The flyback switching power supply designed here will have a flashback problem after power is turned off when the output end is fully loaded.

西里古1992

The flyback switching power supply designed here will have a flashback problem after power is turned off when the output end is fully loaded. [Copy link]

 

 

The above is the main circuit of the power supply, which is built with the UC2844 chip. When the +5V voltage output is fully loaded, a pulse will appear after power is turned off, and the other output windings are also affected by this and there is such a re-powering.

Then we tested the VCC pin supply voltage waveform and DC bus voltage P waveform of the UC2844 chip and found that VCC did not drop continuously after power off, but had a voltage waveform similar to rebound, which happened to coincide with the output voltage recovery pulse. What is the reason for this?

maychang

You posted a thread with the same content in the simulation forum. In the 22nd post of that thread, I asked you: "Do you know the startup process of the UC3842 series chip?" You did not answer.

maychang

On the 23rd floor of that post, I also said: "Understand the process of how the chip starts up, so I can explain to you the waveform on the 18th floor and why it restarts."

西里古1992

maychang posted on 2024-8-19 11:11 You posted a post with the same content in the simulation forum. In the 22nd post, I asked you: "Do you know the startup process of the UC3842 series chip?" ...

The problem I have now occurred after power failure, not during the startup phase. This chip is basically powered by AC rectifiers, then stepped down through several resistors and then supplied to the VCC pin for pre-startup. After the chip is working, it outputs a PWM waveform, which is then powered by pins 7 and 8 of the transformer.

maychang

Sirigu1992 posted on 2024-8-19 11:14 My problem now occurs after power failure, not during the startup phase. This chip basically rectifies the AC power and then steps it down through several resistors to VCC...

It seems that it is known. Then, let's talk about the waveform in the third picture of this post in detail. Wait for me to draw the time in the third picture.

tagetage

I can think of 2 points.

1. Is your load a pure resistor?

2. Many manufacturers make UC2844. If you can't do it, try changing to one from a big manufacturer.

西里古1992

tagetage posted on 2024-8-19 11:54 I can think of two points: 1. Is your load a pure resistor? 2. Many manufacturers make UC2844. If you can't do it, change to one from a big manufacturer.

It is a pure resistive load

maychang

This is the original waveform you posted.

I marked several moments in the figure, namely t1, t2, ... t6. Before t1, the green waveform (DC bus voltage) has been zero for at least 10 seconds. Your flyback switching power supply has completely stopped working.

At time t1, the DC bus voltage rises extremely quickly, reaching about 560V. This can only be because the mechanical switch is turned on again, because the capacity of C22 and C23 is quite large, reaching 150uF. Except for AC mains, other equipment cannot charge such a large capacitor in such a short time.

maychang

Sirigu1992 posted on 2024-8-19 11:14 My problem now occurs after power failure, not during the startup phase. This chip basically rectifies the AC power and then steps it down through several resistors to VCC...

At time t2, the DC bus voltage charges C40 through four 75 kΩ resistors, and VCC starts to rise. At this time, the chip has not started to work, of course, the power switch tube is not working (turned off), and the transformer has no output.

At time t3, C40 is fully charged, the chip starts to work, the power switch tube works, the transformer works, and pins 7 and 8 supply power to C40 through diode D5. Your mechanical power switch may have been turned off because there is a small drop in the DC bus voltage.

From time t3 to time t4, the chip works, the power switch works, and the four secondary terminals of the transformer have normal output. During this period, your flyback switching power supply works by the energy stored in C22 and C23. Since the capacity of C22C23 is quite large, your flyback switching power supply works for another 4.5 seconds from the waveform.

maychang

Sirigu1992 posted on 2024-8-19 11:14 My problem now occurs after power failure, not during the startup phase. This chip basically rectifies the AC power and then steps it down through several resistors to VCC...

At time t4, the energy stored in C22C23 has been almost released, and the voltage at both ends drops rapidly. However, during the period from t4 to t5, VCC is still maintained, the chip is still outputting, and the power switch tube is also intermittently turned on. At time t5, the chip stops working, the power switch tube is turned off, and the transformer has no output (and no longer supplies power to the chip), so the DC bus voltage is maintained at a level slightly less than 200V.

maychang

Sirigu1992 posted on 2024-8-19 11:14 My problem now occurs after power failure, not during the startup phase. This chip basically rectifies the AC power and then steps it down through several resistors to VCC...

From time t5 to time t6, the switching power supply does not work and does not supply power to VCC, so VCC drops rapidly.

But there is still nearly 200V voltage across C22C23, and this voltage will charge C40 through four resistors. But because the voltage across C22C23 is lower than 560V at this time, the slope of the VCC charging voltage rising curve is relatively small. You can see this from the slopes of the curves in sections A and B in the figure.

When VCC rises to a certain level, the chip starts working again, but at this time the DC bus voltage is quite low. Although the power switch is turned on for a while, the voltage of each winding of the transformer is quite low and cannot supply power to VCC, so VCC drops immediately. The smaller slope of the VCC charging curve at point C illustrates this point.

maychang

Sirigu1992 posted on 2024-8-19 11:14 My problem now occurs after power failure, not during the startup phase. This chip basically rectifies the AC power and then steps it down through several resistors to VCC...

Regarding your "flashback" problem, I judged on the 8th floor that it was caused by the mechanical switch being briefly turned on. Whether this is true or not can be discussed later.

西里古1992

maychang posted on 2024-8-19 16:58 From time t5 to time t6, the switching power supply does not work, and does not supply power to VCC, so VCC drops rapidly. However, there is still a voltage of nearly 200V across C22C23. At this time...

OK, I understand the process and reasons. Let me explain again. I powered on the power supply from t1 to t3, and the power supply was turned off from t4 to t6. The power supply was running in the middle, and it was not turned off at the beginning of t1. Is there any solution for this residual voltage to continue charging the capacitor?

西里古1992

maychang posted on 2024-8-19 16:32 At time t4, the energy stored in C22C23 has been almost released, and the voltage at both ends drops rapidly, but from t4 to t5, VCC is still maintained and the chip still outputs...

Yes, teacher, I changed the starting resistors R26, R91 and other four resistors from 75K to 100K, and the capacitance of C40 from 47uF to 100uF, and there is no such problem. I feel that it is caused by increasing the starting capacitor capacity and slowing down the charging current speed. However, a new problem arises that the chip hiccups protection or the machine does not start up after the power is turned on occasionally, so this method is abandoned. Can I change the resistance of the four balancing resistors R100 in parallel with the electrolytic capacitor to be smaller and increase the capacitor discharge speed?

maychang

Sirigu1992 posted on 2024-10-12 09:50 Yes, teacher, I changed the starting resistors R26, R91 and other four resistors from 75K to 100K, and the capacitance of C40 from 47uF to 100uF, and there was no such...

[Can I change the 4 balancing resistors R100 and other resistors in parallel of the electrolytic capacitor to reduce the resistance and increase the capacitor discharge speed?]

Reducing the four resistors such as R100 will increase the power dissipated by the switching power supply during normal operation, reduce efficiency, and increase heat generation. Therefore, reducing the four resistors such as R100 is not advisable.

maychang

Sirigu1992 posted on 2024-10-12 09:50 Yes, teacher, I changed the starting resistors R26, R91 and other four resistors from 75K to 100K, and the capacitance of C40 from 47uF to 100uF, and there was no such...

The 2844 chip does not work after the switching power supply is powered on, and the current consumption is very small, generally less than 1mA. After power-on, C40 is charged by four resistors such as R26, and the voltage at both ends gradually increases. After the voltage at both ends of C40 reaches 16V, the chip starts to work and outputs PWM pulses. At this time, the current consumed by 2844 is much larger, from more than ten mA to tens of mA (so the voltage at both ends of C40 is reduced, but 2844 can still work), and the flyback switching power supply starts to work. After the flyback switching power supply starts to work, C40 is charged by the transformer 7/8 pin winding through the diode D5 rectification to maintain the chip output PWM pulses. After the flyback switching power supply starts to work, the four resistors such as R26 are useless, but for the simplicity of the circuit, there is no need to remove them, anyway, these four resistors do not consume much power.

maychang

Sirigu1992 posted on 2024-10-12 09:50 Yes, teacher, I changed the starting resistors R26, R91 and other four resistors from 75K to 100K, and the capacitance of C40 from 47uF to 100uF, and there was no such...

After the switching power supply is powered off, the AC mains stops charging C22C23 (charging once every 10ms during normal operation), but the switching power supply still outputs power to the load, so the voltage across C22C23 decreases, the switching power supply output also decreases, the winding voltage between the 7/8 pins of the transformer also decreases, and the voltage across C40 also decreases. When the voltage across C40 drops to 10V, the chip stops working. The chip starts working when it is powered by 16V and stops working at 10V, which is determined when the chip is designed.

maychang

Sirigu1992 posted on 2024-10-12 09:50 Yes, teacher, I changed the starting resistors R26, R91 and other four resistors from 75K to 100K, and the capacitance of C40 from 47uF to 100uF, and there was no such...

However, when the chip supply voltage (C40) drops below 10V, C22C23 is not completely discharged, and the voltage across the two ends may be quite high (but it cannot maintain the normal operation of the switching power supply). Therefore, after the switching power supply stops working, C22C23 will charge C40 through four resistors such as R26. When the voltage across C40 reaches 16V, the chip starts to output PWM pulses again, and the switching power supply works. But at this time, the voltage across C22C23 can no longer make the switching power supply output enough voltage to charge C40 through the winding between pins 7/8. When the voltage across C40 drops to 10V, the chip stops working and the switching power supply stops working. If the residual voltage on C22C23 is higher than 16V, C40 will be charged again (of course, it will take much longer to charge to 16V). The voltage across C40 rises to 16V, the chip works again, and the switching power supply works again. This is the reason why your flyback switching power supply has "flashback" after power off.

maychang

Sirigu1992 posted on 2024-10-12 09:50 Yes, teacher, I changed the starting resistors R26, R91 and other four resistors from 75K to 100K, and the capacitance of C40 from 47uF to 100uF, and there was no such...

Obviously, if you increase the capacity of C40, the chip will keep working longer after power off, the switching power supply will work longer, and the capacitors C22C23 will discharge more thoroughly (the voltage drops more), so the "flashback" disappears.

As for the "failure to start up" after changing the four resistors such as C40 and R26, it is probably because you increased the four resistors such as R26, which caused C40 to charge for too long. It should be noted that the chip does not consume no current after power-on, but consumes about 1mA. Your DC power supply is said to be more than 500V. After the four resistors such as R26 are changed to 100 kilo-ohms, the charging current for C40 after power-on is 1.25mA (500V/400 kilo-ohms). If the chip consumes 1.25mA, then your flyback switching power supply will never start up.

西里古1992

maychang posted on 2024-10-12 11:58 Obviously, if you increase the capacity of C40, the chip will maintain working time longer after power off, and the switching power supply will also work longer. Capacitors C22C23 ...

Is there a best of both worlds solution, such as reducing the capacitance of C22 and C23?