AN954

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 applications
that 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 the appliance 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 both with 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 no transformer 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 in
Figure 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
VOUT
IIN
D1
5.1V
IOUT
C2
470 µF
C1
N
.47µ 250V
R1
470 1/2W
IIN is given by:
D2
EQUATION 3:
XC1 =
EQUATION 1:
VHFRMS
≥ IOUT
XC1 + R1
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
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.
DS00954A-page 2
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:
EQUATION 4:
IIN = √ 2VRMS – VZ
1
2  2 πfC1 + R1



IIN =
1
2πfC1
 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.
EXAMPLE 1:
CALCULATE MINIMUM
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
=
=
=
=
EXAMPLE 2:
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.
OBSERVATIONS
Figure 2 shows an oscilloscope plot of VOUT at powerup with a 10 kΩ load on the output (between VOUT and
ground.) The 10 kΩ load draws only 0.45 mA. As a
result, the rise time of VOUT is 280 ms (as fast as
possible for given IIN and C2), ripple is minimal when
VOUT stabilizes at the voltage calculated in Equation 5,
approximately 4.5V.
CALCULATE MAXIMUM
POSSIBLE IIN
Assume maximum values of all components except
VZ and R1. Assume minimum value of VZ and R1.
•
•
•
•
VRMS
VZ
f
C
• R
=
=
=
=
120 VAC
5V
60.1 Hz
C1 = 0.47 µF x 1.20 = 0.56 µF
(assuming ±20% capacitor)
= R1 = 470 x 0.9 = 423 (assuming ±10%
resistor)
IINMAX = 16.0 mA
FIGURE 2:
VOUT AT START-UP WITH 10 KΩ LOAD
 2004 Microchip Technology Inc.
DS00954A-page 3
AN954
If the load is increased, the behavior of the circuit
changes in several ways. Figure 3 shows an oscilloscope plot of VOUT during the same time frame for a
500Ω load. A 500Ω load draws 9 mA at 4.5V. This is
near the 10.4 mA limit calculated in Example 1. The
rise time of VOUT is longer (680 ms) as expected
because not only is IOUT charging C2, but a significant
amount of current is being drawn by the load. VOUT
stabilizes at approximately 4.1V, about four tenths of a
volt below the output voltage calculated in Equation 5.
The ripple on VOUT is more pronounced with the
increased current draw.
FIGURE 3:
VOUT AT START-UP WITH 500 Ω LOAD
If even more current is demanded from the circuit, the
supply will stabilize at a voltage below the desired level.
Figure 4, shows an oscilloscope plot of VOUT during the
same time frame for a 270Ω load. A 270Ω load will
draw approximately 16 mA with an output voltage of
4.5V. This current cannot be provided by the circuit,
therefore, the output voltage is compromised.
FIGURE 4:
DS00954A-page 4
VOUT AT START-UP WITH A 270 Ω LOAD
 2004 Microchip Technology Inc.
AN954
POWER CONSIDERATIONS
Sizing D2:
Determining the power dissipation of the components
in the circuit is a critical consideration. As a general
rule, components should be selected with power
ratings at least twice the maximum power calculated for
each part. For AC components, the maximum RMS
values of both voltage and current are used to calculate
the power requirements.
The maximum RMS current that will flow through D2
was calculated in Example 2. Assuming a 0.7V drop
across the resistor for half the wave, the following
equation (over) approximates the power dissipated in
D2.
EQUATION 8:
Pd2 = IxV = (16.0 mA)(0.7V) = 0.011W
Sizing R1:
The current through R1 is the full-wave current. This
current is equivalent to the line voltage divided by the
impedance of C1.
EQUATION 6:
Pr1 = I2R = (VRMS*2πfC)2R1
= (21.3 mA)2(470Ω x 1.1) = 0.23W
(assuming ±10% resistor)
A 1/8 W rectifier is sufficient for D2.
Sizing C2:
C2 should be rated at twice the voltage of the zener
diode. In this case, a 16V electrolytic capacitor will
work. C2 simply stores current for release to the load.
It is sized based on the ripple that is acceptable in
VOUT. VOUT with decay according to Equation 9.
EQUATION 9:
-t
Doubling this gives 0.46W, so a 1/2W resistor is
sufficient.
Sizing C1:
Assuming a maximum wall voltage of 120 VAC, double
this is 240V. A 250V X2 class capacitor will suffice.
Note:
The class of X2 capacitor is intended for
use in applications defined by IEC664
installation category II. This category
covers applications using line voltages
from 150 to 250 AC (nominal).
Sizing D1:
D1 will be subjected to the most current if no load is
present. Assuming this worst case condition, D1 will be
subjected to approximately the full-wave current once
C2 is charged. This current was calculated when sizing
R1 (see above).
Vout = Vd e
RC
VD was calculated in Equation 5
Advantages and Disadvantages
Advantages of Capacitive Power Supply:
1.
2.
3.
Significantly smaller than a transformer-based
power supply.
More cost effective than a transformer-based or
switcher-based power supply.
Power supply is more efficient than a resistive
transformerless power supply (discussed next).
Disadvantages of Capacitive Power Supply:
1.
2.
Not isolated from the AC line voltage which
introduces safety issues.
Higher cost than a resistive power supply.
EQUATION 7:
Pd1 = IxV = (21.3 mA)(5.1V) = 0.089W
Doubling this exceeds 1/4W, so a 1/2W 5.1V zener
diode is a good choice.
 2004 Microchip Technology Inc.
DS00954A-page 5
AN954
RESISTIVE TRANSFORMERLESS
POWER SUPPLY
A basic resistive transformerless power supply is
shown in Figure 5. Instead of using reactance to limit
current, this power supply simply uses resistance. As
with the capacitive power supply, VOUT will remain
stable as long as current out (IOUT) is less than or equal
to current in (IIN.)
FIGURE 5:
RESISTIVE POWER SUPPLY
L
VOUT
IIN
D1
5.1V
C2
470 µF
IOUT
N
R1
2K 10W
D2
IIN is given by:
EXAMPLE 3:
EQUATION 10:
Assume minimum value of VRMS. Assume maximum
value of VZ and R.
IIN =
VHFRMS
R1
≥ IOUT
Where VHFRMS is the RMS voltage of a half-wave
AC sine wave.
CALCULATE MINIMUM
POSSIBLE IIN
• VRMS = 110 VAC
• VZ
= 5.1V
• R
= R1 = 2 kΩ x 1.1 = 2.2 kΩ (assuming
±10% resistor)
IINMIN
= 34.2 mA
EQUATION 11:
VPEAK – VZ
VHFRMS =
2
=
√ 2VRMS – VZ
EXAMPLE 4:
2
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.
Substituting Equation 11 into Equation 10 results in:
CALCULATE MAXIMUM
POSSIBLE IIN
Assume maximum value of VRMS. Assume minimum
value of VZ and R.
• VRMS = 120 VAC
• VZ = 5V
• R
= R1 = 2 kΩ x 0.9 = 1.8 kΩ (assuming
±10% resistor)
IINMIN
= 45.8 mA
EQUATION 12:
IIN =
√ 2VRMS – VZ
VOUT is the same as given for the capacitive power
supply (see Equation 5).
2R1
The minimum value of IIN should be calculated for the
application while the maximum value of IIN should be
calculated for power requirements.
DS00954A-page 6
 2004 Microchip Technology Inc.
AN954
OBSERVATIONS
The observations for the resistive power supply are
very similar to the capacitive power supply. Please refer
to the “Observations” in Section “Capacitive Transformerless Power Supply” for more details.
Figure 6, Figure 7 and Figure 8 show VOUT at start-up
for the resistive power supply with loads of 10 kΩ, 270Ω
and 100Ω, respectively. These loads correspond to output currents of 0.45 mA, 16 mA and 45 mA, respectively, assuming an output voltage of 4.5V. Clearly VOUT
is not 4.5V in Figure 6 because the current demand
placed on the power supply is too high.
FIGURE 6:
VOUT AT START-UP WITH 10 KΩ LO AD
FIGURE 7:
VOUT AT START-UP WITH 270Ω LOAD
 2004 Microchip Technology Inc.
DS00954A-page 7
AN954
FIGURE 8:
VOUT AT START-UP WITH 100Ω LOAD
When working with an 60 Hz AC source, it is often
desirable to know when the line voltage crosses
Neutral. The crossing, known as zero-cross, can easily
be captured by connecting the node formed by D1, C1
and D2 to an input on the microcontroller. The
waveform observed at this node is shown in Figure 9.
For the resistive power supply, the transition in this
waveform occurs at the zero-cross. For capacitive
supplies, some delay is present due to the in-series
capacitor (C1 in Figure 1).
FIGURE 9:
DS00954A-page 8
FIGURE A: WAVEFORM AT ZERO CROSS NODE
 2004 Microchip Technology Inc.
AN954
POWER CONSIDERATIONS
Sizing C2:
Selecting component power rating in the circuit is a
critical consideration. As a general rule, components
should be sized at twice the maximum power calculated for each device. For the AC components, the
RMS values of both voltage and current are used to
calculate the power requirements.
C2 should be rated at twice the voltage of the zener
diode. In this case, a 16V electrolytic capacitor will
work. C2 simply stores current for release to the load.
It is sized based on the voltage fluctuations that are
acceptable on VOUT. VOUT decays according to
Equation 9.
Sizing R1:
Advantages and Disadvantages
Advantages of Resistive Power Supply:
EQUATION 13:
V2
R
1202

= 8W
 2 kΩ x 0.9
1.
P R 1 = I 2R =


(assuming ±10% resistor)
2.
3.
Disadvantages of Resistive Power Supply:
1.
A 10W resistor builds in 2 watts of safety so it will be
used.
2.
Sizing D1:
With no load, the current through D1 will be
approximately equal to the full wave current through
R1.
Significantly smaller than a transformer-based
power supply.
Lower cost than a transformer-based power
supply.
Lower cost than a capacitive power supply.
3.
Not isolated from the AC line voltage which
introduces safety issues.
Power supply is less energy efficient than a
capacitive power supply.
Loss energy is dissipated as heat in R1.
EQUATION 14:
5.1V 
VRMS
R1
120
 2 kΩ x 0.9


PD1 = Vx1 = Vz
= 0.34W
A 1 W 5.1V zener diode should be used.
Sizing D2:
The maximum RMS current that will flow through D2
was calculated in Example 4. Assuming a 0.7V drop
across the resistor for half the wave, the following
equation (over) approximates the power dissipated in
D2.
EQUATION 15:
PD2 = IxV = (45.8 mA)(0.7V) = 0.032W
A 1/8W diode is a sufficient for D2.
 2004 Microchip Technology Inc.
DS00954A-page 9
AN954
OTHER CONSIDERATIONS
Safety Considerations
Disclaimer: This section does not provide all the information needed to meet UL requirements. UL requirements are
application specific and are not exclusive to the circuit design itself. Some of the other characteristics
that are factors in meeting UL requirements are trace width, trace proximity to one another, and (but not
limited to) other layout requirements. Visit the Underwriters Laboratories Inc. Web page at www.ul.com
for more information.
FIGURE 10:
CAPACITIVE POWER SUPPLY WITH SAFETY CONSIDERATIONS
Fuse
L
VOUT
D1
5.1V
VR1
C2
470µ
IOUT
IIN
C1
N
.47µ 250V
R1
470 1/2W
D2
R2
1M
Figure 10 shows a capacitive power supply with
several UL considerations designed in. A fuse is added
to protect the circuit during an over-current condition.
Adding R2 in parallel with C1 creates a filter that will
attenuate EMI from traveling back onto the line. A
varistor, or MOV, provides transient protection.
FIGURE 11:
Figure 11 shows a resistive power supply with several
UL considerations(1) designed in.
Note 1: User must research applicable UL
specifications that apply to the user’s
specific product. Products must be tested
by a certified lab to make sure all UL
requirements are met.
RESISTIVE POWER SUPPLY WITH SAFETY CONSIDERATIONS
Fuse
L
VOUT
VR1
C3
.047µ
D1
5.1V
R3
3M
C2
470µ
IOUT
IIN
N
R1
1K 5W
DS00954A-page 10
R2
1K 5W
D2
 2004 Microchip Technology Inc.
AN954
As with the capacitive power supply, a fuse and varistor
have been added to provide over current and transient
protection respectively. The 2 kΩ resistor is separated
into two 1 kΩ in-series resistors. Series resistors
should be split into two resistors so that a high voltage
transient will not bypass the resistor. The use of the two
resistors also lowers the potential across the resistors,
reducing the possibility of arcing. C3 and R3 create a
filter which prevents EMI created by the circuit from
migrating onto the Line or Neutral busses.
FIGURE 12:
RESISTIVE POWER SUPPLY WITH BRIDGE RECTIFIER
L
VOUT
D1
5.1V
C2
470µ
IOUT
IIN
N
R1
5K 5W
Bridge Rectification
Advantages of bridge rectifier over half-wave rectifier:
The current output of each of the circuits described can
be increased by 141% with the addition of a low-cost
bridge rectifier. Figure 12 shows what the resistive
power supply looks like with this addition.
1.
2.
3.
Instead of providing current during only one half of the
AC waveform period, current is supplied by the source
during both halves. Equation 16 gives the RMS voltage
for the full wave RMS voltage seen across R1.
Provides 141% more current.
More efficient.
VOUT is more stable.
Disadvantages of bridge rectifier compared to
half-wave rectifier:
1.
2.
More expensive.
VOUT is not referenced to just line or neutral
making triac control impossible.
EQUATION 16:
VFLRMS =
√ 2VRMS – VZ
√2
Substituting into Equation 10 gives an equation for IIN:
EQUATION 17:
IIN =
√ 2VRMS – VZ
√ 2R
 2004 Microchip Technology Inc.
DS00954A-page 11
AN954
CONCLUSION
Transformerless power supplies are instrumental in
keeping costs low in microcontroller-based applications powered from a wall receptacle. Both resistive
and capacitive power supplies offer substantial cost
and space savings over transformer-based and switchbased supplies. Capacitive power supplies offer an
energy efficient solution, while resistive power supplies
offer increased cost savings.
REFERENCES
“Transformerless Power Supply”
TB008, Microchip Technology Inc.
DS00954A-page 12
D’Souza,
Stan,
 2004 Microchip Technology Inc.
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DS00954A-page 13
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DS00954A-page 14
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