This circuit cannot maintain constant current. Please analyze it.

灞波儿奔

This circuit cannot maintain constant current. Please analyze it. [Copy link]

The following is a constant current circuit diagram of a charging circuit. During debugging, it was found that the 5th pin of the dual op amp IC could not be stabilized. After disconnecting R35, it became stable. Please tell me what is the reason.

To facilitate the analysis of teachers, the op amp IC pin diagram is posted below

The 8-pin VCC power supply input of the op amp IC, the 4-pin is GND, the 2-pin voltage is compared with the 3-pin voltage, and the 6-pin voltage is compared with the 5-pin voltage, corresponding to two independent outputs: 1OUT and 2OUT

maychang

There is a line on the top of the electrical schematic and two lines on the right are outside the diagram. Where do these three lines go?

In a constant current circuit, the current on the load must remain constant. Where is the load?

maychang

There is no need to post the pins of LM358. They are clearly shown in the schematic diagram. However, the most important load part is not in the schematic diagram.

chunyang

The circuit is incomplete and cannot be analyzed.

Gen_X

Without seeing your "constant current circuit", I cannot make the correct judgment, let alone help you answer the question.

灞波儿奔

This post was last edited by Baboerben on 2022-2-15 09:05

The circuit diagram has been uploaded, please reply

maychang

Just post the complete electrical schematic, that's all. Only then can your question be answered.

maychang

The upper left part of this figure is a single-ended flyback switching power supply, with control chip U1 and power tube Q1 as the core. Transformer T1 has three sets of secondary: 3-4 pins, 7-6 pins and 7-8 pins, which output three sets of DC voltage respectively.

maychang

From the connection between the optocoupler and the voltage reference U2, the flyback switching power supply has a stable voltage output, and the stable voltage is the one corresponding to pins 7 to 6. The output voltage of this circuit, that is, the voltage across capacitor C4, is applied to the optocoupler light-emitting tube side through a resistor, and the photoelectric tube side of the optocoupler is connected to pin 2 of the control chip to control the duty cycle of the pulse driving Q1. This is a fairly typical circuit.

maychang

On the transformer pins 7 to 8, the output current flows through a small resistor R25 to obtain a small voltage proportional to the output current.

This voltage is applied to the input of U4A and U4B through resistor R28. Both U4A and U4B work in open loop with very high gain. U4A gets the reference voltage from U3 through R21 and R26, and U4B gets the reference voltage from U3 through R22 and R30. Both reference voltages are quite small, and you can calculate the specific value yourself.

maychang

The output of U4B is connected to the output of voltage reference U2 via R35 and D6.

Note that the sampling resistor R25 is connected in series with the transformer pins 7 to 8. Therefore, the action of R25 to U4B is: the voltage across R25 (proportional to the output current of pins 7 to 8) is higher than the voltage reference of U4B (voltage divided by R22 and R30) --> the output of U4B is low --> the current in the optocoupler increases --> the pulse duty cycle of the control chip U1 output decreases --> the output of the switching power supply decreases.

maychang

Judging from the open-loop operation of U4B in the circuit and the values of several capacitors in the circuit, this feedback is unlikely to be stable, that is, it is unlikely to be "constant current" as mentioned in the first post. Its operation is mostly: when the output (C12 ends) is short-circuited or overcurrent occurs, the switching power supply stops outputting, and after the output current decreases, the switching power supply continues to output, then overcurrent occurs, and then stops outputting... This is a "hiccup" type overcurrent protection.

maychang

Back to the question in the first post: "When debugging, I found that the 5th pin of the dual op amp IC was not stable. It became stable after disconnecting R35."

As mentioned above, the function of U4B is to turn off the output of the switching power supply when the output is overcurrent (in extreme cases, short circuit). When U4B is in action, the switching power supply has no output, and U3 certainly has no output, so R22 and R30 cannot divide any voltage (this is what you call pin 5). When the switching power supply resumes working, U3 is powered, and the voltage division of R22 and R30 is normal. Therefore, "the voltage of pin 5 cannot be stabilized."

maychang

After R35 is disconnected, the entire overcurrent protection circuit is cancelled, the switching power supply will not stop working, and of course the voltage at pin 5 of U4B can be stabilized.

maychang

In this circuit, a diode D10 is connected in series in the main output loop. I don't know what the designer's intention is in connecting this diode. My opinion: it is completely redundant.

Gen_X

I believe that D10 is designed to provide a purer "ground" level for the output, so that the output contains less AC high-frequency common mode (this is discovered and learned from actual applications, and it plays a simple isolation role between the AC circuit and the output DC circuit, and its actual role cannot be underestimated).

Small component, big effect, a masterpiece by a master!

Gen_X

In addition, if there is no D10, the sampled value on R25 will be disturbed by the external circuit and the circuit will be unstable.

灞波儿奔

Thanks for the advice

It is indeed interference to the external circuit

Solved. The Y capacitor was directly connected to the ground of 358, which caused interference to 358. It was fixed by removing the Y capacitor. Now the Y capacitor is connected directly to the ground terminal of the transformer in the circuit, and it is normal.

This circuit cannot maintain constant current. Please analyze it. [Copy link]

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