Simple and practical capacitor step-down circuit

The conventional method of converting AC mains power into low-voltage DC is to use a transformer to step down the voltage and then rectify and filter it. When limited by factors such as size and cost, the simplest and most practical method is to use a capacitor step-down power supply. 

The following points should be noted when using capacitors to reduce voltage: 

1. Select an appropriate capacitor based on the current size of the load and the operating frequency of the AC, rather than the voltage and power of the load. 

2. The current-limiting capacitor must be a non-polar capacitor, and electrolytic capacitors must not be used. The withstand voltage of the capacitor must be above 400V. The ideal capacitor is an iron-shell oil-immersed capacitor. 

3 Capacitor voltage reduction cannot be used under high power conditions because it is unsafe. 

4 Capacitive voltage reduction is not suitable for dynamic load conditions.

5 Likewise, capacitive step-down is not suitable for capacitive and inductive loads. 

6. When DC operation is required, try to use half-wave rectification. Bridge rectification is not recommended, and constant load conditions must be met.

This type of circuit is usually used to obtain non-isolated low-current power supplies at low cost. Its output voltage can usually range from a few volts to thirty tens of volts, depending on the Zener voltage regulator used. The amount of current that can be provided is proportional to the current limiting capacitor capacity. When using half-wave rectification, the current (average value) available per microfarad capacitor is: (SI unit)

I(AV)=0.44×V/Zc=0.44×220×2×Pi×f×C =0.44×220×2×3.14×50×C=30000C =30000×0.000001=0.03A=30mA 

If full-wave rectification is used, double the current (average value) can be obtained: 

I(AV)=0.89×V/Zc=0.89×220×2×Pi×f×C =0.89×220×2×3.14×50×C=60000C =60000×0.000001=0.06A=60mA 

Generally speaking, although the full-wave rectification of this type of circuit has a slightly larger current, its stability and safety are worse than that of the half-wave rectification type because it is floating, so it is used less often. 

When using this circuit, you need to pay attention to the following things: 

1. It is not isolated from 220V AC high voltage, please pay attention to safety and be careful to prevent electric shock! 

2. The current-limiting capacitor must be connected to the live wire, the withstand voltage must be large enough (greater than 400V), and a series of anti-surge and insurance resistors and parallel discharge resistors must be added. 

3. Pay attention to the power consumption of the zener tube, and it is strictly forbidden to disconnect the zener tube. 

The simplest capacitor step-down DC power supply circuit and its equivalent circuit are shown in the figure. C1 is a step-down capacitor, generally 0.33~3.3uF. Assuming C1=2uF, its capacitive reactance XCL=1/(2PI×fC1)=1592. Since the on-resistance of the rectifier is only a few ohms, the dynamic resistance of the voltage regulator VS is about 10 ohms, the current limiting resistor R1 and the load resistor RL are generally 100~200, and the filter capacitor is generally 100uF~1000uF, and its capacitive reactance is very Small and can be ignored. If R is used to represent the equivalent resistance of all components except C1, the AC equivalent circuit of the figure can be drawn. At the same time, the condition of According to electrical principles, it can be known that the relationship between the average value of the rectified DC current Id and the average value of the AC current I is Id=V/XC1. If C1 is in uF, then Id is in milliamps. For 220V, 50 Hz AC, Id=0.62C1 can be obtained. 

From this, the following two conclusions can be drawn: (1) When using a power transformer as a rectified power supply, after the parameters in the circuit are determined, the output voltage is constant, while the output current Id changes with the increase or decrease of the load; ( 2) When using a capacitor to step down the voltage as a rectifier circuit, since Id=0.62 C1, it can be seen that Id is proportional to C1, that is, after C1 is determined, the output current Id is constant, but the output DC voltage changes with the size of the load resistor R. The difference varies within a certain range. The smaller RL is, the lower the output voltage is, and the larger RL is, the higher the output voltage is. The value of C1 should be selected based on the load current. For example, the load circuit requires a 9V working voltage and the average load current is 75 mA. Since Id=0.62C1, C1=1.2uF can be calculated. Considering the loss of the voltage regulator tube VD5, C1 can be taken as 1.5uF. At this time, the actual current provided by the power supply is Id=93 mA. 

The voltage stabilization value of the voltage regulator tube should be equal to the working voltage of the load circuit, and the selection of its stable current is also very important. Since the capacitor step-down power supply provides a constant current, which is approximately a constant current source, it is generally not afraid of the load being short-circuited. However, when the load is completely open, a full 93 mA current will pass through the R1 and VD5 loops, so the maximum stability of VD5 The current should be 100 mA. Since RL and VD5 are connected in parallel, while ensuring that RL draws 75 mA of operating current, there is still 18 mA of current passing through VD5, so its minimum stable current must not be greater than 18 mA, otherwise the voltage stabilizing effect will be lost. 

The value of the current limiting resistor cannot be too large, otherwise it will increase the power loss and also increase the voltage resistance requirement of C2. If R1=100 ohms and the voltage drop on R1 is 9.3V, the loss is 0.86 watts, and a resistor of 100 ohms and 1 watt can be used. 

The filter capacitor is generally 100 microfarads to 1000 microfarads, but attention should be paid to the selection of its resistance. As mentioned before, the load voltage is 9V, the voltage drop on R1 is 9.3V, and the total voltage drop is 18.3V. Considering the remaining There is a certain margin, so the C2 withstand voltage is preferably above 25V. 

As shown in Figure 1, C1 is a step-down capacitor, D2 is a half-wave rectifier diode, D1 provides a discharge circuit for C1 during the negative half cycle of the mains power, D3 is a Zener diode, and R1 is the charge discharge resistor of C1 after turning off the power supply. In practical applications, the circuit shown in Figure 2 is often used. When it is necessary to provide a larger current to the load, the bridge rectifier circuit shown in Figure 3 can be used. The unstabilized DC voltage after rectification will generally be higher than 30 volts, and will fluctuate greatly with changes in load current. This is due to the large internal resistance of this type of power supply, so it is not suitable for large current power supply. application occasions. 

Device selection 

1. When designing the circuit, you should first determine the accurate value of the load current, and then refer to the example to select the capacity of the step-down capacitor. Because the current Io provided to the load through the buck capacitor C1 is actually the charge and discharge current Ic flowing through C1. The greater the capacity of C1 and the smaller the capacitive reactance Xc, the greater the charge and discharge current flowing through C1. When the load current Io is less than the charge and discharge current of C1, excess current will flow through the voltage regulator tube. If the maximum allowable current Idmax of the voltage regulator tube is less than Ic-Io, it will easily cause the voltage regulator tube to burn out.

2. To ensure that C1 can work safely, its withstand voltage should be greater than twice the power supply voltage. 3. The selection of the discharge resistor R1 must ensure that the charge on C1 is discharged within the required time. 

Design examples 

In Figure 2, it is known that C1 is 0.33μF and the AC input is 220V/50Hz. Find the maximum current that the circuit can supply to the load. 

The capacitive reactance Xc of C1 in the circuit is: 

Xc=1 / (2 πf C) = 1/ (2*3.14*50*0.33*10-6) = 9.65K The charging current (Ic) flowing through the capacitor C1 is: Ic = U / Xc = 220 / 9.65 = 22mA. 

Usually the relationship between the capacity C of the step-down capacitor C1 and the load current Io can be approximated as: C=14.5 I, where the capacity unit of C is μF and the unit of Io is A. 

Capacitor step-down power supply is a non-isolated power supply. Special attention should be paid to isolation during application to prevent electric shock. 

The unstabilized DC voltage after rectification is generally higher than 30 volts, and will fluctuate greatly with changes in load current. This is due to the large internal resistance of this type of power supply, so it is not suitable for large current power supply. application occasions. The capacitor step-down power supply is a non-isolated power supply. Special attention should be paid to isolation in application to prevent electric shock. The working principle of the capacitor step-down power supply is not complicated. Its working principle is to use the capacitance generated by the capacitor at a certain AC signal frequency. resistor to limit the maximum operating current. For example, under the power frequency condition of 50Hz, the capacitive reactance produced by a 1uF capacitor is approximately 3180 ohms. When an AC voltage of 220V is applied to both ends of the capacitor, the maximum current flowing through the capacitor is approximately 70mA. Although the current flowing through the capacitor is 70mA, no power consumption is generated on the capacitor. If the capacitor is an ideal capacitor, the current flowing through the capacitor is the imaginary part current, and the work it does is reactive power. According to this characteristic, if we connect a resistive element in series with a 1uF capacitor, the voltage obtained across the resistive element and the power consumption it generates completely depend on the characteristics of the resistive element. For example, we connect a 110V/8W light bulb in series with a 1uF capacitor. When connected to an AC voltage of 220V/50Hz, the light bulb is lit and emits normal brightness without being burned, because the 110V/8W light bulb has The required current is 8W/110V=72mA, which is consistent with the current limiting characteristic produced by the 1uF capacitor. In the same way, we can also connect a 5W/65V light bulb and a 1uF capacitor in series to the 220V/50Hz AC power. The same is true for the light bulb. Will be lit, not burned. Because the operating current of a 5W/65V light bulb is also about 70mA. Therefore, capacitive voltage reduction actually uses capacitive reactance to limit current, and the capacitor actually plays the role of limiting the current and dynamically distributing the voltage across the capacitor and load.

Figure 1 shows an actual LED driving circuit using capacitor voltage reduction: Please note that most application circuits do not connect varistor or transient voltage suppression transistor. It is recommended to connect them because the varistor or transient voltage suppression transistor can At the moment of sudden voltage change (such as lightning, starting of large electrical equipment, etc.), the sudden change current is effectively discharged to protect the secondary switch and other transistors. Their response time is generally in the micro-millisecond level.

Circuit working principle:

The function of capacitor C1 is to reduce voltage and limit current: As we all know, the characteristics of capacitor are to pass AC and block DC. When the capacitor is connected to an AC circuit, its capacitive reactance calculation formula is:

XC = 1/2πf C

In the formula, XC represents the capacitive reactance of the capacitor, f represents the frequency of the input AC power supply, and C represents the capacity of the step-down capacitor.

The calculation formula for the current flowing through the capacitor step-down circuit is:

I = U/XC

In the formula, I represents the current flowing through the capacitor, U represents the power supply voltage, and XC represents the capacitive reactance of the capacitor.

In a 220V, 50Hz AC circuit, when the load voltage is much less than 220V, the relationship between current and capacitance is:

I = 69C where the unit of capacitance is uF and the unit of current is mA

The following table shows the comparison between the theoretical current and the actual measured current in a 220V, 50Hz AC circuit.

Resistor R1 is a bleeder resistor. Its function is: when the sine wave is cut off at the maximum peak moment, the residual charge on capacitor C1 cannot be released and will exist for a long time. During maintenance, if the human body comes into contact with the metal part of C1, there will be a strong electric shock. Electric shock may occur, but the presence of resistor R1 can discharge the remaining charge, thus ensuring the safety of man and machine. The resistance of the bleeder resistor is related to the size of the capacitor. Generally, the greater the capacity of the capacitor, the more charge remains. Therefore, the resistance of the bleeder resistor should be smaller. The empirical data is in the table below for reference during design:

The function of D1 ~ D4 is rectification, which is to rectify alternating current into pulsating DC voltage.

The function of C2 and C3 is to filter the rectified pulsating DC voltage into a stable DC voltage.

The function of the varistor (or transient voltage suppression transistor) is to discharge the instantaneous pulse high-voltage voltage in the input power supply to the ground, thereby protecting the LED from being broken down by the instantaneous high voltage.

The number of LEDs in series depends on its forward conduction voltage (Vf). In a 220V AC circuit, the maximum can reach about 80. Component selection: The withstand voltage of the capacitor is generally required to be greater than the peak value of the input power supply voltage. In a 220V, 50Hz AC circuit, you can choose polyester capacitors or paper capacitors with a withstand voltage of more than 400 volts.

D1 ~D4 can choose IN4007.

The withstand voltage of filter capacitors C2 and C3 depends on the load voltage, which is generally 1.2 times the load voltage. Its capacitance depends on the size of the load current.