Transformerless AC to DC power Supply circuits

I often make small projects. However have problems to find a small power supply is not. The problem is not to find a small transformer, However it is big and heavy. Not suitable for my project. But I see in the small appliance production in China.

Do not widely used a transformer. However like to use a capacitor replace. Which it smaller and lighter than a transformer. Today I collect these transformerless power supply circuit, so you can select to use in required. Hopefully it will be useful to you. There are 4 circuits, as follows. (see below!)

Danger! Do not touch any part of these circuits. Because you may be hit by electric shocks.

(1.)Power Supply 9VDC no transformer

This is circuiut power supply 9VDC no transformer,It easy circuit and small size.
Volt output = Volt Zenerdiode. Output Current 100mA (min).


It easy circuit and small size. Power Supply 9VDC no transformer
You may know the reason that make power supply circuit have large-sized as a result , because transformer. Way this out be must not use. We use capacitor replace as a result work also well. In Power Supply 9VDC no transformer circuit . The this base equipment have zener diode perform heal voltage be stable or 9V DC Regulator. but kind this circuit be defective that current not tall sir. The detail is other , see in the circuit sir.

(2.) DC power supply not use transformer

Work cycle is C1 and R1 acts reduced AC Current Source to rectifier current is a DC Volt by bridge diode circuit which D1-D4 acts to change AC to DC with the C2 makes the filtering power DC to smoother will a zener diode acts to reduce voltage is equal to the number of volt zener diode and C3 will DCV voltage output that is more smooth.

circuit diagram of DC-power-supply-not-use-transformer

(3.)Negative Volage Regulator 6V not use Transformer

If friends with look for Mini power supply 6V sizes. That give negative voltage. I begs for to advise this circuit. It has prominent small-sized point and light weight. Because do not use Transformer. But still can use current get 0.07Amp not exceed and when see the circuit. Think use Zener diode control voltage be stable that 6V and have transistor enlarge the trend equal to that want. And still systematically protect through the circuit or Shot Circuit Protect with Transistor as well. The detail is other , see in circuit picture has yes.

(4.) DC Regulator supply 15V non transformer

Friends at don’t like large-sized of a transformer. I begs for to advise 15VDC regulator power supply can give the trend current get low about 80mA not exceed. But enough will usable small-sized circuit has comfortablely. Think I uses Zener Diode be formed help maintain one’s position volt be stable , and use Transistor help enlarge current tallly increasingly. Make this circuit is usable well , But friends as a result must use the carefulness specially , because apply to the level voltage tall directly yes.

Today John Lam send feedback to me, It is very good content. Read below:
The use of transformerless power supply dated back at least 15-20 years ago if not earlier. It was a historic innovation at that time when transformer power supply that is heavy, bulky and cost-ineffective is the mainstream power supply in small electrical and electronic gadgets.

Transformerless power supply are normally for loads that consume a small current ranging from a few mA to a few tenths mA. In recent years, they are also designed to supply larger current like those inside a LED light bulb. Every transformerless power supply is in fact custom design specific to the current load of the application, and is put inside an insulated enclosure. The risk of electric shock is minimal. In comparison to its transformer counterpart, it is more cost-effective in manufacturing and more compact. They are not supposed to be used as a general power supply.

For anyone who are interested in this kind of circuits, treat them for educational purposes, learn how and why the circuits work. One can surely try to build one to suit a specific application, but certainly not using it as a general-purpose power supply.

Since Anand mentioned the first circuit is used for a 0.5W LED night lamp. Let’s see how the circuit works.

R2 is used as a choke to limit inrush current, or limit the current in case both the 0.33uF capacitor and R1 fail before F1 will blow off. R1 is a bleeder resistor for discharging the 0.33uF after plug off. Both R1, R2 and F1 are safety features.

The circuit can theoretically supply 22.9mA current.

The equation used is current, I = V / (Total impedance of R1, R2 and 0.33uF)

Since R1 and 0.33uF are in parallel and then in series with R2, the calculation is a
bit complicated to show here. In addition, the actual calculation also have to take into
consideration the 2 x 0.7V = 1.4V voltage drop of the bridge rectifier. For those interested, google on “how to calculate impedance of RC circuit” to learn more.

Next is the bridge rectifier that rectifies the main AC sinusoidal wave into full-wave pulsating DC. The 470uF electrolytic capacitor smooth the pulsating DC to decrease the ripple voltage. Then the ZD1 regulates the output DC
voltage to 9V.

As mentioned earlier, the circuit can supply 22.9mA current, depending on the color of the LEDs used, with 9V output, it provides sufficient voltage and current for either 3 (white LED) or 5 (red LED) connected in series. That’s how the circuit works.

Power dissipated by R2 = (0.023)^2 * (20) = minimal. Any resistor rated 1/8W up is fine.

AC is 220V that translates into a peak voltage of 312V, so voltage rating of the 0.33uF capacitor should be 350V or higher.

Power dissipated by R1 is minimal (you do the math), the 1W rating specified is fine.

For the bridge rectifier, 1N4007 is rated at 700Vrms reverse voltage and 1A, and is more than sufficient for this circuit. 1n4004-1n4007 are all fine.

Since the zener diode is 9V, the rating of the 470uF electrolytic capacitor can be 15V and up.

When no load is connected, all the current will pass through the zener diode when it has the maximum power dissipation, W = V*I = (0.023) * (9) = 0.225W. The rating can be 1/4W, or 1/2W better. The reason that 1/4W will work fine is that it’s a custom circuit, the LEDs are always in the circuit as a load, current passing through the zener diode is in fact minimal.

Hope you enjoy my writings.


  1. Walter Rinebold November 1, 2016 Reply
  2. Ashutosh December 25, 2016 Reply
  3. raj July 16, 2017 Reply

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