Analysis of non-isolated buck switch power supply circuit for LED lighting

In LED lighting applications, non-isolated BUCK switching power supplies are favored by many customers due to their high efficiency, and they occupy a part of the market share. Represented by SN3910, there are a series of ICs with similar functions on the market, such as BP2808, SMD802, AM850, SN3910, etc. In terms of price, some domestic manufacturers' ICs have dropped to more than two yuan. Therefore, for low-cost applications, it is still a reasonable choice. Next, take SN3910 as an example to illustrate the application circuit and ideas of this type of IC.

Figure 1 Buck LED circuit diagram (BUCK) from

Working principle:

1. When Q1 is turned on, the input current Iin passes through the load LED, inductor L1, Q1 to the negative pole of the input power supply. While the LED is emitting light, the current in the L1 inductor slowly rises and reaches a peak value until Q1 is turned off and L1 stores energy.

2. When Q1 is disconnected, due to the principle that "the voltage across the capacitor cannot change suddenly, and the current flowing through the inductor cannot change suddenly", the current flowing through the L1 inductor passes through the freewheeling diode D1 and the load LED to form a loop. The current in the inductor drops from the peak value to a value (if the value is greater than zero, Q1 is turned on, it works in CCM; if it is equal to zero, Q1 is turned on immediately, it works in BCM; if it is equal to zero, Q1 is not turned on in time, it works in DCM) until Q1 is turned on.

Note: Most buck circuits are designed to work in CCM because of the following two advantages:

1. Working in CCM, the output ripple current is relatively small.

2. Working in CCM, the output current is easier to control: Io=(ILpkh+ILpkl)/2

Here Io is the effective value of the output current, ILpkh and ILpkl are the peak and valley values ​​of the inductor current respectively.

From the DATAsheet, we can see that the internal VCS voltage is 250mV. By setting the resistance from the CS pin to ground, the peak current through the load LED can be set. So how is the constant current achieved?

From the formula Io=(ILpkh+ILpkl)/2, we can know that if ILpkh can be set through RCS, the problem will be solved.

A capacitor from the TOFF pin to the ground is used to set TOFF (off time). In each working cycle, there is the same off time. From the discharge loop of the inductor, it can be seen that if the current through the inductor does not drop to 0 within the TOFF time (CCM), then Io=(ILpkh+ILpkl)/2 can obtain a constant ILpkl, thus achieving the purpose of constant current.

Now we can consider how to set the inductor L1, RCS, and COFF. Let's take a commonly used example to illustrate:

INPUT: 85~265Vac 50/60HZ

OUTPUT: 40V 0.35A (12LED)

First: set TOFF. The following formula can be found in the data sheet.

Assume: COFF=220, then: Toff=11.73uS.

Assuming the output current ripple factor is 0.8, then:

RCS=VCS/IPKH=0.25/(0.35+(0.35*0.8)/2)=0.51R

Calculate the inductance:

L is equal to 40*11.73/(0.35*0.8)=1.676mH