Calculation and selection of power input rectifier bridge and filter capacitor

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Calculation and selection of power input rectifier bridge and filter capacitor [Copy link]

Input rectifier bridge selection

1) The conduction time and gating characteristics of the rectifier bridge

After full-wave rectification, the 50Hz AC voltage becomes a pulsating DC voltage u1, and then passes through the input filter capacitor to obtain the DC high voltage U1. Ideally, the conduction angle of the rectifier bridge should be 180° (the conduction range is from 0° to 180°), but due to the effect of the filter capacitor C, the input current flows through the rectifier bridge to charge C only in a very short time close to the AC peak voltage. The half cycle of 50Hz AC is 10ms, the conduction time of the rectifier bridge tC≈3ms, and its conduction angle is only 54° (the conduction range is 36° to 90°). Therefore, the rectifier bridge actually passes a narrow pulse current. The principle of the bridge rectifier filter circuit is shown in Figure 1(a), and the waveforms of the rectifier filter voltage and the rectifier current are shown in Figures 1(b) and (c), respectively.

Summary:

(1) The above characteristics of the rectifier bridge can be equivalent to a duty cycle of about 30% corresponding to the input voltage frequency.

(2) The primary conduction process of the rectifier diode can be regarded as a "selection pulse", and its pulse repetition frequency is equal to the frequency of the AC power grid (50Hz).

(3) In order to reduce the conducted noise below 500kHz in the switching power supply, two ordinary silicon rectifiers (such as 1N4007) and two fast recovery diodes (such as FR106) are sometimes used to form a rectifier bridge. The reverse recovery time trr of FRl06 is ≈250ns.

2) Parameter selection of rectifier bridge

The isolated switching power supply generally uses a rectifier bridge composed of rectifier tubes, or a finished rectifier bridge can be directly selected to complete the bridge rectification. The full-wave bridge rectifier is referred to as a silicon rectifier bridge. It is a semiconductor device that connects four silicon rectifier tubes into a bridge circuit and then encapsulates them in plastic. It has the advantages of small size, easy use, and good parameter consistency of each rectifier tube. It can be widely used in the rectifier circuit of the switching power supply. The silicon rectifier bridge has 4 lead-out terminals, two of which are AC input terminals and two of which are DC output terminals.

The average value of the maximum rectifier current of the silicon rectifier bridge is divided into various specifications such as 0.5~40A, and the maximum reverse working voltage is divided into various specifications such as 50~1000V. Low-power silicon rectifier bridges can be directly soldered on the printed board, while large and medium-power silicon rectifier bridges must be fixed with screws and a suitable heat sink must be installed.

The main parameters of the rectifier bridge are reverse peak voltage URM (V), forward voltage drop UF (V), average rectified current Id (A), forward peak surge current IFSM (A), and maximum reverse leakage current IR (μA). The reverse breakdown voltage URR of the rectifier bridge should meet the following requirements:

For example, when the AC input voltage range is 85-132V, umax=132V, and UBR=233.3V is calculated by formula (1). A finished rectifier bridge with a withstand voltage of 400V can be selected. For a wide range of input AC voltage, umax=265V, and UBR=468.4V is obtained by the same logic. A finished rectifier bridge with a withstand voltage of 600V should be selected. It should be pointed out that if four silicon rectifier tubes are used to form a rectifier bridge, the withstand voltage value of the rectifier tube should be further improved. For example, 1N4007 (1A/1000V) and 1N5408 (3A/1000V) plastic-encapsulated rectifier tubes can be selected. This is because the price of such tubes is low, and according to the principle of "higher rather than lower" withstand voltage value, the safety and reliability of the rectifier bridge can be improved.

Assuming the input effective current is IRMS and the rated effective current of the rectifier bridge is IBR, IBR should be ≥ 2IRMS. The formula for calculating IRMS is as follows:

In the formula, PO is the output power of the switching power supply, η is the power efficiency, umin is the minimum value of the AC input voltage, cosφ is the power factor of the switching power supply, and cosφ=0.5~0.7 is allowed. Since the rectifier bridge actually passes a narrow pulse current instead of a sinusoidal current (see Figure 1), the average rectified current of the rectifier bridge Id<IRMS, and the IAVG value can generally be calculated as Id=(0.6~0.7)IRMS.

For example, design a 7.5V/2A (15W) switching power supply with an AC input voltage range of 85-265V and a requirement of η=80%. Substituting Po=15W, η=80%, umin=85V, and cosψ=0.7 into equation (2), we get IRMS=0.32A, and then Id=0.65×IRMS=0.21A. In practice, a 1A/600V rectifier bridge is selected to leave a certain margin.

Input filter capacitor selection

1) Selection of input filter capacitor capacity

In order to reduce the output ripple of the rectifier filter, the capacity CI of the input filter capacitor must be selected appropriately. Let the proportional coefficient of the input filter capacitor capacity (μF) required per unit output power (W) be k. When the AC voltage u=85～265V, k=（2～3）μF／W should be taken; when the AC voltage u=230V（1±15％）, k=1μF／W should be taken. The selection method of the input filter capacitor capacity is detailed in Appendix 1, Po is the output power of the switching power supply.

2) Method for accurately calculating the capacity of input filter capacitor

The capacity of the input filter capacitor is an important parameter of the switching power supply. If the CI value is too low, the UImin value will be greatly reduced, while the input pulsating voltage UR will increase. However, if the CI value is too high, the capacitor cost will increase, and the effect of increasing the UImin value and reducing the pulsating voltage will not be obvious. The following introduces the method of calculating the accurate value of CI.

Assume that the minimum value of the AC voltage u is umin. After bridge rectification and CI filtering, the input voltage waveform under the condition of u=umin is shown in Figure 2. This figure is drawn under the conditions of Po=POM, f=50Hz, conduction time of the rectifier bridge tC=3ms, and η=80%. It can be seen from the figure that a primary side pulsating voltage with an amplitude of UR is superimposed on the minimum value of the DC high voltage UImin, which is formed by CI during the charging and discharging process. To obtain the accurate value of CI, it can be calculated as follows:

For example, when the input voltage is in a wide range, umin = 85 V. Take UImin = 90 V, f = 50 Hz, tC = 3 ms, assume Po = 30 W, η = 80%, and substitute them into (3) to obtain CI = 84.2 μF, and the proportionality coefficient CI/PO = 84.2 μF/30 W = 2.8 μF/W, which is just within the allowable range of (2-3) μF/W.

The switching power supply determines the primary side filter capacitor, and the main parameters that affect the power supply performance are the output low-frequency AC ripple and the hold time. The larger the filter capacitor, the higher the Vin (min) on the capacitor, and the higher the power output, but the relative price is also higher.

3) Calculation method of input electrolytic capacitance (with examples):

A. Since the output voltage is 12V and the output current is 2A, the output power is:

Pout=Vo*Io=12.0V*2A=24W

B. Assuming the transformer conversion efficiency is about 80%, the input power of a power supply with an output power of 24W is:

Pin = Pout / efficiency = 24W / 80% = 30W

C. Since the minimum input AC voltage is 90VAC, the DC output voltage is:

Vin=90* √2=127Vdc

Therefore, the load DC current is:

I=Pin/Vin=30W/127VDC=0.236A

If the power level requirement is higher, the following parameters can be considered for calculation; since the minimum input AC voltage is 90VAC, the minimum output DC voltage is:

Vin(min)=90*√2 -30(DC ripple voltage)=97Vdc

Therefore, the maximum load DC current is:

IMAX=Pin/Vin（min）=30W/97VAC=0.309A

D. The design allows a DC ripple voltage of 30VPP, and the capacitor must maintain the voltage for half a cycle t, that is, the half-cycle of the power frequency AC voltage is about 8ms.

T=1/f=1/60=0.0167S=16.7 ms

but:

C=I*t/V=(0.236*8*10^3)/30≈62uF

62uF is between the commonly used capacitors 47-82uF. Due to cost considerations, the actual capacitance selected is 47uF.

E. Since the maximum input AC voltage is 264Vac, the maximum DC voltage is:

V=264*√2=373VDC

In practice, a general-purpose electrolytic capacitor with a withstand voltage of 400Vdc is selected. At this voltage level, the capacitor has a 95% margin.

F. The ripple current value that the capacitor can withstand determines the temperature rise of the capacitor, and thus determines the life of the capacitor. (The maximum ripple current value of the capacitor is related to its volume and material. The larger the volume, the better the heat dissipation and the higher the ripple current value.) Therefore, when selecting a capacitor, it is necessary to consider that the actual ripple current value is less than the maximum ripple current value of the capacitor.

G. The temperature rise of switching source components is generally high, and 105℃ capacitors are usually selected. In special circumstances where the temperature rise cannot be overcome, 125℃ capacitors can be selected.

Therefore, the selection of 47uF, 400v, 105℃ electrolytic capacitor can meet the requirements (in actual use, the installation structure size, body size, heat dissipation environment, etc. should also be considered).