Constant current drive circuit design: The LED drive circuit based on PT4115 is shown in Figure 1. The circuit can use Atmega8 microcontroller as the controller and set two input interfaces. The input voltage of the circuit can be DC or AC. The PWM signal is added to the DIM terminal of PT4115 to realize LED dimming, and a dimming button is set.
Figure 1 LED driver circuit based on PT4115
Improvement scheme to reduce circuit power consumption
To reduce the power consumption of the circuit feedback resistor, the most direct way is to reduce the resistance of the feedback resistor. Under the same current, according to the resistor power consumption formula, it can be known that the power consumption can be reduced by the same proportion as the resistance of the resistor. However, the reduction of the resistance of the feedback resistor means that the feedback voltage value is insufficient and the circuit will not work. Therefore, it is necessary to design an amplifier circuit to ensure that the feedback voltage remains unchanged while reducing the resistance of the feedback resistor. The amplifier circuit designed in this paper is a differential amplifier circuit, as shown in Figure 4. The voltages Vf+ and Vf- at both ends of the feedback resistor Rf are added to the non-inverting input and the inverting input of the operational amplifier respectively.
Figure 4 Feedback differential amplifier circuit
From Figure 4, the calculation formula of VK is:
(4)
From formula (4), we can know that by setting different coefficients, we can get the corresponding signal amplification factor. In Figure 4, R3=R2=10kΩ, R1=R4=100kΩ, then from formula (4), we can know:
(5)
For the energy-saving effect shown in Figure 4, a set of comparative calculations can be done. Assuming there is a conventional Buck LED drive circuit, the range of its drive current is (0, 1A), and the corresponding feedback voltage Vk range is (0, 800mV), so the size of its feedback resistor is 0.8Ω. When the drive current is 1A, the power consumption of the feedback resistor is 0.8W; from formula (5), when the range of Vk is (0, 800mV), the range of (Vf+, Vf-) is (0, 80mV), so the equivalent feedback resistor R is 0.08Ω; when the drive current is 1A, the power consumption of the equivalent feedback resistor is 0.08W, which is about 1/10 of the power consumption of the conventional Buck drive circuit. The circuit is a linear amplification, allowing a large range of operating current changes, that is, a large dimming range, and the detection signal can be accurately and smoothly transmitted to the output end.
The experiment verified that after the feedback voltage Vk was processed by the differential amplifier circuit shown in Figure 4, the feedback path loss of the LED drive circuit was significantly reduced, which was consistent with expectations. The total output efficiency of the LED drive circuit was above 95.5%. After the feedback path loss was reduced, the total output efficiency was increased by about 3.5% on average.
Improvement scheme for improving current control accuracy
From the above analysis, it can be seen that in order to improve the accuracy of the LED control drive circuit, the key is to improve the sampling accuracy of the feedback current. The direct and effective way to improve the sampling accuracy of the feedback current is to reduce the fluctuation range of the forward current. Therefore, the solution to improve the current control accuracy is mainly implemented in the design of the feedback control circuit of the drive circuit. The improvement scheme of this article is to introduce a dual-integral circuit in the feedback control unit based on the peak current control method. The dual-integral feedback circuit is shown in Figure 5.
Figure 5 Dual integral feedback circuit
In FIG. 5 , Vk is a feedback voltage signal. If combined with the improvement measures for reducing power consumption, Vk is a sampling signal after differential amplification, and Vout is output to the LED driving circuit.
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