Generating low voltage DC (like 5V, 6V, 9V, 12V, 15V, etc.) from the 220V or 110V AC mains is very useful and necessary in the field of electronics. Low voltage DC is used in electronics circuits, LED bulbs, toys, and may household electronics items.

Generally, batteries are used to power electronics devices, but they need to be replaced from time to time, which is not cost-effective.

The alternative way to generate the DC from Mains AC supply is a capacitive power supply. It consumes less space, so the gadget becomes handy and also cost-effective.

But, the question arises do you know

*how does a capacitive power supply works*,*how to calculate it*, and*capacitive power supply circuit diagram?*Let's learn about**Capacitive power supply**and**Circuit diagram with working explanation**.## What is a Capacitive power supply?

Capacitive power supply (CPS) is also called a transformerless capacitive power supply, and capacitive dropper. This type of power supply uses the capacitive reactance of a capacitor to reduce the mains voltage to a lower voltage to power the electronics circuit.

The circuit is a combination of a voltage dropping circuit, a full-wave bridge rectifier circuit, a voltage regulator circuit, and a power indicator circuit.

There are many methods to convert the mains AC voltage to lower DC voltage. One of the most common methods is the use of a step-down transformer to reduce 230V AC to a lower value AC, then rectified and made ripple-free DC output voltage.

A transformer-based power supply is efficient in providing sufficient current. But, it consumes much space, makes the gadget bulky, and the cost is also high.

An ideal solution to power low current demanding logic circuits and microprocessor circuits is the capacitive power supply. It is small, easy to use, consumes less space (so the gadget becomes handy), and also cost-effective.

### Is Capacitive power supply Safe for Us?

No! The capacitive power supply is not safe for us.

Because, when this power supply is on no-load, no current flowing through the circuit, and no voltage drop in the capacitor. Otherhand, there is no isolation from the mains. So, if we touch the circuit, we will get an electric shock from it.

## How to select a Voltage dropper capacitor for a Capacitive power supply?

Selection of the voltage dropping capacitor for capacitive power supply, some technical knowledge, and practical experience requires to get the desired voltage and current output.

An ordinary capacitor will not do the same job since the mains spikes will make holes in the dielectric, and the capacitor will fail to work. Then the device will be destroyed by the rushing current from the mains supply.

The voltage dropper capacitor is known to us X-rated capacitor. This is specified for reducing AC mains voltage. The X-rated capacitor is designed for 250, 400, 600 VAC, and higher voltage versions are also available.

The important parameters to be considered while selecting a dropping capacitor are given below.

- Reactance (X).
- Effective Impedance (Z).
- The Mains Frequency (50-60 Hz).

The

*Reactance (X)*of the*Voltage dropping capacitor (C)*in the*Mains frequency (f)*can be calculated using the following given formula:**X = 1 / (2πfc)**

For example, the reactance of a 0.68uF capacitor running in the mains frequency 50Hz will be:

X = 1 / {2 x 3.14159 x 50 x 0.68 x ( 1 / 1,000,000)} = 4681.03169 Ohms 0r 4.64 Kilo ohms.

Where X is the reactance of the capacitor, f is the 50 Hz frequency of AC mains supply and C is the value of the capacitor in Farads.

X = 1 / {2 x 3.14159 x 50 x 0.68 x ( 1 / 1,000,000)} = 4681.03169 Ohms 0r 4.64 Kilo ohms.

Where X is the reactance of the capacitor, f is the 50 Hz frequency of AC mains supply and C is the value of the capacitor in Farads.

We know, 1 microfarad is equal to 1/1,000,000 farads.

Hence, 0.68 microfarad is {0.68 x (1/1,000,000)} farads.

Therefore, the reactance of the capacitor appears as 4681.03169 Ohms or 4.44 Kilo Ohms.

To get current (I) divide mains voltage by the reactance in kilo ohm.

**I = V / X**

That is 230 / 4.64 = 49.569 mA.

Therefore if a 0.68 uF capacitor rated for 230 V is used, it can deliver around 49 mA current to the circuit. But, this current is not sufficient for many electronics circuits.

An important parameter effective impedance (Z) of the capacitor is determined by taking the load resistance (R).

Impedance can be calculated using the following given formula:

**Z = √ R + X**

## Circuit diagram of Capacitive power supply

Schematic of capacitive power supply circuit shown below.

## Working principle of Capacitive power pupply circuit

The working principle of the capacitive power supply is simple.

From the Capacitive power supply circuit diagram we can observe the circuit is a combination of four different circuits.

- Voltage dropping circuit.
- Full-wave bridge rectifier circuit.
- Voltage regulator circuit.
- Power indicator circuit.

### Voltage Dropping Circuit

A capacitive power supply has a voltage dropping capacitor (C1), this is the main component in the circuit. It is used to drop the mains voltage to lower voltage. The dropping capacitor is non-polarized so, it can be connected to any side in the circuit.

A resistor (R1) also connected parallel with this capacitor for removes the stored current from the capacitor when the circuit is unplugged from the mains supply. This resistor is called the

*Bleeder resistor.*If the Bleeder resistor is not connected in parallel with the voltage dropper capacitor, there is a chance for a fatal shock if anyone touched the circuit.

A resistor (R2) is used to protects the capacitive power supply circuit from inrush current at source on. This resistor can be replaced by a fuse.

### Full-Wave Bridge Rectifier Circuit

A full-wave bridge rectifier comprising 1N4007 diodes D1 through D4 is used to rectify the low voltage AC from the capacitor C1 and, this process is called Rectification.

Capacitor (C2) is used to removes ripples from DC. It may be seen that very little ripple is left in the output.

### Voltage Regulator Circuit

The required output DC voltage can be regulated using a suitable Zener diode. The Zener output voltage is not seriously affected by this resistor (R3), and the output remains as a stable reference voltage.

The limiting resistor (R3) is important. Otherwise, the Zener diode will be destroyed. Even if the supply voltage varies, the resistor (R3) will take up any excess voltage.

### Power Indicator Circuit

A LED is used as a power ON indicator, and resistor (R4) used for LED protection.

very nicely explained. thank you

ReplyDeleteAfter the voltage dropping circuit, what is the the input voltage given to bridge rectifier? I observed this voltage to be around 150 volts still. Naturally bridge output is high and how can it be a charging voltage for 4 volts battery?

ReplyDeleteThat would be with the circuit unloaded. If it was loaded and drawing current, the voltage DC should be a lot less. It still doesn't hurt to have a zener to regulate a more stable reference voltage. However, word of warning, if you do NOT know what you're doing, do NOT build or test this circuit. It can be potentially lethal.

DeleteI love the way you explain each and every component and its role. Really its very clear as a person who has very basic knowledge of electronics can understand. Thank you, keep posting 👌👏

ReplyDeleteHello all. I am a Chartered Electrical Engineer. Hobbyists - Please do NOT build circuits like this. The writer acknowledges that you might receive a shock. Please allow me put this more starkly. You could easily kill yourself playing around with a lash-up like this, connected to the mains. If you are asking questions here, please don't even try.

ReplyDeleteElectrical safety is based on i) barriers between people and live parts and ii) separation between high voltage parts and low voltage parts. Even if the circuit here were put into a plastic box (providing a barrier), it would still be potentially lethal as there is no "electrical separation" between the input and the output. There is a conductive path from input to output. It is possible for the output to be at 240V above earth potential.

Any low voltage supply for any sort of consumer equipment where it is possible to touch anything that is connected to the low voltage side (e.g. connectors, wires, etc) MUST have a so-called "isolated" supply that provides an electrical separation barrier between the mains input and the low voltage side.

An isolation barrier requires the use of certified transformers, opto-couplers and so on. It also involves construction that maintains specified physical clearances between the high voltage and low voltage parts. International electrotechnical standards set out the requirements. Otherwise, equipment cannot be deemed safe and cannot be approved or sold.

Please heed this message. Don't build circuits like this or connect them to the mains. It could very easily be lethal.

I've got these cicuits all over my house - they are in the LED lights which are all CE marked. These lights are certainly no more dangerous than the incandescent lamps that they replaced. These cicuits are fine if used appropriately.

DeleteThis type of ps are suitable for low power gadgets whose enclosures are sealed plastic.most of the china made emergency lites rechargeable toys are of this type.However life is more precious than money.

ReplyDeletewhat decides in the circuit about output voltage?

ReplyDeleteThe Zener diode connection.

DeleteNicely explained. However, a word of caution to readers of this blog. If you do not know what you're doing, or do not possess a basic understanding of electronics or what this circuit is doing, do NOT build or test it. You are playing with AC mains voltage here which can be potentially lethal. Best case scenario, if there's a wiring fault with the circuit, hopefully the circuit breaker of the power point the circuit is plugged in to will trip. The other issue is, this type of circuit is NOT isolated at the output from the input. So, a shock hazard, and potentially fatal shock hazard exists if you come in to contact with any part of the circuit when operating. Remember, the is going to be a fairly high DC voltage potential before the Zener. If you're doing this at home, you're doing so at your own risk.

ReplyDeleteHello, I would like to inquire about the Zener diode. From what I've learned in creating a regulator with a Zener diode, a series resistor is needed to dissipate the power clipped by the Zener diode. However, in the above circuit, I don't see the presence of this resistor. Could the x-rated capacitor used be replacing this resistor? Thanks

ReplyDelete