# Simple transformerless power supply circuits

I often make the small projects. They require the small power supply.  But I cannot find the small transformers. The normal transformer is big and heavy, not suitable for my project.

But I  look around my home in a small appliance in China. Even, in most LED light bulbs. They use the transformerless power supply circuit.

They use a capacitor instead of a bigger transformer. Thus, power supplies are smaller and lighter.

Today, we will learn these transformerless power supply circuit.  So you can select as you want.

Hopefully, it will be useful to you. There are three circuits, as follows.

SEE  Below!

## The AC main

First, we need to know the AC MAINS is a high voltage of 220V or 110V. It has a much higher voltage rating than a battery. And also have different signal waveforms.

It is called an alternating current (AC). Usually generated by rotating a coil in a magnetic field.

The mains is 50Hz or 60Hz (in USA).

## Danger!

Do not touch any part of these circuits. Because you may be hit by electric shocks. Though it makes a low voltage. We cannot touch it all.  Because it does not use an isolated transformer.

### Why AC Main is dangerous?

Our body can only handle 60V to 80V. So, any overvoltages that can cause instant death.

### AC mains Measurement

In normal we know voltage in 0.707 times the peak voltage. It is called the RMS (Root-Mean-Square) voltage. And, the peak voltage (or current) is 1.41 times the RMS value.

For example, the RMS voltage 220V is 311 Vp-p. It is very high voltage.

The AC MAINS is dangerous. Because the voltage is too high.
It is actually 311v for the 220V AC. or 345V for 240V.

Look at image

The Line rises 311v then falls 311v below “ground” 50 times per second (50Hz of frequency). Then, this will produce a CURRENT FLOW through your body and it will kill you very quickly.

## Basic DC Power supply circuit

Look at below. It is transformer power supply circuit.

It is basic of unregulated supply circuit,12V 0.2A. Also, we called half rectifier circuit.

We use the transformer to reduce from high AC voltage to lower voltage. See in its symbol. There is the insulation is between the primary and secondary.

And has the two lines indicate a magnetic circuit that exists between the two windings.

Power transformer Separate the coils clearly. We are therefore quite safe for the electric shock. But if using a capacitor instead of a small transformer

## CAPACITOR-FED POWER SUPPLY

If the Neutral connected to 0v of the supply. It is no problem.

But what happens if the wires are reversed.
The line will connect to the 0V like the hole of the power-outlet on the wall.

If you touch it. You will get a shock.

### Dangers of CAPACITOR-FED POWER SUPPLY

Let’s learn the dangers of a CAPACITOR-FED POWER SUPPLY. And How it works.

#### What is the output voltage?

In normal circuit with the load connecting, the output voltage
of capacitor-fed power supply only drops to 12V or 35V. is

But…

When the load is removed, the supply rises to 180v or 311V or 340v. That’s another reason why they are so DANGEROUS.

## Easy capacitor calculation

All our calculations are made with multiples of 0.1uF capacitors.
This makes the calculations easy.
A 0.1uF capacitor will pass 7mA when connected to a bridge. Or, 3.5mA if only a single diode( half rectifier).

And, all values are halved for 110V AC.

For example. You use 0.33uF. It will pass 7mA x 0.33uF = 23.1mA

### Simplest AC mains LED display circuit

Here are the simplest AC mains LED display circuits. Or it is a CAPACITOR-FED power supply that requires a diode and a red LED.

These two items are called the LOAD.

The capacitor passes(charges) current in one direction when the mains is rising. And then, it passes current(discharges) in the opposite direction when the mains is falling.

It is a sine waveform as above.

When the mains is rising and the output of the power supply rises. And when it is 1.7v. The red LED turns ON and this voltage does not rise any more.

So, now the capacitor will store or charge the voltage about 309V. (AC main – VLED).

When the voltage of the mains falls. The output of the power supply will be negative. And when it is 0.7V negative. the diode prevents the voltage from falling.

Then, the capacitor discharges and starts to charge in the opposite direction to 309V.

The red LED is the LOAD in one direction and the diode is the load in the other direction.

## Half-wave transformerless power supply

Look at the diagram below. It is the basic half-wave capacitor-fed power supply is shown in the diagram.

Each 0.1uF of capacitance will deliver 7mA RMS.
In the half-wave supply, the capacitor delivers 3.5mA RMS. Because the current is lost in the lower diode when it discharges the capacitor.

### Using a Zener Diode

See the circuit. This is only one Zener diode instead of two diodes, previous. It is a clever design.

Why?

A Zener diode effectively breaks down in both directions.
On the top, It is 12V Zener breaks down on the cathode. And
In the opposite direction, it breaks down at 0.7V.

The load will get a maximum of 12V. And, the Zener will discharge the capacitor. To get ready for the next cycle.

#### How it works

The output current is 16mA. Because the capacitor is 0.47uF.

When we connect a load. Some current will be pulled from the Zener and will flow through the LOAD.

It has an interesting point.

• Reduce the resistance of the load. Then more current will flow through the load. Until it reaches 16mA. All the current from the capacitor will flow through the load only. No current to the Zenner diode.
• Make the more load until the voltage across it down to 11V, 10V, 9V …. But the current will remain at 16mA. Eventually, the voltage will reduce to 1v at 16mA.
• But if No load, all the current from the capacitor will pass through the Zener Diode.

### What is the Rating of the Zener diode?

The Zener diode has a wattage rating like a resistor. This is the amount of heat it will get rid of without getting too hot. If It is overheated. Eventually, it can get damaged. We should pick and set the circuit in rightly.

Often we see 500mW and 1W.

We can find out the wattage dissipation easily.

Its power is V x I

• First, V is the Voltage of the Zener diode. It is 12V.
• Second, I =?

For each 0.1uF, the circuit will deliver 3.5mA
Suppose the capacitor is 0.47uF = 16mA

The wattage dissipation for the Zener diode will be 12×16 = 200mW.
We can use A 500mW. It will not get too hot.

### Half-wave cap-fed power supply with electrolytic filter

It like other DC power supply. If we need a low ripple voltage. We need to add electrolytic capacitor filter.

Look:

We need to add a diode to prevent the electrolytic discharging in the second-half of the cycle.

We can see that the Half-wave transformerless power supply has the advantage of being simple, but it gives a low current. We should choose a better bridge circuit. Read on.

## Using Special capacitor

We must use a special type of capacitor. And it must be rated at 400V AC type. And must be constructed with materials and insulation the not blow up.

These special types of capacitors have the identification X2.
Any capacitor will work. But some will short-circuit or blow-up for no apparent reason.

Because the capacitor is charging and discharging 100 or 120 times per second.

There is a certain amount of stress on the foil and insulation. why it must be strongly constructed.

Although there is theoretically no energy lost in the capacitor, it will heat up a small amount due to losses.

The charging and discharging are grouped as ripple current and this current always causes a small amount of heating.

When the circuit is turned ON. We do not know, the mains voltage is zero, a small positive value, or a full 311V.

If it is 311V. In first, a very high current will flow to charge the capacitor. This will damage the LED.

What can we help it?

Limit this current. We add a 470 ohms resistor in series with the AC line.

### 6 LED Display with AC main

See this circuit: 6 LED Display for AC main.
If we can add more LEDs to the circuit. They will ALL glow.

We cannot add hundreds of LEDs. Because as we add another LED, the voltage across the combination increases by 1.7V.

And when the total becomes 311v. NONE of the LEDs will illuminate.

That is because there is zero volt difference between the mains voltage and the LED voltage.

It is not a good design using a single-diode. Because the LEDs are only illuminated for each half-cycle.

LEDs turn ON and OFF very quickly and they will appear to flicker, too. It is a better circuit if use a BRIDGE.

## Bridge transformerless power supply

It works like a normal bridge DC power supply circuit. A bridge is a set of 4 diodes. The output waveform is called Pulsating DC or “DC with Ripple”.

In bridge circuit, we can use lower capacitor. Because it is a full-wave rectifier.

Why?

If using 0.1uF the output current is 7mA. If we use 0.47uF. The output current is? (0.47uF x 7mA) / 0.1uF = 32.9mA

The bridge will deliver 2 pulses of energy during each cycle. And this will result in 100 blinks each second(50Hz).
And if we add more LEDs. They will all illuminate.

#### Eliminate the flicker

If we want to eliminate the flicker. The output needs an electrolytic capacitor. This will store the energy during a peak and deliver it when the mains voltage is low.

See the waveform on the circuit. The voltage remains high enough to keep the LED constantly light.

### 100 White LEDs on AC mains

Here is 100 white LEDs Display on AC main. This circuit is easy and very clever. Because No rectifier diodes are needed. We use the LEDs are the rectifiers.

How?

We need to use at least 50 LEDs in each string and the 1K resistor. To prevent them from being damaged via a surge. If the circuit is turned on at the peak of the waveform.

The resistor is provided to take a heavy surge current through one of the strings of LEDs if the circuit is switched on when the mains is at a peak.

Though we can add more LEDs to each string, the current will drop a very small amount until eventually, when you have 90 LEDs in each string, the current will be zero.

For 50 LEDs in each string, the total characteristic voltage will be 180V. Each LED requires 3.3V to 3.6V.

Each LED will take less than 7mA peak during the half-cycle they glow.

Anyway, Look at the 1K resistor. It will drop 7v. Because the RMS current is 7mA (7mA x 1,000 ohms = 7v).

And its Wattage is 7V x 7mA = 49mW

You must have LEDs in both directions to charge and discharge the capacitor.

### 5 LED Display with best Cap-fed power supply circuit

This circuit is the best cap-fed power supply for 5 LED display.

It uses 4 diodes(Bridge diodes) to produce the best current from the 0.22uF capacitor and an electrolytic to smooth out any flickering.

### 38 LED LAMP using transformerless supply

This is an example of practical use in LED lamps. It is a 38 LED lamp using a capacitor-fed power supply to illuminate 38 white LEDs.

The total voltage across the LEDs is 38 x 3.6 = 138v. The 0.33uF capacitor will deliver about 20mA. At the power about 4.4 watts (220V x 20mA)

### Fixed voltage regulated transformerless power supply

Here is a DC voltage regulated transformer power supply circuit.

See in the circuit. This clever design uses 4 diodes in a bridge to produce a fixed voltage power supply capable of the current 35mA.

All diodes (every type of diode) are Zener diodes. They all break down at a particular voltage. The fact is, a power diode breaks down at 100v or 400v and its Zener characteristic is not useful.

But if we put 2 Zener diodes in a bridge with two ordinary power diodes, the bridge will break-down at the voltage of the Zener.

This is what we have done. If we use 18v Zeners, the output will be 17v4 regulated power supply.

When the incoming voltage is positive at the top, the D1-left Zener provides an 18v limit. And, the other Zener (D2) produces a drop of 0.6v.

This allows the right Zener to pass current just like a normal diode.

The output is 17v4. The same with the other half-cycle.

The current is set by the value of C1 and C2 capacitors(in parallel). From the Bridge rectifier, the current is 7mA for each 0.1uF. So, we have 1u capacitance. So, the circuit will supply 70mA. but it will only deliver 35mA before the output drops.

The C1 and C2 capacitors should comply with the X1 or X2 class.

The R1 resistor, 10 ohms is a safety-fuse resistor.

The problem with this power supply is this will kill you as the current will flow through the diode and be lethal. if you need to touch the negative rail (or the positive rail) and any earthed device such as a toaster to get killed.

The only solution is enclosed with this circuit in a box with no outputs.

### 9V DC regulated transformerless Power Supply

This is  9V DC power supply no transformer, It is an easy circuit and small size.

From the principle above. We try to set this circuit.

The output voltage is the same as the voltage drop across the Zener diode-ZD1.

The current can be found 7mA current for 0.1uF capacitor. It should be 70mA. But some current drop R4 through R6 (in parallel). The output is lower 35mA that fixed voltage at 9V. So, we can use this instead of 9V battery.

If you like this circuit look at: AC Siren without transformer

### Conclusion

We can see that transformerless power supplies are very useful and popular. Especially in LED bulbs. But I would like to emphasize on safety. Always come first.

Note:
Although I used to use this type of power supply circuit. In appliances made from China.

Many friends are interested in it. So I learned from many places. I found Mr. Colin Mitchell to describe it very easy to understand.
Thanks. Source http://www.talkingelectronics.com/

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