Design and production of ultra-bright LED with one battery

Publisher:平安幸福Latest update time:2011-12-24 Source: 电子发烧友网Keywords:Battery  LED Reading articles on mobile phones Scan QR code
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Design and production of ultra-bright LED with one battery

1. Circuit Design
The voltage of a nickel-metal hydride battery is only 1.2V, while the ultra-bright LED requires a working voltage of more than 3.3V to ensure sufficient brightness. Therefore, it is necessary to find a way to increase the voltage. Common boost circuits generally have two forms, namely high-frequency oscillation circuits and electromagnetic induction boost circuits. For boost circuits, there are two circuits to choose from. As shown in Figures 1 and 2.
Light up a super bright LED with one battery
The circuit in Figure 1 uses a small pulse transformer. The power tube VT3 amplifies the high-frequency oscillation signal and adds it to L1 to directly boost the voltage through the transformer T.
Figure 2 uses the self-inductance high voltage of the inductor to achieve voltage boost. When the oscillation signal is input to the base of VT3, VT3 will periodically saturate and cut off. When saturated, the inductor L is energized, and the electrical energy is converted into magnetic energy and stored in L. At this time, the diode is cut off, and the energy stored in C3 is used to supply power to the load; when VT3 is cut off. The inductor will generate a self-inductance electromotive force with a positive bottom and a negative top. The diode VD is turned on, and the self-inductance electromotive force is superimposed on the power supply electromotive force to charge the capacitor C3 and supply power to the load. Because the two electromotive forces are in series. A voltage higher than the power supply can be obtained, and the specific size is mainly determined by the ratio of the load and the current passing through the inductor L when VT3 is saturated.
Both circuits can increase 1.2V to more than 3.3V. If the first circuit is wound with a positive feedback coil on the transformer, the oscillation circuit can be eliminated, making the circuit more concise. However, the calculation using this circuit is more complicated. The output power is difficult to adjust, and the winding of the transformer is also somewhat troublesome. The second type only requires a small inductor. There is no large requirement for the inductance. By adjusting the driving current of the inductor, the output voltage can be easily adjusted. The second circuit is used here. The oscillation circuit
uses the circuit shown in Figure 3. Although there are many oscillation circuits that can work normally at a voltage of 1.2V, it has been proved in practice that the circuit in Figure 3 is easy to make and simple to calculate. The success rate is high. The oscillation frequency is also easy to determine. Moreover, by adjusting the size of R4, the amplitude of the signal can be adjusted without affecting the signal frequency. Therefore, this circuit is used to generate a high-frequency square wave pulse to prepare for the boost circuit. In this way, the circuit design is completed, which is composed of Figures 2 and 3.
Light up a super bright LED with one battery

2. Calculation parameters
Regarding the calculation of circuit parameters, the key lies in power. After the inductor is energized, the stored electric energy is E=LI2/2. Let f be the frequency of the square wave. The switch tube will be turned on f times within 1a. In this way, the electric energy stored in the inductor per second is W=f×E. Let the efficiency of converting this energy to the load be η, then the output power is P=η×W+Po, Po is the power directly supplied by the power supply to the load (because the power supply and the self-inductance high voltage are superimposed. This must be considered).
Now make an estimate. It takes about 100mW to drive an LED. The Po of the power supply is about 20mW. In order to ensure the supply, calculate according to P=100mW. Take η=80%, and then find an inductor of several hundred uH, such as 500 uH: On the other hand, according to the law of conservation of energy. 3.3V is about 3 times of 1.2V. Due to efficiency issues. The driving current of the inductor is about 3-4 times the working current of the LED, so it is taken as 120mA. It can be calculated that the oscillation frequency is about 34kHz, so R=2kΩ, C=0.01 uF can meet the requirements. When determining the parameters, the frequency can be high rather than low, and the inductance is better to be large rather than small, so as to ensure that the output power is large enough and there is enough adjustment space.

Components table
Components model illustrate
R1 0.1-2 kΩ 500Ω is appropriate. A larger value is conducive to the saturation of VT1. A smaller value is conducive to the symmetry of the circuit.
R2 2 kΩ
R3 2 kΩ
R4 100 Ω The specific value depends on the drive current
C1 0.01 uF
C2 0.01 uF
C3 10-100uF electrolytic capacitor
VT1
VT2
9014 Any NPN tube with β≥ 40 can be used. It is best if the two tubes have the same model.
VT3 8050 Pcm ≥1W. Icm ≥1A. Medium power NPN tubes with β ≥40 are all acceptable.
Inductance L A few hundred uH is better to be big than small
Diode VD 1N5819 Germanium tube is the best. It has a small voltage drop. If it is not available, silicon tube is also OK.
Battery No.7

3. Production
Since the circuit is simple, the components are on a 2×2cm board. As long as the operation is correct, the circuit can work after the power is turned on. Do not connect the LED first, use a multimeter to measure the output voltage, at this time, adjust the size of R4, the larger the R4, the smaller the output voltage. Vice versa, when the output voltage is around 3.2V, you can connect the LED, and then adjust the size of R4 to make it bright enough. Note that the voltage across the LED should not exceed 3.6V, otherwise it may burn the LED. In this way, the circuit is debugged.

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