ST AN1476 APPLICATION NOTE
AN1476
APPLICATION NOTE
LOW-COST POWER SUPPLY FOR HOME APPLIANCES
INTRODUCTION
In most non-battery applications, the power to the microcontroller is supplied by using a stepdown transformer, which is then rectified, filtered and regulated. However, in many smaller
low-cost applications, the cost of the transformer becomes the key factor in the system. Under
these circumstances, the step-down transformer is normally not used in order to reduce the
cost as well as the size. The power supply is a simple one-way rectifier with very few components. The output voltage is regulated by using a 5.6V zener diode. Despite its simplicity and
low cost, it is still able to deliver enough current to the microcontroller and application circuits.
The purpose of this application note is to present the basic principle and cost analysis of the
various power supply circuits for home appliance applications.
AN1476/0904 1/8
1
LOW-COST POWER SUPPLY FOR HOME APPLIANCES
1 BASIC CIRCUITS
1.1 TRANSFORMER POWER SUPPLY
Figure 1. Transformer Power Supply Diagram
T1
N
1
3
4
2
L
4
1N4003*4
3
2
100nF
C1
C2
470uF/35V
U2
78L05
IN
GND
OUT
C3
10nF
VCC
C4
100uF/16V
Figure 1. describes how to obtain a 5V DC voltage from the AC power line. In this circuit, the
AC voltage drops down on the transformer’s secondary winding. A rectifier bridge with 4 diodes is used to convert the alternating AC voltage to a continuous DC voltage supply. A filter
capacitor is added after the rectifier bridge in order to decrease the DC voltage ripple. The
78L05 triple terminal voltage regulator provides a very stable output and high current.
The advantages of this solution are:
■ Power Supply is isolated from the AC line voltage,
■ Power Supply can deliver high current (up to 100 mA for 78L05),
■ The ripple of DC voltage will be small.
However, the disadvantages of this solution are also obvious:
■ Much more expensive than transformerless power supplies,
■ Power supply size is bigger due to the transformer and other components.
2/8
2
LOW-COST POWER SUPPLY FOR HOME APPLIANCES
1.2 CAPACITIVE POWER SUPPLY
Figure 2. Capacitive Power Supply Diagram
N
D1
V
Iin
R
L
470/0.5W
5.6V
D2
1N4148
C1
470nF/400V
Cs
220uF/16V
Iout
Vdd
MCU
Vss
Figure 2. describes the capacitive power supply. In order to have a constant voltage across
capacitor Cs, the average value of the input current (I
of the output current (I
). Current through this capacitor (Cs) flows during only the positive
OUT
) must be equal to the average value
IN
half cycle of the supply and during the negative half cycle it flows through the zener diode.
Input current (I
) is a half-wave current, whose value is given by the following equation (R and
IN
Cs can be disregarded as the impedance offered by these is negligible in comparison to C1 ):
I
INav
V
--------------------------------------=
peak
2πfC1⋅⋅
π
Here Vpeak/PI = Average(dc equivalent) voltage in one cycle.
π.f.C1 = impedance offered ( neglecting R and Cs) . The voltage rating of C1 should be
2.
greater than Vpeak as it is charged up to Vpeak.
Resistance and Capacitance(Cs) significance:
Cs has to be charged upto 5.6 V. So the voltage rating should be more than 5.6V. This capacitor reduces ripples from the output supply. As one tries to withdraw more and more current,
the ripples will be increased. The high value of Cs would reduce the ripples from the supply up
to a certain limit.
Resistace(R) limits the current through the zener diode. The value of R should be such that
Vpeak/R is more than the current limit of the zener.
To ensure the delivery of enough current in the worst conditions (Vpeak, f and C at minimum),
the average output current, sunk by the MCU, must match the following conditions (capacitor
3/8