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Touch Dimmer Switch for Lamps: 4-Step 110/220VAC Control Circuit – With PCB

Electronic Circuits — Mon, 28 Mar 2022 21:42:00 +0000

110/220VAC 4 Step Control Touch Dimmer Switch Lamp Circuit with PCB

Enhance Your Lighting: Build a 110/220VAC 4-Step Touch Dimmer Switch Circuit for Lamps with PCB Control

This type of circuit is widely used in lampshades that are sold in residential lighting stores, it activates an incandescent bulb through touch on the device’s casing.

It works as a dimmer with 4 pre-set levels, activated by touching a part of the sensor with your finger, which can be a metallic point, or a metallic housing, etc. The entire 4-level brightness control circuit is based on the LS7237 IC.

🧷 LS7237 IC Description

LS7237 is an 8-Pin monolithic, MOS integrated circuit designed to control the brightness of an incandescent lamp, as shown in Figure 2 above. The output of LS7237 triggers a Triac connected in series with a lamp.

Fig. 2 – Pinout LS7237

The lamp brightness is determined by controlling the output conduction angle (Triac triggering angle) in relation to the AC line frequency.

⚠️ Caution!⚠️

We are not responsible for any type of event. If you do not have enough experience to assemble this circuit, do not do it, and if you do, when testing, be sure to have the proper protections and be accompanied by someone else.

🤔 How the Circuit Works

The circuit works as follows: when a touch is made to the board, it causes the brightness of the lamp to change in specified steps as follows:

LEVEL – – BRIGHTNESS(% Rated Wattage)
Off ……………….. 0
Night Light ……. 9
Mood Light …… 29
Medium ………… 66
Maximum ……… 99

After AC power-up, the output comes up in the OFF state. Following that, every time the Touch Plate is touched, the output steps to the next level of brightness. The next step following the maximum brightness is the OFF state, initiating a new sequence.

🛠️ Features

PLL synchronization allows use as a Wall Switch
Provides brightness control of an incandescent lamp with a touch plate or mechanical switch
Can control speed of shaded pole and universal AC motors
Controls the “duty cycle” from 23% to 88% (conduction angles for AC half-cycles between 45˚ and 158˚, respectively)
Operates at 50Hz/60Hz line frequency
Extension input for remote activation
+12V to +18V DC Power Supply (VSS – VDD)
8-Pin Plastic DIP, 8-Pin SOIC

🔌 The Schematic Circuit

The 110/220VAC 4 Step Control Touch Dimmer Switch Lamp Circuit diagram is shown in Figure 3 below.

Is a moderately simple circuit to assemble, with few external components, however one must have at least basic to advanced experience to assemble this circuit, if you are not experienced enough, ask someone more experienced to help you.

Fig. 3 – 110/220VAC 4 Step Control Touch Dimmer Switch Lamp Circuit

⚡ 110Vac or 220Vac

This circuit was designed to work with 220Vac voltage, for use in 110Vac power network, following components must be replaced in component list.

R1, R2 …. 2.7MΩ 1/4W (red, violet, red, gold)
R6 ……….. 270Ω 1W (red, violet, brown, gold)
C5 ……….. 330nF / 400V polyester Capacitor
L1 ………. 60µH (RFI Filter)

🧮 Components List

Semiconductors
- U1 ………… LS7237 Integrated Circuit
- Q1 ………… BT136 Triac Transistor
- D1 ………… 1N4007 Diode
- DZ1 ……… 1N4744 15V/1W Zener Diode
Resistors
- R1, R2 …. 4.7MΩ 1/4W (yellow, violet, green, gold)
- R3 ……….. 1.8MΩ 1/4W (brown, gray, green, gold)
- R4 ……….. 1.5MΩ 1/4W (brown, yellow, green, gold)
- R5 ……….. 100Ω 1/4W (brown, black, brown, gold)
- R6 ……….. 1KΩ 2W (brown, black, red, gold)
Capacitors
- C1 ……….. 680pF / 400V Polyester Capacitor
- C2 ………… 47nF / 400V Polyester Capacitor
- C3 ……….. 470pF/ 400V Ceramic Capacitor
- C4 ……….. 47µF / 25V Electrolytic Capacitor
- C5 ……….. 220nF / 400V Polyester Capacitor
- C6 ……….. 150nF /400V polyester Capacitor
Miscellaneous
- L1 ………. 120µH (RFI Filter)
- P1, P2 …. 2-pin PCB soldering terminal blocks
- F1 ………. 3A/250V Soldering fuse
- Others …. PCB, heat sink, power supply, wires, etc.

🖨️ Printed Circuit Board – Download

We are offering PCB – Printed Circuit Board, in GERBER, PDF and PNG files, for you who want to do most optimized assembly, either at home.

If you prefer in a company that develops board, you can be downloading and make files in Download option below.

Fig. 4 – PCB 110/220VAC 4 Step Control Touch Dimmer Switch Lamp Circuit

📥 Files to Download, Direct Link:

Click Here to Download Files

✨ Our Gratitude and Next Steps

We sincerely hope this guide has been useful and enriching for your projects! Thank you for dedicating your time to this content.

Your Feedback is Invaluable:

Have any questions, suggestions, or corrections? Feel free to share them in the comments below! Your contribution helps us refine this content for the entire ElCircuits community.

If you found this guide helpful, spread the knowledge!

🔗 Share This Guide

Best regards,
The ElCircuits Team ⚡

O post Touch Dimmer Switch for Lamps: 4-Step 110/220VAC Control Circuit – With PCB apareceu primeiro em Electronic Circuits.

How to Wire LEDs for 110V or 220V: 6 Practical Circuits with Formulas

Electronic Circuits — Sat, 26 Mar 2022 14:22:00 +0000

Fig. 1 – How to Wire LEDs in 110V or 220 Volts – 6 Different Circuits! Formulas & Calculations!

Illuminating Insights: Wiring LEDs for 110V or 220V – Explore 6 Distinct Circuits with Formulas and 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! ⚠️

As simple as the circuits presented are, it is important to know that the circuit is connected to direct mains voltage, this is extremely dangerous, an oversight or design error, can cause irreversible damage.

Be cautious when handling electrical voltage, if you have no electronics/electrical experience, do not do this circuit.

If you are experienced, do it with caution, and always have someone nearby, do not handle mains-connected equipment when you are alone.

We are not responsible for any damage that happens to you or others.

⚙️ The working voltage

In our country the working voltage is 110VAC, if your electrical network is 220VAC, just substitute the working voltage of your region in the formula.

It is necessary to know that our grid voltage has peak voltages, as showing in Figure 2 below, and for our safety we will use the peak to peak voltage (VPP) in our calculations.

Fig. 2 – Peak to Peak 110Vac Calculation – VPP

🧮 The calculation is determined by the mathematical equation:

V_P = V_AC * √(2)

As our power grid is 110V_AC:

V_P = 110 * 1.414
V_P = 155.54V_AC

If you use the 220V_AC power grid:

V_P = 220V * 1.414
V_P = 311.08V_AC

💡 The LED

We will use a white LED, which in its specifications is 3.2V for 20mA, or 0.02A.

1️⃣ Determine the resistor resistance:

To determine the resistor resistance for the circuit, using the Ohms Law formula:

V = R * I

V = Voltage

R = Resistance

I = Current

2️⃣ Determine the Resistor’s Power:

And to determine the resistor’s power we will also use Ohms’ Law:

P = R * I²

P = Resistor Power

R= Resistor Value

I = Current passing through the resistor.

3️⃣ Determine Capacitive Reactance:

Capacitive reactance is the opposition that a capacitor presents to the flow of current in AC circuits.

Capacitive reactance is represented by the notation Xc, and is expressed in ohms. To determine the capacitive reactance Xc, we will use the equation:

X_C = 1 / (2 π * F * C)

X_C = Capacitive Reactance in Ω

π = 3.14 – Constant

F = AC Frequency in Hz

C = Capacitance in F

🧐 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.

Fig. 3 – Wire LED in 110v or 220V Circuit 1 – ELC

Designing the Circuit

We need to determine the resistance to be used, for this we will use the ohms law formula:

General Formula:

V = R * I

Applying the formula to our circuit:

R = (V_S – V_L) / I

V_S = Peak mains voltage, which is 155.54Vac
V_L = Voltage of the LED, which is 3.2V
I_L = The LED current, which is 0.02A

Then:

R = (155.54 – 3.2) / 0.02
R = 152.34 / 0.02
R = 7.617Ω

As we know, when it comes to electronic components, there is the tolerance of the components that make up the circuit, such as the tolerance of the resistor, the LED, and the variations “tolerance” coming from the Power grid.

For this reason, we give a tolerance margin of more or less 40% more in the load resistor, that is:

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

We finish here the development of our circuit 1, the calculated values we will have:

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

We need to determine the capacitive reactance to be used, for this we will use the Ohms’ Law formula, it is exactly the formula used to figure out the resistance of R1 in the previous circuits.

Remember: The V and I values are effective, so we will use the RMS voltage, not the V_PP voltage.

General Formula:

X_C = (V_S – V_L) / I_L

Applying the formula to our circuit:

X_C = (V_S – V_L) / I
V_S = Mains voltage, which is 110Vac
V_L = Voltage of the LED, which is 3.2V
I_L = The LED current, which is 0.02A

Then:

X_C = (110 – 3.2) / 0.02
X_C = 106.8 / 0.02
X_C = 5.340Ω or 5.3K

Since we have already found out the X_C 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 = V_S / X_C
V_S = Main voltagem in RMS
X_C = Capacitive Reactance in Ω

Applying the formula to our circuit:

I = (V_S – V_L) / X_C

V_S = Mains voltage RMS, which is 110Vac
V_L = Voltage of the LED, which is 3.2V

X_C = Capacitive Reactance, which is 5.340Ω

Then:

I = (110 – 3.2) / 5.340
I = (106.8) / 5.340
I = 0.02A
I = 20mA

Knowing the resistance X_C and current I values in the circuit, we need to determine the capacitance of the capacitor. We will do this as follows below:

General Formula:

C = 1 / (2 π * F * X_C)

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 * X_C)

C = Capacitance that we need to know
π = Is a constant 3.14
X_C = 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)

✔️ 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)

✔️ 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.

Fig. 8 – Wire LED in 110v or 220V Circuit 6 – ELC

🔌 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 = (V_S – V_L) / I
V_S = Mains voltage, which is 110Vac
V_L = Voltage of the LED, which is 3.2V
I_L = 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)

✔️ 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…

✨ Our Gratitude and Next Steps

We sincerely hope this guide has been useful and enriching for your projects! Thank you for dedicating your time to this content.

Your Feedback is Invaluable:

Have any questions, suggestions, or corrections? Feel free to share them in the comments below! Your contribution helps us refine this content for the entire ElCircuits community.

If you found this guide helpful, spread the knowledge!

🔗 Share This Guide

Best regards,
The ElCircuits Team ⚡

O post How to Wire LEDs for 110V or 220V: 6 Practical Circuits with Formulas apareceu primeiro em Electronic Circuits.

25A Solid State Relay (SSR) Circuit for 110/220V with BTA24-600 + PCB

Electronic Circuits — Mon, 28 Feb 2022 23:07:00 +0000

25A – 110/220V Solid State Relay (SSR) Circuits using Triac BTA24-600 With PCB

Efficient Power Control: Building 25A – 110/220V Solid State Relay (SSR) Circuits with Triac BTA24-600 and PCB

Solid State Relays (SSR) are nothing new, but making your own SSR is priceless! With it we can connect to digital control devices such as; Arduino, PIC, ESP, CLP, Raspberry, etc.

Besides that, we can spend a fraction of the price of a commercial relay if we make our own SSR relay, and still have in our hands control that if something goes wrong, we can repair it without too much trouble.

🧲 The Solid State Relay

A Solid State Relay is similar to switching relays, they all function as a switch that is controlled by an input voltage or current, isolated from the output.

The basic difference is that Solid State Relay, or (SSR), has no moving parts, but uses the electrical and optical properties of solid-state semiconductors.

Electromechanical Relays (EMR), on the other hand, use coils, magnetic fields, springs, and mechanical contacts to operate and switch through a supply voltage.

Our circuit uses few external components, and is easy to assemble, the components are easy to acquire in the market, and basically there are two:

The isolation circuit formed by the MOC3041optical isolator.
The power control formed by the BA24-600 TRIAC, they will be explained just below.

⚠️ CAUTION!!! ⚠️

This circuit works directly connected to the 110/220V electrical network, and has a high power load, any carelessness, or wrong connections, error in the project, or any other occasion, can lead to irreversible damage. We are not responsible for any type of event. If you do not have enough experience to assemble this circuit, do not do it, and if you do, when testing, be sure to have the proper protections and be accompanied by someone else.

📚 BTA24 Description

Available either in through-hole or surface-mount packages, the BTA24 TRIAC is suitable for general purpose AC switching. They can be used as an ON/OFF function in applications such as static relays, heating regulation, induction motor starting circuits… or for phase control operation in light dimmers, motor speed controllers. The snubber-less versions (BTA/BTB…W and T25 series) are specially recommended for use on inductive loads, thanks to their high commutation performances. By using an internal ceramic pad, the BTA series provides voltage insulated tab (rated at 2500V RMS) complying with UL standards.

🧷 6-Pin DIP Zero-Cross Opto-isolators Triac Driver Output

(400 Volts Peak)

The MOC3041, MOC3042 and MOC3043 devices consist of gallium arsenide infrared emitting diodes optically coupled to a monolithic silicon detector performing the function of a Zero Voltage Crossing bilateral triac driver. They are designed for use with a triac in the interface of logic systems to equipment powered from 115/220 Vac lines, such as solid–state relays, industrial controls, motors, solenoids and consumer appliances, etc.

Simplifies Logic Control of 115 Vac Power
Zero Voltage Crossing
dv/dt of 2000 V/μs Typical, 1000 V/μs Guaranteed

We are using the MOC3041 optical isolator, because of this model, having a zero-crossing SSR accepts triggering at any time, but delays the triggering of AC load until the next time the AC voltage passes zero volts. This is useful to eliminate RFI (Radio Frequency Interference) and to prevent a large current from flowing to the load almost instantly.

🔌 Solid State Relay Circuit

The 25A – 110/220V Solid State Relay (SSR) Circuits using Triac BTA24-600 circuit diagram is shown in Figure 2 below. It uses a 25 Amp TRIAC BTA24-600, this is enough to handle loads up to a little over 5.500W, obviously using a isolated heat sink.

Fig. 2 – 25A – 110/220V Solid State Relay (SSR) Circuits using Triac BTA24-600

💡 Basic Components Function

Diode D1, is used for reverse voltage protection, it inhibits reverse voltage.
Resistor R1 of 240 ohm, limits the input current to the internal LED of the MOC.
Resistor R2 of 330 ohms for 1W, it serves to limit the supply current to the MOC‘s internal DIAC.
Resistor R3 of 56 ohm, prevents any di/dt current when the TRIAC is off, eliminating false triggering.

🧮 Components List

Semiconductors
- Q1 …. BTA24-600 Triac
- U1 …. MOC 3041 opto-isolator
- D1 …. 1N4007 Diode
Resistor
- R1 …. 240Ω (red, yellow, brown, gold)
- R2 …. 330Ω (orange, orange, brown, gold)
- R3 …. 56Ω (green, blue, black, gold)
Miscellaneous
- P1, P2 …. 2-pin PCB soldering terminal blocks (Optional)
- Others …. PCB, heat sink, wires, etc.

🖨️ Printed Circuit Board – Download

We are offering the PCB – Printed Circuit Board, in GERBER, PDF and PNG files, for you who want to do the most optimized assembly, either at home. If you prefer in a company that develops the board, you can download the files in the Download option below.

Fig. 3 – PCB – 25A – 110-220V Solid State Relay (SSR) Circuits using Triac BTA24-600

📥 Files to Download, Direct Link:

Click Here to Download the Files

✨ Our Gratitude and Next Steps

We sincerely hope this guide has been useful and enriching for your projects! Thank you for dedicating your time to this content.

Your Feedback is Invaluable:

Have any questions, suggestions, or corrections? Feel free to share them in the comments below! Your contribution helps us refine this content for the entire ElCircuits community.

If you found this guide helpful, spread the knowledge!

🔗 Share This Guide

Best regards, The ElCircuits Team ⚡

O post 25A Solid State Relay (SSR) Circuit for 110/220V with BTA24-600 + PCB apareceu primeiro em Electronic Circuits.

3500W AC Dimmer for 110V/220V with TRIAC TIC246 + PCB

Electronic Circuits — Fri, 25 Feb 2022 16:45:00 +0000

3500W Dimmer for 110V or 220V using TRIAC TIC246M with PCB

This is a dimmable load controller circuit, its operation is based on control of the sine cycle, keeping it off during a small period of wave. And work only for a specific part of wave keeping the load with part of wave controlled by period, similar to PWM.

With this type of circuit we can control the intensity of an incandescent light, ceiling fans, resistive load, among others, through decay of the cycle regulated through a potentiometer.

The Dimmer Circuit

The 3500W Dimmer for 110V or 220V using TRIAC TIC246M with PCB circuit diagram is shown in Figure 2 below.

It uses a 16 Amp TRIAC TIC246, this is enough to handle loads up to a little over 3500W, obviously with heat sink.

Fig. 2 – Schematic Circuit 3500W Dimmer for 110V or 220V using TRIAC TIC246M

If you need to increase the power of the circuit, you can replace the thyristor in the circuit. Using the simple formula of Ohms Law, P = V * I, through current and voltage, we can find the power of the circuit with each of the Thyristors:

TIC246 = 16A:
- At 110V => P = 110 * 16 = 1760W
- At 220V = P = 220 * 16 = 3,520W
TIC256 = 20A:
- At 110V => P = 110 * 20 = 2,200W
- At 220V => P = 220 * 20 = 4,400W
TC266 = 25A:
- At 110V => P = 110 * 25 = 2750W
- At 220V => P = 110 * 25 = 5,500W

CAUTION!!!

How it works

When we connect AC mains to the circuit, there is a charging of capacitor C4 through the voltage set in Trimpot RP1. When biased, there is a sending of this voltage to the DIAC through the current limiting resistor R3.

The DIAC is a bidirectionally biased diode, and is triggered when it reaches its breakdown voltage, about 30V, as it is connected to the Gate of the TRIAC. As soon as it reaches its breakdown voltage, both positive and negative pulses are activated in the Gate of the TRIAC.

However, this also charges the capacitor with reverse voltage from the negative half-cycle, and in this charging time the TRIAC stays open until the cycles compound.

This is repeated with each cycle of the AC sine wave signal from the grid, which maintains its drive and cut cycle repeatedly, leading to an output voltage lower than that of the input.

The network formed by capacitor C1 and coil L1 works as a filter to inhibit RF spurious propagation through the power network. While R1 and C2 are employed for transient reductions.

The network formed by C4 and R5 in parallel with the TRIAC, serves to prevent the TRIAC from burning out, because when the dimmer is controlling inductive loads, reverse voltage spikes are formed at the moment of switching.

Thus, the capacitor absorbs the generated overvoltage and the resistor limits the discharge current from the capacitor onto the TRIAC.

The resistor R4 connected in parallel, is used to decrease the ohmic rating of the variable resistor RV1, since the applicable value for RV1 is 150k ohms.

Since it is not easy to find this variable resistor, we made an association of resistors to get it close to 150k ohms.

The network formed by capacitor C1 and coil L1 works as a filter to inhibit RF spurious propagation through the power network.

The L1 coil consists of a small ferrite rod, 1/4″ diameter and 11/4″ long, wound with 55 turns of 28 SWG enameled copper wire.

You can be using a ferrite coil from a PC power supply to make your coil, or you can be buying a commercial 40uH coil.

Components List

Semiconductors

U1 ………….. TIC246 Triac *See Text
D1 ………….. DB-3 DIAC Diode

Resistor

R1 ……………. 56Ω (green, blue, black, gold)
R2 ……………. 2K2Ω (red, red, red, gold)
R3 ……………. 5K6Ω (green, blue, red, gold)
R4 ……………. 390Ω (orange, white, brown, gold)
R5 ……………. 250KΩ Potentiometer

Capacitor

C1, C2, C4 …. 100nF / 600V Polyester Capacitor
C3 …………….. 47nF Ceramic/Polyester Capacitor

Miscellaneous

P1, P2 ……… 2-pin PCB soldering terminal blocks
F1 ……………. Fuse 15A with soldering terminal blocks
L1 ……………. 40uH Inductor *See Text
Others ……… PCB, heat sink, wires, etc.

Printed Circuit Board

We are offering PCB – Printed Circuit Board, in GERBER, PDF and PNG files, for you who want to do the most optimized assembly, either at home.

If you prefer in a company that develops the board, you can download the files in the Download option below.

Fig. 3 – PCB – 3500W Dimmer for 110V or 220V using TRIAC TIC246M with PCB

Files to download, Direct Link:

Click Here to Download Files

✨ Our Gratitude and Next Steps

We sincerely hope this guide has been useful and enriching for your projects! Thank you for dedicating your time to this content.

Your Feedback is Invaluable:

Have any questions, suggestions, or corrections? Feel free to share them in the comments below! Your contribution helps us refine this content for the entire ElCircuits community.

If you found this guide helpful, spread the knowledge!

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The ElCircuits Team ⚡

O post 3500W AC Dimmer for 110V/220V with TRIAC TIC246 + PCB apareceu primeiro em Electronic Circuits.

12V to 220V 500W Inverter 60Hz with IR2153D IC + PCB

Eng. Jemerson — Sun, 06 Feb 2022 14:35:00 +0000

You know that moment when you are at home tired from work, ready for bed, and suddenly the power goes out? Yes my friends, it is a moment that we don’t want to happen, but we know it happens.

The best thing in these moments is to have something that can supply our power outage problem… With this we present a simple circuit, easy to build and very cheap.

I present to you a simple circuit to build, whose purpose is precisely to provide AC power to feed a fan, the lights, and some electronic equipment, with a 12V battery.

Integrated Circuit IR2153D

The IR2153D is an improved version of the popular IR2155 and IR2151 gate driver ICs, and incorporates a high voltage half-bridge gate driver with a front end oscillator similar to the industry standard CMOS 555 timer.

The IR2153D provides more functionality and is easier to use than previous ICs. A shutdown feature has been designed into the CT pin, so that both gate driver outputs can be disabled using a low voltage control signal.

In addition, the gate driver output pulse widths are the same once the rising under voltage lock out threshold on VCC has been reached, resulting in a more stable profile of frequency vs time at startup.

Noise immunity has been improved significantly, both by lowering the peak di/dt of the gate drivers, and by increasing the under-voltage lockout hysteresis to 1V.

Finally, special attention has been payed to maximizing the latch immunity of the device, and providing comprehensive ESD protection on all pins.

Features

Integrated 600V half-bridge gate driver
15.6V zener clamp on Vcc
True micropower start up
Tighter initial deadtime control
Low temperature coefficient deadtime
Shutdown feature (1/6th Vcc) on CT pin
Increased under-voltage lockout Hysteresis (1V)
Lower power level-shifting circuit
Constant LO, HO pulse widths at startup
Lower di/dt gate driver for better noise immunity
Low side output in phase with RT
Internal 50nsec (typ.) bootstrap diode (IR2153D)
Excellent latch immunity on all inputs and outputs
ESD protection on all leads
Also available LEAD-FREE

Circuit Works

In Figure 2, below, we can see the schematic diagram of 12V to 220V 600Hz 500W inverter, the circuit works in a simple and direct way, when feeding the circuit the IR2153D IC starts operating, and triggers a square wave in the GATEs of the output MOSFETs transistors.

Fig. 2 – Schematic Circuit 12V to 220V 60Hz 500W Inverter using IR2153D with PCB

This triggering is done by cycle, when triggering the HO output, pin 7 is at HIGH, and the MOSFETs are activated, in the next cycle the work, the HO output is turned off, and the LO output is activated, that is, pin 5 is set to HIGH, and this cycle repeats.

This causes an oscillation in the secondary of the transformer, generating a magnetic field that will be passed to the primary of the transformer, which is the output, thus generating an output voltage of 110V or b at a frequency of 50Hz or 60Hz, this frequency is adjusted in the trimpot.

Transformer

The transformer is a network transformer with secondary windings with 10V center tape, and should have a power according to the consumption power, or load that you will use.

Power Supply – Safety Voltage

The power supply must have enough current to provide the circuit’s consumption demand. The supply voltage should be in the 9 – 14V range.

If the supply voltage drops too low and falls below 9V, the IR2153D circuit will shut down, preventing damage to the battery or battery bank, or to the inverter circuit.

Efficiency and Consumption

The battery, or batteries bank, must provide a sufficiently high current, according to the consumption of your device, for example, for a 100W consumption of the inverter, you should take into account a battery that supplies this power.

Considering that the average efficiency factor of this equipment is approximately 80%, we will consider that for an average consumption of 100W, we will use a basic account for this:

Power in W of the load * 1.2 (20% efficiency loss) = Power in W of the Inverter

So:

100W of the load x 1.2 = 120W total

So let us now use ohms law to formulate our account:

General formula:
- P = V * I

A consumption of 120W with a battery voltage of 12V, we would be left with:

I = P / V
I = 120 / 12
I = 10A

For a 100W load we would have a consumption of 10A per hour.

Components List

Semiconductors
- U1 ………. IR2153D Integrated Circuit
- Q1 to Q6 …. IRF3205 N-Channel Power Mosfet
Resistor
- R1 ……….. 47KΩ (yellow, violet, orange, gold)
- RP1 ……… 10KΩ Trimpot
Capacitor
- C1 ………. 47nF Ceramic Capacitor
- C2 ………. 100nF Ceramic Capacitor
- C3 ………. 4.700uF / 35V Electrolytic Capacitor
Miscellaneous
- F1 ………. 20A – 250V soldering Fuse
- P1 ………. 2-pin PCB soldering terminal blocks
- P2 ………. 3-pin PCB soldering terminal blocks
- Others …. Printed Circuit Board, heat sink, wires, etc.

Printed Circuit Board

In Figure 3, we provide the PCB – Printed Circuit Board, in GERBER, PDF and PNG files. These files are available for free download, on the MEGA server, in a direct link, without any bypass.

All to make it easier for you to do a more optimized assembly, either at home, or with a company that prints the board. You can download the files in the Download option below.

Fig. 3 – PCB – 12V to 220V 60Hz 500W Inverter using IR2153D

Files to Download, Direct Link:

Click on the link beside: GERBER, PDF and PNG files

✨ Our Gratitude and Next Steps

We sincerely hope this guide has been useful and enriching for your
projects! Thank you for dedicating your time to this content.

Your Feedback is Invaluable:

Have any questions, suggestions, or corrections? Feel free to share them
in the comments below! Your contribution helps us refine this
content for the entire ElCircuits community.

If you found this guide helpful, spread the knowledge!

🔗 Share This
Guide

Best regards,

The ElCircuits Team ⚡

O post 12V to 220V 500W Inverter 60Hz with IR2153D IC + PCB apareceu primeiro em Electronic Circuits.

Arquivo de Electrical - Circuits & Systems! - Electronic Circuits

Touch Dimmer Switch for Lamps: 4-Step 110/220VAC Control Circuit – With PCB

Enhance Your Lighting: Build a 110/220VAC 4-Step Touch Dimmer Switch Circuit for Lamps with PCB Control

🧷 LS7237 IC Description

⚠️ Caution!⚠️

🤔 How the Circuit Works

🛠️ Features

🔌 The Schematic Circuit

⚡ 110Vac or 220Vac

🧮 Components List

🖨️ Printed Circuit Board – Download

📥 Files to Download, Direct Link:

How to Wire LEDs for 110V or 220V: 6 Practical Circuits with Formulas

Illuminating Insights: Wiring LEDs for 110V or 220V – Explore 6 Distinct Circuits with Formulas and Calculations!

⚠️ CAUTION! ⚠️

⚙️ The working voltage

🧮 The calculation is determined by the mathematical equation:

As our power grid is 110VAC:

If you use the 220VAC power grid:

💡 The LED

1️⃣ Determine the resistor resistance:

2️⃣ Determine the Resistor’s Power:

3️⃣ Determine Capacitive Reactance:

🧐 Knowing all the formulas that we will use in our circuits, let’s start with the simplest to the most complex.

✅ 1° Circuit:

Designing the Circuit

General Formula:

Applying the formula to our circuit:

Then:

We now need to determine the power of the resistor, for this we will use the ohms law formula:

General Formula:

Then:

🆗 Project Finished – Circuit 1

✔️ Advantages:

❌ Disadvantages:

✅ 2° Circuit:

Designing the Circuit

🆗 Project Finished – Circuit 2

✔️ Advantages:

❌ Disadvantages:

✅ 3° Circuit

Designing the Circuit

🆗 Project Finished – Circuit 3

✔️ Advantages:

❌ Disadvantages:

✅ 4° Circuit:

Designing the Circuit

General Formula:

Applying the formula to our circuit:

Then:

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:

Applying the formula to our circuit:

Applying the formula to our circuit:

🆗 Project Finished – Circuit 4

✔️ Advantages:

❌ Disadvantages:

✅ 5° Circuit:

Designing the Circuit

🆗 Project Finished – Circuit 5

✔️ Advantages:

❌ Disadvantages:

✅ 6° Circuit:

🔌 Designing the Circuit

General Formula:

Applying the formula to our circuit:

Then:

General Formula:

🆗 Project Finished – Circuit 6

✔️ Advantages:

❌ Disadvantages:

25A Solid State Relay (SSR) Circuit for 110/220V with BTA24-600 + PCB

Efficient Power Control: Building 25A – 110/220V Solid State Relay (SSR) Circuits with Triac BTA24-600 and PCB

🧲 The Solid State Relay

⚠️ CAUTION!!! ⚠️

📚 BTA24 Description

🧷 6-Pin DIP Zero-Cross Opto-isolators Triac Driver Output

(400 Volts Peak)

🔌 Solid State Relay Circuit

💡 Basic Components Function

As our power grid is 110V_AC:

If you use the 220V_AC power grid:

Since we have already found out the X_C 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: