Sunday, September 25, 2016

Small LED Powered By 220 Volts AC Without Transformer



The following video shows that small LED with a maximum voltage of about 3 vdc, it is supplied with a high voltage of 220 vac with a very simple circuit without transformer. One 1N4007 diode is used as a rectifier. One 330 kiloohms resistor is used as current limiter.



The following schematic has circuit components which are connected in series.


CAUTION: all parts of the circuits contained in this article are connected to high voltage, either directly or indirectly. Do not touch circuit while being connected to the high voltage. Use equipment that are well insulated.

As the above schematic:
VD is the voltage on rectifier diode 1N4007
VL is the voltage on LED
VR is the voltage on resistor 330 kiloohms
VAC is a supply voltage of 220 vac


When current is flowing in 1N4007 diode, it will be a voltage of 0.6 volts (VD) between the anode and the cathode pins. Diode is needed because LED cannot stand high reverse voltage. Usually the maximum allowable reverse voltage is only 5 volts, if more than that will immediately damage the LED. Shown in the video, LED is blinking at a frequency of 50-60 hertz. This happens because the diode only allows half wave of alternating electric current. If you want to reduce LED blinking, you need to install a capacitor to stabilize the current, and use a diode bridge as a full wave rectifier.

LED that is used in video has diameter of 5 mm. This LED has a maximum forward voltage of about 3 vdc. There is also an LED that has only 2.2 vdc maximum voltage. Maximum forward current is usually around 30 mA. Therefore, the internal resistance of the LED can be calculated by using Ohm's law, with a maximum forward voltage of 3 volts:


LED resistance = Voltage / Current = 3 volts : 30 mA = 3 volts : (30 / 1000) amperes = 100 ohms


In the above schematic, VAC is 220 volts grid voltage, so the voltage between resistor pins is:

Resistor voltage = supply voltage - diode voltage - LED voltage = VR


VR = VAC - VD - VL =  220 - 0.6 - 3 = 216.4 volts


It is known that the maximum LED current is 30 mA. Because it is assembled in series, then the current is also flow through the resistor with an equal amperage. Therefore the resistor value can be calculated by Ohm's law:

Resistance = resistor voltage / resistor current = 216.4 volts / 30 mA = 216.4 volts : (30 / 1000) amperes = 7,213.33 ohms.


The above resistor value is the smallest value for maximum LED brightness, if the resistance is reduced then LED could be broken because over voltage.

In the video, the resistor has much higher value for safety, and to avoid LED light blinds the camera. With 330 kiloohms resistor, white LED is still very bright. If LED brightness is too low, the value of  resistor can be reduced from 330 kiloohms carefully.

In choosing resistor size, it is necessary to consider the power capacity of the resistor. For a 330 kiloohm resistor, the power calculation is:


Resistor power = resistor voltage x current = resistor voltage x resistor voltage / resistance

Resistor power = VR x VR / 330.000 = 216.4 x 216.4 / 330.000 =  0.14 watts


Since the current is intermittent or half-wave current, the actual power of the resistor is half of the power above, ie 0.07 watts. So, resistors with 1/4 or 1/2 watts power are widely available in the market, those resistors can be used safely.

Note that LED tends to keep the voltage values ​​at the anode and cathode remains stable. So the internal resistance of the LED will change according to the current through it. The  bigger the current, the lower the internal resistance of the LED. The smaller the current, the higher the internal resistance of the LED.

As seen in the video, although the current through the LED is very small about 0.67 mA, but the voltage on the LED pins still reach 1.06 volts. The voltage is measured by DC voltmeter, not AC voltmeter, because the current through the LED is direct current. Then the internal resistance of LED with such a small current is:


Resistance = 1,06 volts / 0,67 mA = 1,06 volts / 0,00067 amperes = 1.582,09 ohms

The actual value of the resistance and the voltage of the LED in the test in the video are higher from the value above, because only half the wave of alternating voltage is used to turn on the LED.

Theoretically, if there are many LEDs connected in series, then resistor is no longer needed to limit current. Example calculation: it is known LED with a maximum voltage of 3 volts DC, for 220 volt supply voltage will need LED as much as 220/3 = 73.3 pieces, which are connected in series. Of course that amount is the minimum amount, for safety reason then it should be more LEDs that are connected in series. And the rectifier diode is still needed for safety, only one (1x) piece.


The following schematic describes the circuit with a diode connected in parallel with the LED.


In the above circuit, the LED will light up when the current flows from left to right. LED will be off when current flows from right to left through diode, not through LED. The disadvantage of this circuit is that there is a power consumption when current flows from right to left, although the LED is not on.

The VL voltage will only occur when the LED is on. And the VD voltage will only occur when the LED is off. Since the current is alternating current, it will be difficult to measure the LED voltage (VL) accurately. The voltage on the resistor (VR) is alternating voltage.



In the above circuit, the resistor is replaced by a capacitor. When alternating current is applied, the capacitor will work like a resistor. The capacitor resistance is called reactance, because the resistance to the capacitor is the reaction of the capacitor to the alternating current. In order to obtain the correct reactance value, then the size of the capacitor can be calculated by the formula:


Capacitor reactance = Xc = 1 / (2 . Pi . f . C )


If the reactance of the capacitor is tuned equal to the resistance as in the previous circuit, ie 330 kilo ohm, and the frequency (f) of alternating voltage is 50 hertz. Then the size of the capacitor is:


C = 1 / (2 . Pi . f . Xc) = 1 / (2 x 22/7 x 50 x 330.000) = 9.65 x 10 ^ (-9) = 9.65 nanofarads


Or it can be rounded up to 10 nanofarads. But the problem is that this capacitor must be withstand to high voltage, reaching at least 250 volts. And this capacitor must also be of the type of capacitor for alternating current (AC), which is usually more expensive than direct current (DC) capacitors.

Theoretically the capacitor in this circuit will not generate heat. Because the capacitor discharges the electrical energy it holds, not in the form of heat energy. So there is no need for power calculations. But in practice the dielectric material inside the capacitor can generate heat, as the effect of the voltage changes. This often happens at high frequencies. Then we need a capacitor with a good dielectric, and of course the price is more expensive.

If the LED light is not bright enough, then the capacitor value is raised carefully. The nature of the capacitor with the alternating current is the inverse of the resistor. On the resistor, the higher the value the greater the resistant. In the capacitor, the higher the value the smaller the resistance to the alternating current, as can be calculated on the above reactance formula.

Video of testing LED with capacitor to connect to 220 VAC.



Sometimes LED is very sensitive and can not withstand reverse voltage. For that reason we need to add a rectifier diode, as the following scheme.



Seen in the above circuit, a rectifier diode is added in series to LED. So the LED is completely shielded from reverse voltage.

Due to the above considerations, of all suggested circuits, the topmost circuit is the best and recorded in video.








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