One of the main problems to be solved in an electronic circuit design is the production of low voltage DC power supply from AC to power the circuit. The conventional method is to use a step-down transformer to step down the 230 V AC to a desired level of low AC voltage. The most suitable and low-cost method is to use a Voltage Drop Capacitor in series with the phase line.
Drop capacitor selection and circuit design require some technical knowledge and hands-on experience to obtain the desired voltage and current. An ordinary capacitor will not work, as the device will be destroyed by the intense current of the electrical network. Network spikes will create holes in the dielectric and the capacitor will not work. The X-rated capacitor specified for AC mains use is required to reduce AC voltage.
Fig. 1: Image of a capacitor
400 Volt X Rated Capacitor
Before selecting the dropout capacitor, it is necessary to understand the working principle and operation of the dropout capacitor. The X-rated capacitor is designed for 250, 400, 600 VAC. Higher voltage versions are also available. Effective Impedance (Z), Retance (X) and Network Frequency (50 – 60 Hz) are the important parameters to be considered while selecting the capacitor. The reactance (X) of the capacitor (C) at mains frequency (f) can be calculated using the formula
X = 1 / (2 ¶fC)
For example, the reactance of a 0.22 uF capacitor operating at the mains frequency of 50 Hz will be 0r 14.4 Kilo ohms. The reactance of the 0.22 uF capacitor is calculated as That is, 1 microfarad is equivalent to 1/1,000,000 farads. Therefore, 0.22 microfarads equals 0.22 x 1/1,000,000 farads. Therefore, the capacitor's rectance appears as 14,475.97 Ohms or 14.4 K Ohms. To obtain the current, I divide the volt of the network by the rectance in kilo ohms. That's 230/14.4 = 15.9 mA.
The effective impedance (Z) of the capacitor is determined by taking the load resistance (R) as an important parameter. Impedance can be calculated using the formula
Z = v R + X
Suppose the current in the circuit is I and the grid voltage is V, then the equation looks like
I = V/X
The final equation becomes like this
I = 230 V / 14. 4 = 15.9 mA.
Therefore, if a 0.22 uF capacitor rated for 230 V is used, it can supply about 15 mA of current to the circuit. But this is not enough for many circuits. Therefore, it is recommended to use a 470 nF capacitor rated for 400 V for such circuits to provide the required current.
X-Rated AC Capacitors – 250V, 400V, 680V AC
Table showing the types of X-rated capacitors and the no-load output voltage and current
Fig. 3: Table showing X-rated capacitor types and no-load output voltage and current
Rectification
Diodes used for rectification must have sufficient peak inverse voltage (PIV). Peak reverse voltage is the maximum voltage a diode can withstand when it is reverse biased. The 1N 4001 diode can withstand up to 50 Volts and 1N 4007 has a tolerance of 1000 Volts. The important characteristics of general purpose rectifier diodes are given in the table.
Fig. 4: Table showing the characteristics of general purpose rectifier diodes
Therefore, a suitable option is a 1N4007 rectifier diode. Typically, a silicon diode has a forward voltage drop of 0.6 V. The current rating (forward current) of rectifier diodes also varies. Most 1N series general purpose rectifier diodes have a current rating of 1 amp.
Fig. 5: Diode Image
DC smoothing
A smoothing capacitor is used to generate ripple-free DC. The smoothing capacitor is also called a filter capacitor and its function is to convert the half-wave/full-wave output of the rectifier into smooth DC. Rated power and capacitance are two important aspects to consider when selecting the smoothing capacitor. The rated power must be greater than the no-load output voltage of the power supply. The capacitance value determines the amount of ripples that appear in the DC output when the load receives current. For example, a full-wave rectified DC output obtained from a 50 Hz AC network operating on a circuit that draws 100 mA current will have a ripple of 700 mV peak to peak across the filter capacitor rated at 1000 uF. The ripple that appears in the capacitor is directly proportional to the charging current and inversely proportional to the capacitance value. It is best to keep ripple below 1.5V peak to peak under full load condition. Therefore, a high value capacitor (1000 uF or 2200 uF) rated at 25 volts or more must be used to obtain a ripple-free DC output. If the ripple is excessive, it will affect the operation of the circuit, especially the RF and IR circuits.
Voltage regulation
The Zener diode is used to generate a regulated DC output. A Zener diode is designed to operate in the reverse breakdown region. If a silicon diode is reverse biased, it reaches a point where its reverse current suddenly increases. The voltage at which this occurs is known as the “Avalanche or Zener” value of the diode. Zener diodes are specially made to exploit the avalanche effect for use in 'reference voltage' regulators. A Zener diode can be used to generate a fixed voltage by passing a limited current through it using the series resistor (R). The Zener output voltage is not seriously affected by R and the output remains a stable reference voltage. But the limiting resistor R is important, without which the Zener diode will be destroyed. Even if the supply voltage varies, R will absorb any excess voltage. The R value can be calculated using the formula
R = Vin – Vz / Iz
Where Vin is the input voltage, the output voltage Vz and the current Iz through the Zener
In most circuits, Iz is maintained at 5mA. If the supply voltage is 18 V, the voltage that must fall across R to obtain the 12 V output is 6 volts. If the maximum allowable Zener current is 100 mA, then R will pass the maximum desired output current plus 5 mA. Then the value of R appears as
R = 18 – 12/105 mA = 6/105 x 1000 = 57 ohms
The Zener power rating is also an important factor to consider when selecting the Zener diode. According to the formula P = IV. P is the power in watts, I is the current in Amperes and V is the voltage. Therefore, the maximum power dissipation that can be allowed in a Zener is the Zener voltage multiplied by the current flowing through it. For example, if a 12 V Zener passes 12 V DC current and 100 mA, its power dissipation will be 1.2 Watts. Therefore, a 1.3W Zener diode must be used.
LED indicator and diagram
LED indicator
The LED indicator is used as a power indicator. A significant voltage drop (about 2 volts) occurs across the LED when it passes direct current. The forward voltage drops of various LEDs are shown in Table.
Fig. 6: Table showing forward voltage drops of various LEDs
A typical LED can pass a current of 30 to 40 mA without destroying the device. The normal current that provides sufficient brightness for a standard red LED is 20 mA. But this can be 40 mA for blue and white LEDs. A current limiting resistor is needed to protect the LED from excess current flowing through it. The value of this series resistor must be carefully selected to avoid damage to the LED and also to obtain sufficient brightness at 20 mA current. The current limiting resistor can be selected using the formula
R = V/I
Where R is the resistor value in ohms, V is the supply voltage and I is the allowable current in Amps. For a typical red LED, the voltage drop is 1.8 volts. Therefore, if the supply voltage is 12 V (Vs), the voltage drop across the LED is 1.8 V (Vf) and the allowable current is 20 mA (If), then the value of the series resistor will be
Vs – Vf / Se = 12 – 1.8 / 20 mA = 10.2 / 0.02 A = 510 Ohms.
A suitable available resistor value is 470 Ohms. But it is advisable to use 1K resistor to increase the lifespan of the LED even if there is a slight reduction in brightness. As the LED consumes 1.8 volts, the output voltage will be 2 volts lower than the Zener value. Therefore, if the circuit requires 12 volts, it is necessary to increase the Zener value to 15 volts. The table below is a ready-made calculation for selecting the limiting resistor for various versions of LEDs at different voltages.
Fig. 7: Table showing the ready calculator for limiting resistor selection for various versions of LEDs at different voltages.
Circuit Diagram
The diagram shown below is a simple transformer without a power supply. Here nominal 225K (2.2uF) 400 volt capacitor X is used to step down 230 volts AC. Resistor R2 is the bleed resistor that removes the stored current from the capacitor when the circuit is disconnected. Without R2, there is a possibility of fatal shock if the circuit is touched. Resistor R1 protects the circuit from inrush current when switching on. A full-wave rectifier comprising D1 to D4 is used to rectify the low AC voltage of capacitor C1 and C2 and removes DC ripples. With this design, about 24 volts at 100 mA of current will be available at the output. This 24 volt DC can be regulated to the required output voltage using a suitable 1 watt Zener. It is best to add a safety fuse on the phase line and a MOV on the phase and neutral lines as a safety measure if there is a voltage spike or short circuit in the mains.
Caution: Construction of this type of power supply is only recommended for people experienced or competent in handling AC networks. Therefore, do not attempt this circuit if you have no experience in handling high voltages.
The disadvantage of capacitor power supply includes
1. No galvanic isolation from the electrical network. Therefore, if the power supply section fails, it can damage the gadget.
2. Low current output. With a capacitor power supply. The maximum available output current will be 100 mA or less. Therefore, it is not ideal for operating heavy current inductive loads.
3. The output voltage and current will not be stable if the AC input varies.
Careful
Great care must be taken when testing the power supply using a dropping resistor. Do not touch any point on the PCB as some points are at mains potential. Even after turning off the circuit, avoid touching the points around the falling capacitor to avoid electric shock. Extreme care must be taken when constructing the circuit to avoid short circuits and fires. Sufficient spacing must be provided between components. The high value smoothing capacitor will explode if connected in reverse polarity. The drop capacitor is not polarized so it can be connected in any direction. The power supply must be isolated from the remaining part of the circuit using insulators. The circuit must be housed in a metal case without touching any part of the PCB in the metal case. The metal box must be properly grounded.
Circuit diagrams
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