Sr Ing

AN954
Transformerless Power Supplies: Resistive and Capacitive
Author: Reston Condit Microchip Technology Inc.

INTRODUCTION
There are several ways to convert an AC voltage at a wall receptacle into the DC voltage required by a microcontroller. Traditionally, this has been done with a transformer and rectifier circuit. There are also switching power supply solutions, however, in applicationsthat involve providing a DC voltage to only the microcontroller and a few other low-current devices, transformer-based or switcher-based power supplies may not be cost effective. The reason is that the transformers in transformer-based solutions, and the inductor/MOSFET/controller in switch-based solutions, are expensive and take up a considerable amount of space. This is especially true in theappliance market, where the cost and size of the components surrounding the power supply may be significantly less than the cost of the power supply alone. Transformerless power supplies provide a low-cost alternative to transformer-based and switcher-based power supplies. The two basic types of transformerless power supplies are resistive and capacitive. This application note will discuss bothwith a focus on the following: 1. 2. 3. A circuit analysis of the supply. The advantages and disadvantages of each power supply. Additional considerations including safety requirements and trade-offs associated with half-bridge versus full-bridge rectification.

Warning: An electrocution hazard exists during experimentation with transformerless circuits that interface to wall power. There is notransformer for power-line isolation in the following circuits, so the user must be very careful and assess the risks from line-transients in the user’s application. An isolation transformer should be used when probing the following circuits.

 2004 Microchip Technology Inc.

DS00954A-page 1

AN954
CAPACITIVE TRANSFORMERLESS POWER SUPPLY
A capacitive transformerless power supply is shown inFigure 1. The voltage at the load will remain constant so long as current out (IOUT) is less than or equal to current in (IIN). IIN is limited by R1 and the reactance of C1. Note: R1 limits inrush current. The value of R1 is chosen so that it does not dissipate too much power, yet is large enough to limit inrush current.

FIGURE 1:

CAPACITIVE POWER SUPPLY
L IIN D1 5.1V C2 470 µF IOUT VOUTC1 N .47µ 250V R1 470 1/2W IIN is given by: D2

EQUATION 3:
XC1 = VHFRMS ≥ IOUT XC1 + R1

EQUATION 1:
IIN =

1 2πfC1

Where f is the frequency (i.e., United States: 60 Hz, some countries: 50 Hz). Substituting Equation 2 and Equation 3 into Equation 1 results in:

Where VHFRMS is the RMS voltage of a half-wave AC sine wave and XC1 is the reactance of C1.

EQUATION 2: VHFRMS = VPEAK– VZ √ 2VRMS – VZ = 2 2

EQUATION 4:
 

Where VPEAK is the peak voltage of the wall power, VRMS is the rated voltage of wall power (i.e., United States: 115 VAC, Europe: 220 VAC) and VZ is the voltage drop across D1.

IIN = √ 2VRMS – VZ 1 2  2 πfC1 + R1 

DS00954A-page 2

 2004 Microchip Technology Inc.

AN954
The minimum value of IIN should be calculated for the application,while the maximum value of IIN should be calculated for the power requirements of individual components. VOUT is given by:

EQUATION 5: VOUT = VZ – VD
Where VD is the forward voltage drop across D2. Assuming a 5.1V zener diode and a 0.6V drop across D2, the output voltage will be around 4.5V. This is well within the voltage specification for PIC® microcontrollers.

EXAMPLE 1:

CALCULATEMINIMUM POSSIBLE IIN

Assume minimum values of all components except VZ and R1. Assume maximum value of VZ and R1. 110 VAC 5.1V 59.5 Hz C1 = 0.47 µF x 0.8 = 0.38 µF (assuming ±20% capacitor) • R = R1 = 470 x 1.1 = 517 (assuming ±10% resistor) • IINMIN = 10.4 mA • • • • VRMS VZ f C = = = =

OBSERVATIONS
Figure 2 shows an oscilloscope plot of VOUT at powerup with a 10 kΩ load on the output...