Low power LED driver power supply circuit design

One-resistance buck

1. Principle and application of resistance to voltage reduction

Capacitor bucking is actually using capacitive reactance current limiting, and the capacitor actually acts as a limiting current and dynamically distributing the voltage across the capacitor and the load.

2. Pay attention to the following points when using capacitor step-down

According to the current of the load and the working frequency of the alternating current, the appropriate capacitor is selected instead of the voltage and power of the load. The current limiting capacitor must use a non-polar capacitor and cannot use an electrolytic capacitor. Moreover, the withstand voltage of the capacitor must be above 400V, and the most ideal capacitor is a polypropylene metal film capacitor. Capacitor buck can not be used for high power conditions, generally used for low power applications below 5W. Capacitor buck is not suitable for dynamic load conditions. Capacitor buck is not suitable for capacitive and inductive loads. It is suitable for driving in LED power supply. Voltage application.

3. The basic circuit of the RC capacitor is as shown in (Figure 1).

C1 is a step-down capacitor, VD1, 2, 3, 4 are bridge rectifier diodes, VD5 is a Zener diode, and R1 is the charge bleeder of C1 after the power is turned off.

4. Device selection

When designing the circuit, the exact value of the load current should be measured first, and then refer to the example to select the capacity of the step-down capacitor. Since the current Io supplied to the load through the step-down capacitor C1 is actually the larger the capacity of the charge and discharge current Ic.C1 flowing through C1, the smaller the capacitive reactance Xc is, the larger the charge and discharge current flowing through C1 is. When the load current Io is less than the charge and discharge current of C1, the excess current will flow through the Zener diode. If the maximum allowable current Idmax of the Zener diode is less than Ic-Io, the regulator tube will burn out. In order to ensure reliable operation of C1, the withstand voltage selection should be greater than twice the supply voltage. The bleeder resistor R1 must be selected to vent the charge on C1 for a specified period of time.

5. Actual parameter calculation method

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 capacitor C1 is:

Ic = U / Xc = 220 / 9.65 = 22mA.

Two linear drive circuit

1. Typical circuit such as (Figure 2)

2. Working principle

R3 is a constant current resistor. The voltage drop of R3 is used to control the switch of TL432. The switch of 432 is used to control the conduction of Q1 to achieve the output constant current. The purpose of selecting 432 is to reduce the value of 432 by 1.21V. Loss on R3. The current constant current value is 1.21/R3, and the R1 selection is selected according to the amplification factor of Q1.

3. Application considerations

This circuit is recommended for single-voltage input, output current, low-current LED power supply, such as bulb, T-tube, etc. It is generally recommended that the output current be at 100mA. At the same time, the closer the output voltage is to the input, the better, so that the voltage drop of Q1 is too large, resulting in excessive loss and low efficiency. Therefore, the use of LEDs is also preferably used in series.

Three constant current diode driving circuit

1. Typical circuit such as (Figure 3, Figure 4)

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