Working principle and calculation of capacitive buck

working principle of capacitive Buck
the working principle of capacitive buck is not complicated. The principle of his work is to limit the maximum operating current by capacitive resistance generated by the capacitance at a certain AC signal frequency. Example
For example, in 50Hz of power frequency conditions, a 1uF capacitance generated by the capacitive resistance of about 3180 ohms. When the 220V AC voltage is added to both ends of the capacitor, the maximum flow through the capacitor
the large current is about 70mA. Although the current flowing through the capacitor is 70mA, it does not generate power on the capacitor, and if the capacitor is an ideal capacitor, the capacitance
The flow is the imaginary part of the current, which is reactive power. According to this characteristic, if we cascade a resistive element on a 1uF capacitor, then both ends of the resistive element
The resulting voltage and the power dissipation it generates depend entirely on the characteristics of the resistive element. For example, we put a 110v/8w bulb in series with a 1uF capacitor,
on the AC voltage of the 220v/50hz, the bulb is lit and emits normal brightness without being burnt. Because the 110v/8w bulb requires a current of 8W/110V=72MA,
It is consistent with the current-limiting characteristics of the 1uF capacitor. Similarly, we can also connect the 5w/65v bulb with the 1uF capacitor to the 220v/50hz AC, the bulb also
will be lit up without being burnt. Because the operating current of the 5w/65v bulb is also about 70mA. Therefore, capacitive buck is actually the use of capacitive current limiting. and the capacitor actually
Up to a role that limits current and dynamically distributes capacitors and voltages on both ends of the load.
The following points should be noted when using capacitive buck:
1 Select the appropriate capacitance based on the current size of the load and the operating frequency of the AC, rather than on the voltage and power of the load.
2 Current-limiting capacitors must be of an nonpolar capacitor, and no electrolytic capacitors can be used. and the voltage of the capacitor must be above 400V. The ideal capacitance is the iron-Shell Oil immersion power
and the content.
3 Capacitor buck cannot be used for high power conditions because it is unsafe.
4 capacitive buck is not suitable for dynamic load conditions.
5 Similarly, capacitance Buck does not fit the capacitive and inductive loads.
6 when DC operation is required, use half-wave rectification as far as possible. Bridge-type rectification is not recommended. And to meet the conditions of a constant load.
The above is a brief introduction of the working principle of capacitive buck. I once again raised a question a few days ago, that is, only using resistors and capacitors to form what circuit, further
What circuit can be composed of only one resistor and one capacitor. This can be an answer, interested to think about what else can be formed circuit. In fact, resistance,
capacitors and inductors are the basic components of electronic circuits, and they are well known for their characteristics and flexible to use.
The use of capacitive buck circuit is a common low-current power supply circuit, because of its small volume ﹑ low cost ﹑ current relatively constant and other advantages, but also used in led flooding
in the dynamic circuit.
Figure one is a practical led driver circuit with capacitive buck: Note that most application circuits do not have a varistor or transient voltage suppressor transistor connected ,
It is recommended to connect, because the varistor or the transient voltage suppressor transistor can effectively discharge the mutation current in the moment of voltage mutation (such as lightning ﹑, large electric equipment start, etc.) to
protection of two-level and other transistors, their response time is generally in the micro-millisecond level.
How the circuit works:
the function of the capacitor C1 is buck and current limiting: as we all know, capacitance is the characteristic of the AC ﹑ DC, when the capacitor is connected to the AC circuit, its capacitance resistance calculation formula
are:
1th Page
XC = 1/2πf C
The,XC in the formula indicates that the capacitance of the capacitor ﹑f indicates the frequency of the input AC power ﹑ C represents the capacity of the buck capacitor.
the current calculation formula for the flow through the capacitive Buck Circuit is:
I = U/xc
in the formula I represents the current flowing through the capacitance ﹑u means that the supply voltage ﹑XC represents the capacitance resistance in the 220v﹑50hz AC circuit, when the load voltage is much less than 2
at 20V, the current-to-capacitance relationship is:
I = 69C
where the capacitance is in units of uf, current in units of Ma
The following table is a comparison between the theoretical current and the actual measured current in the 220v﹑50hz AC circuit
The resistor R1 is the discharge resistor, which acts as: when the sine wave is cut off at the maximum peak time, the residual charge on the capacitor C1 cannot be released and will persist for a long time, as
Fruit Body contact with C1 metal parts, there is a strong electric shock possible, and the presence of resistance R1, the residual charge can be released off, thus ensuring the safety of ﹑ machine. Discharge resistor
the resistance and capacitance of the size of the general capacitance of the larger, the residual charge will be more, the resistance of the discharge resistor will be selected smaller. Experience data is provided in the following table for design
Time Reference:
the function of D1 ~ D4 is rectification, which is the commutation of alternating current to the pulsating DC voltage.
the function of C2﹑C3 is filtering, which is to filter the pulsating DC voltage after rectification to a stable DC voltage .
the role of varistors (or transient voltage suppressor transistors) is to discharge the instantaneous high voltage voltages in the input power supply to the ground, thereby protecting the LEDs from the instantaneous high voltage breakdown.
The number of LEDs in series depends on its positive guide voltage (VF), which can reach up to 80 in 220V AC circuits.
Component Selection: The capacitance of the general requirements of the pressure is greater than the peak input supply voltage, in the 220v,50hz AC circuit, you can choose to withstand the pressure of more than 400 kv polyester power
or paper dielectric capacitance.
D1 ~d4 can choose IN4007.
the withstand voltage of the filter capacitor C2﹑C3 is based on load voltages, typically 1.2 times times the load voltage. The capacitance capacity depends on the load current size.
2nd Page

principle of capacitive buck power supply and its relative calculation----suitable for low unit price and low anti-jamming circuit (slightly supplementary)
The conventional method of converting AC mains to low-voltage DC is to use the transformer to lower the rectifier filter, when the volume and cost factors
limit, the simplest and most practical method is to use a capacitive buck power supply.
first, the circuit principle
The basic circuit of the capacitive buck simple power supply is 1,c1 to buck capacitor, the D2 is a half-wave rectifier diode, the D1 in the mains negative
half postmenstrual to C1 provide discharge circuit, D3 is a voltage regulator diode, R1 for power off after the C1 charge discharge resistor. In Practical Applications
The circuit shown in Figure 2 is often used. When it is necessary to provide a large current to the load, the bridge-type integer shown in Figure 3 can be used
flow circuit.
The non-regulated DC voltage after rectification will generally be higher than 30 volts and will fluctuate greatly with the change of load current, which is
for this kind of power supply resistance is caused by large, it is not suitable for high-current power applications.
second, the device selection
1. When designing the circuit, the exact value of the load current should be measured, and then a reference to the example to select the Capacitance
volume. Because the current IO that is provided to the load through the buck capacitor C1 is actually a charge-discharge current flow through the C1 IC. The more C1 capacity
, the smaller the tolerance of XC, the greater the charge and discharge current flowing through the C1. When the load current IO is less than the charge-discharge current of C1,
residual current will flow through the regulator tube, if the maximum allowable current of the regulator pipe Idmax less than ic-io easily caused by voltage regulator
Tube burnt.
2. In order to ensure reliable operation of the C1, its pressure selection should be greater than twice times the supply voltage.
3. The choice of the discharge resistor R1 must be guaranteed to release the charge on the C1 within the required time.
three, design examples

Figure 2, the known C1 is 0.33μf, the AC input is 220v/50hz, the circuit can supply the maximum current load.
C1 in the circuit of the Tolerance XC is:
xc=1/(2πf C) = 1/(2*3.14*50*0.33*10-6) = 9.65K
the charge current (Ic) flowing through the capacitor C1 is:
Ic = U/XC = 220/9.65 = 22mA. (valid value of current)
The capacitance of the Buck capacitor C1 is usually approximated by the relationship of the load current IO: c=14.5 I, where c is the capacity unit
is the unit of Μf,io is a.
capacitive Buck-type power supply is a non-isolated power supply, in the application of special attention to isolation, to prevent electric shock
principle and calculation formula of capacitive Buck power supply
This type of circuit is often used for low cost access to non-isolated small current supply. The output voltage is usually a few volts to three
10 Volts, depending on the Zener regulator used. The current size can be supplied in proportion to the capacitance limit. When using half-wave rectification,
the current (average) per micro-capacitor can be obtained as: (International standard 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 a full-wave rectifier is used, the double current (average) is:
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, this kind of circuit full-wave rectification, although the current slightly larger, but because the float, stability and security than the half-wave rectification type is worse,
so use less.
when using this circuit, the following considerations apply:
1, not and 220V AC high-voltage isolation, please pay attention to safety, to prevent electric shock!
2, the current limit capacitance must be connected to the FireWire, the pressure must be large enough (greater than 400V), and a series of anti-surge impact and safety resistors and
discharge resistor.
3, pay attention to the Zener tube power consumption, the zener tube is forbidden to operate.
The use of capacitive buck circuit is a common low-current power supply circuit, because of its small volume ﹑ low cost ﹑ current relative constant
advantages, and is often used in the LED driver circuit.
Figure one is an actual led driver circuit with capacitive buck: Note that most applications do not have a varistor connected
or transient voltage suppressor transistor, it is recommended to connect, because a varistor or transient voltage suppressor transistor can mutate in voltage
instantaneous (such as lightning ﹑ large electric equipment start, etc.) effectively discharge the mutation current, thereby protecting the two level and its
it transistors, their response time is generally in the micro-millisecond level.
How the circuit works:
the function of the capacitor C1 is buck and current limit: As we all know, the characteristic of capacitance is through AC ﹑ DC, when the capacitance is connected to the AC
in the circuit, its tolerance calculation formula is:
XC = 1/2πf C
The,XC in the formula indicates that the capacitance of the capacitor ﹑f indicates the frequency of the input AC power ﹑ C represents the capacity of the buck capacitor.
the current calculation formula for the flow through the capacitive Buck Circuit is:
I = U/xc
in the formula I represents the current flowing through the capacitance ﹑u means that the supply voltage ﹑XC represents the capacitance tolerance
in the 220v﹑50hz AC circuit, when the load voltage is much less than 220V, the current-to-capacitance relationship is:
I = 69C (this formula is the same as c=14.5i, the unit is different) where the unit of capacitance is the unit of uf, current
to Ma
The following table is a comparison between the theoretical current and the actual measured current in the 220v﹑50hz AC circuit
The resistor R1 is the discharge resistor, which acts as: when the sine wave is cut off at the maximum peak moment, the residual charge on the capacitance C1 cannot
release, will be long-standing, in the maintenance of the body when exposed to C1 metal parts, there is a strong electric shock may, and resistance R1
exist, can discharge the residual charge, so as to ensure the safety of the ﹑ machine. The resistance of the discharge resistor is related to the size of the capacitance, generally
the larger the capacitance, the more residual charge, the resistance of the discharge resistor will be selected smaller. Experience data is provided in the following table for design time
Reference:
the function of D1 ~ D4 is rectification, which is the commutation of alternating current to the pulsating DC voltage.
the function of C2﹑C3 is filtering, which is to filter the pulsating DC voltage after rectification to a stable DC voltage .
the effect of a varistor (or transient voltage suppressor transistor) is to drain the instantaneous pulse voltage in the input power supply from the
While the protection LED is not instantaneous high voltage breakdown.
The number of LEDs in series depends on its positive guide voltage (VF), which can reach up to 80 in 220V AC circuits.
Component Selection: The capacitance of the general requirements of the pressure is greater than the peak input supply voltage, in the 220v,50hz AC circuit, you can
Select a polyester capacitor or paper dielectric capacitance that is more than 400 volts in pressure.
D1 ~d4 can choose IN4007.
the withstand voltage of the filter capacitor C2﹑C3 is based on load voltages, typically 1.2 times times the load voltage. Its capacitive capacity is dependent on the load
the size of the current is determined.
The following circuit diagrams are other forms of capacitive buck Drive circuitry for design-time reference:
in the figure II circuit, SCR and R3 form a protective circuit, when the current flowing through the LED is greater than the set value,SCR conduction a
The current of the circuit, so that the led works in a constant current state, so as to avoid the LED due to instantaneous high pressure and damage.
in the figure three circuit,,c1﹑r1﹑ Varistor ﹑L1﹑R2 constitute the primary filter circuit of the power supply, can filter the input instantaneous high pressure,
C2﹑R2 consists of a buck circuit,c3﹑c4﹑l2﹑ and a varistor composed of a rectifier filter circuit. This circuit uses double-filter power
the road, can effectively protect the LED not by the instantaneous high pressure breakdown damage.
The power supply with transformer is large in volume, which is difficult to be used in some small volume production. The small transformer-free power supply introduced in this paper can provide
3~15v voltage, maximum current 150mA, to meet the needs of small electronic equipment power supply.
circuit, 220V by D2 rectifier C1 Filter, as the Q1 of the conduction drive voltage, when 220V is half a week start, but W sliding end
when the upper voltage is not large enough, the Q2 is in the cut-off state, the voltage on the C1 is R4 plus the Q1 gate to enable the Q1 conduction, 220V positive half-week warp
D1, R5, Q1 to the capacitor C2 fast charging. When the voltage on the W sliding end is sufficient to enable the D3 and Q2 to pass, the Q1 gate loses voltage and
deadline. Adjust the W to adjust the charging time to the C2, and adjust the output voltage. Due to the extremely short conduction time of the Q1, the C2 selected a large
capacitance to ensure a smoother output voltage.
The R5 in the circuit is a current-limiting resistor, which reduces the peak value of C2 charge currents. The regulator tube D5 is designed to prevent Q1 from being damaged by high gate voltages.
D4 is used as output protection, when the C2 voltage is too high D4, Q2 conduction, so that the Q1 cutoff. Because the circuit does not work during the negative half-week of the mains power,
Increased output current, can be added to the input rectifier bridge, so that the positive and negative half of the power supply can be used, so that the output current can be increased by 80mA,
It can also improve the smoothness of the output voltage. In practical application, the potentiometer W, R3 can be replaced with a fixed resistor. The stability of the output voltage must be
when high, three-terminal voltage regulator IC can be added.
This circuit is simple and works as long as the weld is correct. The circuit does not have isolation measures, the use of the power of the L-line, n-line do not connect the wrong.

Working principle and calculation of capacitive buck