Basic Electronics

How to Run DC 12V LED Bulb or MCPCB LED lights on 220Vac


DC 12V LED Bulb-How to Run DC 12V LED Bulb or MCPCB LED lights on 220Vac- In this article, you will learn, how to run 12v DC MCPCB “Metal Core Printed Circuit Board” LED light bulbs on 220Vac. Metalcore printed circuit board (MCPCB) is the vital component that provides electrical and thermal support to ensure the light-emitting diode “LED” performs well.

Learn how to repair 110/220Vac Led Bulb:

DC 12V LED Bulb

Running 12V DC MCPCB LED light bulbs on 12V batteries or Solar Panels is very simple. All you need to do, is connect the two wires + and – of the DC 12V LED Bulb with the + and – terminals of the 12V battery.

But, what if you need to run these DC 12V LED Bulbs on 220 Vac?

This is why I am writing this article, to help you understand, how to run different voltage and wattage MCPCB LED light Bulbs on 220Vac. We will go through all the designing steps. We will use different formulas for designing our own driver circuit for the MCPCB LED light bulbs. We will start with the very basics, so that; at the end, you can design a driver circuit for any type of LED light bulb.

DC 12V LED Bulb

I am going to start with a 3V led, which you are quite familiar with, you will find these types of LEDs in almost 90% of Electronic devices, first we will control this LED.

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Let’s say we have a 5V and 500mA power supply and we want to power up this 3V led. For the driver circuit designing, you should know about the LED electrical specifications. The type of LED I am going to use is 3V and 10mA.

Supply voltage = Vin = 5V

Vled = 3V

Iled = 10mA

From the above data, you can see the supply voltage is greater than the LED voltage. If you directly connect this LED with the 5V power supply, it will completely damage the 3V LED.

DC 12V LED Bulb

DC 12V LED Bulb

To safely power up this LED, we will need to drop the excess voltage across a resistor. We can find the resistor value using the ohm’s law V = IR.

R = V / I

We simply subtract the led voltage from the input supply voltage and then divide it by the LED current.

R = ( 5 – 3) / 10mA

R = 2 / 10mA

R = 2000 / 10

R = 200 ohm

So, in order to protect this LED from the damage and to safely run this led on a 5v DC power supply we will need to connect a 200 ohm’s resistor in the series with the 3V LED. In practice we slightly use a large value resistor, which further ensures, the LED remains safe. From my practical experience I have been using 330 ohm resistor which you can see in my previous projects. So, the final circuit,

DC 12V LED Bulb

Now, let’s take this to another level. What if we control the same LED using a 220Vac power supply?

Input Supply voltage = Vin = 220 Vac

Frequency of the AC supply = 50 Hz

Vled = 3V

Iled = 10mA

This time the input supply voltage is much greater than the LED voltage, which can damage the LED in no time. To protect this led from any damage; we will have to convert 220Vac to 3volts.

Let’s find the value of a resistor which we can use to drop the excess voltage.

As you know V = IR

R = ( 220 – 3 ) / 10mA

After solving this, we get

R = 21.7K ohm.

Now let’s find the resistor wattage, this is important as more current can overheat the resistor, which may result in the permanent damage of the LED. We have a formula

P = VI

P = 217 x 10mA

P = 2170 / 1000

P = 2.170 Watts

We will need a resistor value of 21.7K which can dissipate 2.170 watts. We can select the nearest value 2.5 Watt.

DC 12V LED Bulb

Instead of using this resistor we can also use a nonpolar capacitor. The value of the nonpolar capacitor can be calculated using the following formula.

C = 1 / (6.28 x F x R)

6.28 is a constant

F = frequency = 50 Hz

R = 21.7K ohm

Putting these values in the above formula.

C = 1 / ( 6.28 x 50 x 21.7 K)

C = 146nF

If you look in the table given below, you will find that the nearest value is 150nF. So we can use 150nF.

DC 12V LED Bulb

As you can see in the list the code of the 150nF is 154. So the circuit will become.

DC 12V LED Bulb

If you switch OFF the power, the capacitor remains charged and results in an electric shock when you touch the capacitor. So to discharge the capacitor we connect a large value resistor in parallel of the capacitor. I usually use 1Mega ohm in these types of circuits. So the circuit will become.

DC 12V LED Bulb

Running this LED directly on the AC voltage will reduce the led life span. To make it durable, we can connect a bridge rectifier to convert AC into DC voltage.

DC 12V LED Bulb

This circuit gives pulsating DC, in order to get a smooth DC voltage we will need to use an electrolyte capacitor. I am going to select a 100uf 50V electrolyte capacitor, you can use any other capacitor like for example 10uf 25V. This capacitor should be connected at the output side of the bridge rectifier on the + and – legs. Now, this LED will run perfectly on the 220Vac.

DC 12V LED Bulb

Now the last thing that we really need to take care of is the overvoltage protection. What if the voltage is increased above 220Vac?

Let’s say the voltage is increased to 240Vac or even higher, if this happens, it will damage the LED and the driver circuit. To protect the entire circuit from overvoltages we can add a MOV (Metal Oxide Varistor). A Varistor is nothing but a voltage-dependent resistor. Its electrical resistance decreases with the increase in applied voltage. This is exactly what we need as we want to protect our LED and driver circuit from the high transient voltages.

The MOV is installed at the AC input legs of the bridge rectifier. The MOV should be carefully selected. You should know exactly at which voltage you want to activate the voltage protection.

In this regard, I called my friend Engr. Kashif, who is currently doing a job as the assistant manager at Power Distribution Center. He told me that the standard system voltage is 220Vac, whereas out power network system on LV side fluctuates from 190 to 235Vac, this fluctuation depends on the weather conditions and nature of the load in use; he also added that in very rare cases the voltage increases to 240 Volts.

So, as per the above information, now I have to select the MOV.

The question is how we know the voltage range of a MOV? This is very simple. Certain part numbers are printed on the MOVs, like for example the following three types.

DC 12V LED Bulb




DC 12V LED Bulb

The MOV information is in its part number. In the MOV 471KD07,

K = 10%

L = 15%, and

M = 20%

So, the 471KD07 MOV has 10% tolerance.

The letter D in the 471KD07 MOV, represents that this MOV is of the type disc, if you see the letter S, it means the MOV is of the type square.

07 represents the size in mm. The 471KD07 MOV is 7mm.

The three digits number 471 represents the voltage. The last number 1 represents the number of zero’s. So this is 470Volts which is the maximum voltage range.

I will go for 471KD07 MOV. Don’t get confused with 470 Volts. This MOV is suitable for 220V AC line protection in typical electronic devices against voltage surges and spikes, such as those generated by inductive switching transients.

You may be thinking the MOV says 470V, so how it is suitable for 220V applications?

Let me explain
The AC mains voltage 220V RMS. So peak voltage is 308V = 220V x 1.4 (VPeak = VRMS × √2)

This MOV 7D471K is having 300V AC RMS rating. So it’s suitable for 220V AC mains applications. The final circuit will be.

DC 12V LED Bulb

So far we have covered all the designing steps. Now we will do the designing for the MCPCB DC 12V LED Bulb 160mA board.

DC 12V LED Bulb

Now we will design our own driver circuit so that we can run this MCPCB DC 12V LED Bulb on 220Vac. We will go through the same designing steps.

Vin = 220Vac

Vled = 12V

Iled = 160mA

As you can see the input supply voltage is greater than the LED voltage. So as usual, we will drop the excess voltage across the resistor. As you know V = IR.

R = ( 220 – 12 ) / 160ma

R = (208 x 1000) / 160

R = 3466 ohm

R = 3.466 K ohm

As this LED bulb needs a lot of current, we will need to find the resistor wattage, to find out if the resistor can dissipate this much power.

As we know

P = VI

P = 208 X 160ma

P = 33.28 watt.

It seems quite impractical to use such a high wattage resistor. So instead of using the resistor, we can use a nonpolar resistor that can easily handle this much current. To find the value of the capacitor we are going to use the same formula, which we have already used above.

C1 = I / (6.28 x F x Rc1)

6.28 is the constant.

F = 50Hz

Rc1 is the resistance we calculated = 3.466 K ohm

C1 = 1 / (6.28 x 50 x 3466)

C1 = 1 / 1088324

C1 = .0000009188

C1 = 918.8nf

The nearest is 1000nf which you can find in the list give above. The code for 1000nf is 105. So the circuit will be.

DC 12V LED Bulb

If you compare this circuit with the circuit above, you will find only the value of the nonpolar capacitor is changed, while everything else remains the same.

To further protect the driver circuit you can add a fuse with the MOV.

Caution: while working on such high voltage circuits make sure you wear protective gloves. Such a high voltage can kill you!!! Don’t touch the circuit while the supply is ON.

Engr Fahad

My name is Shahzada Fahad and I am an Electrical Engineer. I have been doing Job in UAE as a site engineer in an Electrical Construction Company. Currently, I am running my own YouTube channel "Electronic Clinic", and managing this Website. My Hobbies are * Watching Movies * Music * Martial Arts * Photography * Travelling * Make Sketches and so on...


  1. Wow Shahzada Fahad the very true essence of explanation ends here You should be in Pakistan teaching the students who are interested in electronics.Mind blowing your method of teaching.May you find success in life always.
    Sorry I couldn’t get from where you got 160 mA in the last circuit

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