Fig. 1  How to Wire LEDs in 110V or 220 Volts  6 Different Circuits! Formulas & Calculations! 
Today we will show you 6 different ways to wire 3mm or 5mm LEDs, which are low voltage DC components, into a 110V or 220V AC voltage grid!
We can use LEDs in several ways connected to 110V or 220V power grid, knowing that some types of connections bring advantages over others, and that each type has its characteristics that best fit each specification.
We will use some basic formulas to calculate the components in our circuit, for this we will use the capacitive reactance formula, and the Ohms Law formula.
So let's start by showing the basic formulas that we will use with the models we made in this post.
We'll apply the basic formulas as needed, so we'll start first by determining the supply voltage.
CAUTION!
The working voltage
Fig. 2  Peak to Peak 110Vac Calculation  VPP 
The calculation is determined by the mathematical equation:
 VP = VAC * √(2)
As our power grid is 110VAC:
 VP = 110 *1.414
 VP = 155.54VAC
If you use the 220VAC power grid:
 VP = 220V * 1.414
 VP = 311,08VAC
The LED
 We will use a white LED, which in its specifications is 3.2V for 20mA, or 0.02A.
Determine the resistor resistance:
 V = R * I
Determine the Resistor's Power:
 P = R * I²
Determine Capacitive Reactance:
 XC = 1 / (2 π * F * C)
Knowing all the formulas that we will use in our circuits, let's start with the simplest to the most complex.
1° Circuit:
This model is the simplest we have, and it is very often used in cheap electrical extensions of those Chinese products, and also as a Pilot lamp in equipment,...
The circuit presented has only one resistor R1, which limits the current that passes through the LED, and is connected in series with the LED, as we can see in Figure 3 below.

Designing the Circuit
General Formula:
 V = R * I
Applying the formula to our circuit:
 R = (VS  VL) / I
VL = Voltage of the LED, which is 3.2V
IL = The LED current, which is 0.02A
Then:
 R = (155,54  3.2) / 0.02
 R = 152,34 / 0.02
 R = 7.617R
 7.617Ω + 40% = 3.047Ω
 7.617Ω + 3.047Ω = 10.6638Ω or 10.66KΩ
 That is, the value of the closest commercial resistor, knowing that we always take the closest one with the highest value is 12KΩ.
We now need to determine the power of the resistor, for this we will use the ohms law formula:
General Formula:
 P = R * I²
Then:
 P = 12.000 * 0.02²
 P = 4.8W
Project Finished  Circuit 1
 LED1 ....... 3.2V / 20mA Light Emitting Diode
 R1 ............ 12K / 5W Resistor for 110V. (27K to 220V).
Advantages:
 It is a Simple and easy circuit to assemble
 Only 2 components
Disadvantages:
 Voltage dissipation will be on the resistor (Joule effect)
 Consumption higher than necessary
 Circuit operating in half wave, LED half off
 Short LED lifetime, reverse voltage on LED
 Low Efficiency
2° Circuit:
This model is still quite simple, it is a circuit widely used also in cheaper electrical extensions of those Chinese products...
The circuit presented has a resistor R1, which limits the current that passes through the LED, and a Diode, which polarizes the AC voltage coming from the power grid, which is connected in series with the LED, as we can see in Circuit 2.1 in Figure 4 below. .
We also have Circuit 2.2, which is the same circuit, but we have added a 2.2uF capacitor that serves to minimize the ripple voltage in the circuit.
Fig. 4  Wire LED in 110v or 220V Circuit 2  ELC 
Designing the Circuit
Just like the previous circuit, the calculations are the same, we already calculated the resistance, after all the process, it was 12K with 5W of power.
Project Finished  Circuit 2
Here we finish the development of our circuit 2, the calculated values are
 LED1 ....... Light Emitting Diode 3.2V / 20mA
 D1 ............ 1N4007 Diode
 C1 ............ 2.2uF / 25V Electrolytic Capacitor (Optional)
 R1 ............ 12K / 5W resistor for 110V. (27K at 220V).
Advantages:
 It is a simple and easy circuit to assemble
 Only 3 or 4 components
 Safer circuit for LED lifetime
Disadvantages:
 Voltage dissipation will be on the resistor (Joule effect)
 Consumption higher than necessary
 Circuit operating in half wave, LED half off
 Low Efficiency
3° Circuit
This model, different from the previous one, adopts a rectifier bridge, this implies that the energy that comes to the LED, is no longer a half wave, but a full wave, which gives more brightness to the LED.
The circuit presented has a resistor R1, current limiter, and a bridge of diodes, which polarize the AC voltage that comes from the grid, and feeds the LED, as we can see in Circuit 2.1 in Figure 5 below.
We also have Circuit 2.2, which is the same circuit, but we add a 2.2uF capacitor that serves to minimize the ripple voltage in the circuit.
Fig. 5  Wire LED in 110v or 220V Circuit 3  ELC 
Designing the Circuit
Just like the previous circuit, the calculations are the same, we already calculated the resistance, after all the process, it was 12K with 5W of power.
Project Finished  Circuit 3
Here we finish the development of our circuit 2, the calculated values are
 LED1 ....... Light Emitting Diode 3.2V / 20mA
 D1 ............ 4 x 1N4007 Diode, or a diode bridge any model
 C1 ............ 2.2uF / 25V Electrolytic Capacitor (Optional)
 R1 ............ 12K / 5W resistor for 110V. (27K at 220V).
Advantages:
 It is a simple circuit to assemble
 Only 3 or 4 components
 Full wave, which gives more brightness to the LED.
 Safer circuit for LED lifetime
Disadvantages:
 Voltage dissipation will be on the resistor (Joule effect)
 Consumption higher than necessary
 Low Efficiency
4° Circuit:
This model is simple, but works in a more efficient way, since the heat dissipation is no longer tied to the Current Limiting Resistor, which dissipated all the voltage in the previous circuits.
This circuit is widely used in Mosquito Bats, rechargeable flashlights, i.e. cheaper Chinese products.
In this circuit we replaced the current limiting resistor with a capacitor. When a capacitor is connected to an AC source, it allows current to flow in a circuit.
With the process of successive charge and discharge of a capacitor, it gives rise to a resistance, in the passage of current in the circuit, and this resistance is called capacitive reactance. With these properties, we can use the capacitor in our circuit as a resistor.
In the case of the capacitor, all of this energy is used, because the capacitor needs to charge and discharge, it "holds" the energy and therefore does not consume it, making the circuit much more efficient.
We also have Circuit 4.2, which is the same circuit, but we have added a 2.2uF capacitor that serves to minimize the ripple voltage in the circuit. The complete circuits are in Figure 6 below.
Fig. 6  Wire LED in 110v or 220V Circuit 4  ELC 
Designing the Circuit
General Formula:
 XC = (VS  VL) / IL
Applying the formula to our circuit:
 XC = (VS  VL) / I
 VS = Mains voltage, which is 110Vac
 VL = Voltage of the LED, which is 3.2V
 IL = The LED current, which is 0.02A
Then:
 XC = (110  3.2) / 0.02
 XC = 106.8 / 0.02
 XC = 5.340Ω or 5.3K
Since we have already found out the XC reactance which is 5.340Ω or 5.3KΩ, we can now calculate the current that this capacitor will supply to our circuit. We will use the same formula as in Ohms' Law:
General Formula:
 I = VS / XC
 VS = Main voltagem in RMS
 XC = Capacitive Reactance in Ω
Applying the formula to our circuit:
 I = (VS  VL) / XC
VL = Voltage of the LED, which is 3.2V
 I = (110  3.2) / 5.340
 I = (106.8) / 5.340
 I = 0.02A
 I = 20mA
 C = 1 / (2 π * F * XC)
Since capacitance is usually not expressed in Farad but in a submultiple, we will use the rewritten formula, so that we can use the capacitor value in µF and make our calculations easier.
Applying the formula to our circuit:
 C = 106 / (2 π * F * XC)
C = Capacitance that we need to know
π = Is a constant 3.14
XC = Capacitive Reactance, which is 5.340Ω
F = Main frequency, which is 60Hz
Then: C = 106 / (2 * 3.14 * 60 * 5.340)
 C = 106 / (6.28 * 60 * 5.340)
 C = 106 / (376.8* 5.340)
 C = 106 / (2,012.112)
 C = 0.4969uF or 497nF
That is, the value of the closest commercial capacitor, knowing that we always take the closest one with the highest value is 560nF.
Project Finished  Circuit 4
We finish here the development of our circuit 4, the calculated values we will have: LED1 ....... Light Emitting Diode 3.2V / 20mA
 D1 ........... 1N4007 Diode
 C1 ............ 560nF / 250V Polyester Capacitor
 C2 ............ 2.2uF / 25V Electrolytic Capacitor (Optional)
 LED1 ....... Light Emitting Diode 3.2V / 20mA
 D1 ........... 1N4007 Diode
 C1 ............ 560nF / 250V Polyester Capacitor
 C2 ............ 2.2uF / 25V Electrolytic Capacitor (Optional)
Advantages:
 No consumption of excess heat energy (Joule effect)
 It is simple circuit to assemble
 Only 3 or 4 components
 High Efficiency
 Safer circuit for LED lifetime
Disadvantages:
 Circuit operating in half wave, LED half off
 High current in the initial steady state of the capacitor, causing those "pops" and sparks in the socket.
5° Circuit:
This model is a more complete and improved circuit, because it brings with it a diode bridge, improving efficiency even more, since the LED will no longer work for half a wave period, but for a full wave period.
This circuit is widely used in small luminaires, even in LED lamps, rechargeable flashlights, or in commercial products.
This circuit is the junction of circuits 3 and 4, thus forming an efficient circuit, with good LED brightness, with full wave, it is almost the perfect circuit.
The circuit presented has a diode bridge, which polarizes the AC voltage coming from the mains, which is connected in series with the LED, as we can see in Circuit 5.1 in Figure 7 below.
We also have Circuit 5.2, which is the same circuit, but we add a 2.2uF capacitor that serves to minimize the ripple voltage in the circuit.
Fig. 7  Wire LED in 110v or 220V Circuit 5  ELC 
Designing the Circuit
First you need to determine the capacitive reactance, which was already done in the previous circuit, the capacitance is 5.340Ω or 5.3K.
Project Finished  Circuit 5
We finish here the development of our circuit 5, the calculated values we will have: LED1 ....... Light Emitting Diode 3.2V / 20mA
 D1 ........... 4 x 1N4007 Diode, or a diode bridge, any model
 C1 ........... 560nF / 250V Polyester Capacitor
 C2 ........... 2.2uF / 25V Electrolytic Capacitor (Optional)
 LED1 ....... Light Emitting Diode 3.2V / 20mA
 D1 ........... 4 x 1N4007 Diode, or a diode bridge, any model
 C1 ........... 560nF / 250V Polyester Capacitor
 C2 ........... 2.2uF / 25V Electrolytic Capacitor (Optional)
Advantages:
 It is a simple circuit to assemble
 Only 3 or 4 components
 Safer circuit for LED lifetime
 No consumption of excess heat energy (Joule effect)
 High Efficiency
 Circuit operating in full wave, LED always on
Disadvantages:
 High current in the initial steady state of the capacitor, causing those "pops" and sparks in the socket.
6° Circuit:
This model is more complete and, like the previous circuit, is more efficient. This circuit is widely used in small light fixtures, some LED lamps, rechargeable flashlights, and in some commercial products.
The circuit presented is identical to circuit 5, with the only difference that we put a resistor R1, that serve to limit the capacitor inrush current. A diode bridge, which polarizes the AC voltage coming from the mains, which is connected in series with the LED, as we can see in Circuit 6.1 in Figure 8 below.
We also have Circuit 6.2, which is the same circuit, but we have added a 2.2uF capacitor that serves to minimize the ripple voltage in the circuit.

Designing the Circuit
First you need to determine the resistance R1, the resistor value was chosen to limit the worst case inrush current to about 100mA which for safety, will be 5 times the current draw of the circuit, which will drop to less than 20mA in a millisecond as the capacitor charges.
In this case, we use ohms' law to figure out what resistor we will use.
General Formula:
 V = R * I
Applying the formula to our circuit:
 R = (VS  VL) / I
 VS = Mains voltage, which is 110Vac
 VL = Voltage of the LED, which is 3.2V
 IL = Inrush current, which is 0.1A or 100mA
Then:
 R = (110  3.2) / 0.100
 R = 106 / 0.100
 R = 1,068Ω or ~1KΩ
Now, we need to determine the power of the resistor, for this we will use the ohms law formula:
General Formula:
 P = R * I²
Then:
 P = 1,068 * 0.02²
 P = 0.427W
That is, the value of the closest commercial power resistor, knowing that we always take the closest one with the highest value is 1/2W.
Project Finished  Circuit 6
We finish here the development of our circuit 5, the calculated values we will have: LED1 ....... Light Emitting Diode 3.2V / 20mA
 D1 ........... 4 x 1N4007 Diode, or a diode bridge, any model
 R1 ........... 1KΩ / 1/2W Resistor
 C1 ........... 560nF / 250V Polyester Capacitor
 C2 ............ 2.2uF / 25V Electrolytic Capacitor (Optional)
 LED1 ....... Light Emitting Diode 3.2V / 20mA
 D1 ........... 4 x 1N4007 Diode, or a diode bridge, any model
 R1 ........... 1KΩ / 1/2W Resistor
 C1 ........... 560nF / 250V Polyester Capacitor
 C2 ............ 2.2uF / 25V Electrolytic Capacitor (Optional)
Advantages:
 It is a simple circuit to assemble
 Only 3 or 4 components
 Safer circuit for LED lifetime
 No consumption of excess heat energy (Joule effect)
 High Efficiency
 Circuit operating in full wave, LED always on
Disadvantages:
 Can be better, by putting a resistor in parallel as a capacitor, to discharge it...
If you have any questions, suggestions or corrections, please leave them in the comments and we will answer them soon.
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