UC3842 detailed knowledge points of application circuits

Source: Internet
Author: User


Figure 1 shows the UC3842 internal diagram and pin diagram. UC3842 adopts a controllable modulation method with fixed working frequency pulse width, with a total of 8 pins. The functions of each pin are as follows:

① The foot is the output end of the error amplifier, and the external resistance element is used to improve the gain and frequency characteristics of the error amplifier;

② The foot is the feedback voltage input end. The foot voltage is compared with the 2.5 V reference voltage at the same phase end of the error amplifier to generate an error voltage to control the pulse width;

③ The foot is the current detection input. When the detection voltage exceeds 1 V, the pulse width is reduced so that the power supply is in intermittent operation;

④ The timing end. The operating frequency of the internal oscillator is determined by the external resistance time constant. f = 1.8/(RT × CT );

The fifth step is the public ground;

6. The foot is the push-pull output end, and the inside is the totem column. The rising and dropping time is only 50ns. The driving capacity is ± 1A;

The pin is a DC power supply end, which has the function of under-and over-voltage locking and the chip power consumption is 15 MW;

The foot is the output end of the 5 V reference voltage and has a load capacity of 50mA.


5 linear photocoupler is equivalent to a variable resistance. The larger the current, the smaller the resistance.

UC3842 feet 2 directly grounded to make the error amplifier (1, 2 feet) saturated

The error amplifier is a comparison amplifier with a front-level amplifier and a rear-level amplifier output end controlled by a constant current source of each 1mA

Figure 5 the magnification of the error amplifier does not make sense here the amplifier saturated output 6 V output end 1 Pin connected to the photocoupler and then grounded

It's a shunt. I personally don't think it's a current feedback circuit on the Forum. 6

This current feedback circuit is not a current feedback circuit because there is a ground resistance missing at the foot 2 end. You can refer to the books and use the current negative feedback circuit comparison.


Tl431 and pc817 have been mentioned in previous blog posts.

Step 3 is used for current sampling: the smaller the sampling resistance, the larger the protection current, the larger the output protection power, the larger the allowable output power.


The above experience from the Forum: http://www.21dianyuan.com/bbs/5312.html


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