One of the major challenges during a circuit design is to generate low-voltage DC from AC to power the circuit. There are many methods to convert AC voltage into DC. The most common method is the use of a step-down transformer to reduce 230V AC to a lower-value AC. This is then rectified and made ripple-free by using a smoothing capacitor.

Even though the transformer-based power supply is efficient in providing sufficient current, it consumes much space and makes the gadget bulky. Cost of the transformer is also high. To power low-current-demanding logic circuits and microprocessor circuits, transformerless power supply is an ideal solution. These power supplies have distinct advantages and disadvantages. These are cost-effective and consume less space, so the gadget becomes handy. But the low current efficiency makes them non-ideal to satisfy high current requirements of most circuits. Moreover, there is no isolation from the mains.

The two basic types of transformerless power supplies are capacitive and resistive. Capacitive type is more efficient since its heat dissipation and power loss are very low. Resistive-type power supply, on the other hand, dissipates more heat, so the power loss is quite high. If a circuit requires very low current of a few milliamperes, transformerless power supply is an ideal solution. Before designing a transformerless power supply, some facts about AC dropping through a capacitor or resistor are to be considered.

**Design considerations**

If a non-polarised capacitor and a resistor are put in series with the AC power line, constant current can be maintained through the resistor, provided that the reactance of the capacitor is greater than the resistance of the series resistor used.

The current flowing through resistor R depends on the value of capacitor C. That is, a higher-value capacitor delivers more current to the circuit. Current flow through dropping capacitor C depends on its reactance, and the value of the current passing through the capacitor is represented as:

IRMS = VIN /X

where ‘X’ is the reactance of the capacitor and VIN is the line voltage (230V).

Before selecting the capacitor, it is necessary to understand how a dropping capacitor behaves when it passes current. The capacitor designed to operate in AC voltage belongs to the ‘X’ category with operating voltage ranging from 250 volts to 600 volts. If the mains frequency is 50 Hz, the reactance (X) of the capacitor is:

This article is an example of what we get when a professor of zoology tries to write articles on Electronics!

Lot of nonsense has been written in this article to confuse hobbyists and students.

For example, 230V/14.4 = 15.9 mA? It is simple Ohms Law, and the answer is not mA but AMPERES, yes, the answer should be 15.9 AMPERES! Mr Mohan Kumar, just try to connect a 14.4 ohm resistor across the 230V mains and you shall get your answer….It seems that the author has never actually done any practical projects in electricity or electronics himself.

You are right. Teoretical is one thing and practical work is the other. The difference between these two is big.

And mentioning theoretical work earlier, in this article the author doesn’t seem to work a lot on theoretical knowledge either 😀

Chinmoy Mitra what you have told is wrong. Mohan kumar is right since the reactance of capacitor for 0.22uF capacitor is 14.4 K ohm and thus the 230/14.4*10000=15.9mA.

Hello sir, I need help about smps circuit it is ac 175 to 230v input and output 24v 2.5amp. Dc so pls give m circuit design. In my mail id. Thank u.

DON’T EVER TRY OUT ANY CIRCUIT by the so called Dr.Mohan Kumar!

Ref: http://www.talkingelectronics.com/projects/SpotMistakes/SpotMistakesP14.html

QUOTE “Professor D.Mohankumar must be the WORST electronics engineer I have ever come across. Here is another of his untested circuits:” Unquote.

Dr.Kumar, plse stop…. U are a source of EMBARRASSMENT to YOUR COUNTRY.