection and debugging

When purchasing LEDs, you must ask about the actual operating voltage and current of the light-emitting diode, and observe whether the white light it emits is bright under the specified voltage and current (view from the top of the tube (do not linger on it, otherwise it will damage your eyes). Some tube parameters are very discrete, such as its forward and reverse resistance. The current value of the LED has a large deviation under the standard voltage. If you want to use this type of tube because of the low price [except for single use). And if multiple tubes are required to be connected in series, you must connect a "voltage-equalizing resistor" with an appropriate and equal value in parallel to each LED. A relatively close marking method is to mark C1. For this purpose, the effect of C2 on the rectified voltage in Figure 1 is ignored, and many complex factors under AC conditions are ignored. It is changed to DC operation. The initial value of C1 is first calculated and then corrected through experiments.

As shown in the attached figure: three LEDs are connected in series as the load (i.e. the light source of the night light). Since the working voltage VLED = 3.3V and the working current ILED = 218~20mA of each LED have been determined at the time of purchase, the lower limit is taken as 18mA for safety reasons. (Experiments show that the brightness of the LED is basically constant when the working current changes from 16~20mA). This requires that the capacitor step-down rectifier source must provide 10V working voltage (3.3V×3) and 18mA working current to the load. Cl is 0.27uF, with a withstand voltage of 400V.

Since there is no capacitor of this specification, we have to temporarily choose 0.33uF/400V as C1 for circuit experiment.

Solder all components according to the diagram, and check carefully before power-on debugging: ④ Connect a multimeter DB (i.e., milliampere gear, first large and then small) and a potentiometer of about RP=500Ω (adjust to the middle value first) in series at the position shown in the figure; ② Turn on the power switch to connect the AC power supply; ③ Observe the ammeter indication value and the LED lighting situation. If the ammeter indication value is only a few milliamperes at this time, the LED will certainly not light up. You can adjust RP to the direction of small resistance to make the ammeter indication value rise. (Conversely, turn it in the direction of large resistance) When RP is adjusted to the normal and stable lighting state of the LED and the ammeter indication value is 18mA, the debugging is over. Remove the ammeter and potentiometer, and use a multimeter to measure the resistance value of RP, and replace it with an equivalent resistor in series (as a current limiting resistor). When it is found that the resistance value of "RP" has been adjusted to "0" but the current still cannot go up, for example, only 12mA, it means that the capacitive reactance of C1 is too large, that is, the capacity of C1 is too small. It is necessary to add a capacitor in parallel appropriately (pay attention to the withstand voltage value). It can be estimated as 5mA for every increase or decrease of 0.1μF. Note that after the Cl capacity is corrected, for safety reasons, the potentiometer RP still needs to be connected in series for debugging according to the above method. Actual experiments show that when C1=0.33uF, even when "RP" is adjusted to "O", the indication value of the ammeter is only 18.6mA, which is relatively ideal, so there is no need to connect a current limiting resistor in series. At this time, the actual measured V4=9.3V. There is a difference of 0.7V with 10V. 0.7V can be regarded as a safe margin. Even when the input voltage of ~220V rises to ~230V, its output terminal V4 will not exceed 10V, and there is also a 10V voltage regulator "clamp" to check it at the end!

If the night light is calculated at 0.3W, it will only consume about one kilowatt-hour even if it is used 24 hours a day for five months. It fully meets the energy-saving requirements of "green" lighting currently advocated in my country.