March 3, 2013 at 03:47LED standby lighting
Often, in different places, lighting of the type “if only not complete darkness” is required. For example, the staircase of an apartment building, where you do not need to look for needles on the floor, but only minimal light is enough to not stumble or be able to get into the keyhole with a key. Usually, in such cases, either a 20-40 watt “Ilyich bulb” or “economical” 7-9 Watt is screwed in. An incandescent lamp tends to burn out often, and “housekeepers” simply steal corny (I have stolen five pieces from my stairwell for 3 years, I screwed up a trifle, but it’s unpleasant) . If you need an economical and protected from theft (well, let's say, more protected than just a light bulb) light source, then read on.
For illumination, we will use powerful 0.5 Watt LEDs of 7.62mm caliber (yes, the same one) , they are bright enough and they don’t need an external additional radiator, unlike the more powerful fellow “stars” of 1-3 Watt. My design uses four-legged white LEDs of 7.62mm 100mA with a scattering angle of 140 °. We take the voltage drop on the LED ~ 3.3V. We will feed from the 230V network. According to Ohm's law, the value of the quenching resistance should be (230V-3.3 * 3) /0.1A=2200ohm. The power dissipated on it, respectively, will be more than 20 watts. A resistor, with such parameters, has a very impressive size and, moreover, will be very hot. We will go the other way and use a capacitor as resistance.
Classic quenching circuit.
It is known from the course of electrical engineering that a capacitor in an AC circuit has a reactance of Xc = 1 / (2πfC), where f is the frequency, C is the capacitance of the capacitor. To get the capacitor resistance in the region of 2200 Ohm at a frequency of 50 Hz, the capacitance should be C = 1 / (2 * 3.14 * 50 * 2200) = 0.0000014Farad. or ~ 1.4uF. This is a very rough calculation, where the presence of a rectifier bridge and a smoothing capacitor in the circuit is not taken into account. We will make a margin of safety, taking a current of 75% of the calculated one (the brightness of the LEDs will be enough, and their operation mode will become more sparing), and take a capacitor with a capacity of 1 μF. Brightness will be sufficient even at 0.68uF.
ATTENTION! As suppressors, I recommend using only special noise suppressing capacitors of class X2, for a voltage of at least 250 volts.Typically, such capacitors have a rectangular shape and a lot of all sorts of badges-certificates on the case. Using improper capacitors may cause a fire!
A 220 ohm resistor reduces the inrush current through the capacitor when turned on. Indeed, a discharged capacitor, at the time of switching on, has a very small resistance and, through the entire circuit, for a split second, a very large current flows. Additionally, to protect the LEDs from inrush currents at the time of switching on and during operation, an electrolytic capacitor and a powerful zener diode are included in the circuit.
Layout "on weight":
For the manufacture you will need:
ATTENTION! During operation of the device, all circuit elements are under voltage dangerous to life! Observe safety precautions and caution! Even when completely disconnected from the mains, the capacitor retains charge for a long time. Touching its findings can cause an unpleasant electric shock. In parallel with the quenching capacitor, you can connect a resistor with a resistance of 500KΩ - 1MΩ, it will discharge the capacitor when turned off.
This device illuminates my stairwell for three months and did not cause any complaints.
Materials used:
1) Biryukov S. Calculation of a power supply network with a quenching capacitor - Radio magazine for 1997, Nr. 5, p. 48 to 50.
For illumination, we will use powerful 0.5 Watt LEDs of 7.62mm caliber (yes, the same one) , they are bright enough and they don’t need an external additional radiator, unlike the more powerful fellow “stars” of 1-3 Watt. My design uses four-legged white LEDs of 7.62mm 100mA with a scattering angle of 140 °. We take the voltage drop on the LED ~ 3.3V. We will feed from the 230V network. According to Ohm's law, the value of the quenching resistance should be (230V-3.3 * 3) /0.1A=2200ohm. The power dissipated on it, respectively, will be more than 20 watts. A resistor, with such parameters, has a very impressive size and, moreover, will be very hot. We will go the other way and use a capacitor as resistance.
Classic quenching circuit.
It is known from the course of electrical engineering that a capacitor in an AC circuit has a reactance of Xc = 1 / (2πfC), where f is the frequency, C is the capacitance of the capacitor. To get the capacitor resistance in the region of 2200 Ohm at a frequency of 50 Hz, the capacitance should be C = 1 / (2 * 3.14 * 50 * 2200) = 0.0000014Farad. or ~ 1.4uF. This is a very rough calculation, where the presence of a rectifier bridge and a smoothing capacitor in the circuit is not taken into account. We will make a margin of safety, taking a current of 75% of the calculated one (the brightness of the LEDs will be enough, and their operation mode will become more sparing), and take a capacitor with a capacity of 1 μF. Brightness will be sufficient even at 0.68uF.
ATTENTION! As suppressors, I recommend using only special noise suppressing capacitors of class X2, for a voltage of at least 250 volts.Typically, such capacitors have a rectangular shape and a lot of all sorts of badges-certificates on the case. Using improper capacitors may cause a fire!
A 220 ohm resistor reduces the inrush current through the capacitor when turned on. Indeed, a discharged capacitor, at the time of switching on, has a very small resistance and, through the entire circuit, for a split second, a very large current flows. Additionally, to protect the LEDs from inrush currents at the time of switching on and during operation, an electrolytic capacitor and a powerful zener diode are included in the circuit.
Layout "on weight":
For the manufacture you will need:
- small distribution box (housing)
- 3 LEDs 0.5 Watt 100mA
- diode bridge for a voltage of at least 400V and a current of 1-2A)
- 5W Zener diode 14-15 Volts (such a margin will not hurt)
- 100μF electrolytic capacitor for 100V voltage
- capacitor (class X2) 0.68-1uF at a voltage of at least 250V
- 1-2 watts resistor to 150-200 ohms.
- 1-2 amp fuse (for every fireman)
- block (terminal block) on two contacts
ATTENTION! During operation of the device, all circuit elements are under voltage dangerous to life! Observe safety precautions and caution! Even when completely disconnected from the mains, the capacitor retains charge for a long time. Touching its findings can cause an unpleasant electric shock. In parallel with the quenching capacitor, you can connect a resistor with a resistance of 500KΩ - 1MΩ, it will discharge the capacitor when turned off.
This device illuminates my stairwell for three months and did not cause any complaints.
Materials used:
1) Biryukov S. Calculation of a power supply network with a quenching capacitor - Radio magazine for 1997, Nr. 5, p. 48 to 50.