I would like to ask if L1, C1, and C2 in the figure form a π-type filter circuit.

西里古1992

I would like to ask if L1, C1, and C2 in the figure form a π-type filter circuit. [Copy link]

 

What is the specific purpose? What does the 0.8A on L1 mean? Isn't it already rectified by the rectifier bridge and filtered by the electrolytic capacitor in front?

maychang

In the figure, L and N usually refer to the live wire and neutral wire of the AC mains. The two wires after the bridge rectifier are generally not marked with L and N.

maychang

In the figure, L1 is marked as "12uH, about 0.8A". If the rectifier bridge is rectifying the industrial frequency AC voltage, then the 12uH inductor will not play any filtering role at all - the inductance is too small, and the inductive reactance at the industrial frequency is insignificant.

maychang

In the figure, a 24V voltage regulator is connected in parallel to the positive and negative output ends of the rectifier bridge, which is absolutely not allowed. If the instantaneous value of the voltage across the capacitor marked "electrolytic capacitor" in the figure is greater than 24V, then this voltage regulator is easy to be damaged. If the instantaneous value of the voltage across the electrolytic capacitor is always less than 24V, then this voltage regulator is useless. Therefore, it is speculated that this circuit is randomly drawn. Then L1 may be added at random.

ddllxxrr

Yes, it is to make the voltage more stable and cleaner

西里古1992

maychang posted on 2018-8-25 11:33 In the figure, L and N usually refer to the live wire and the neutral wire of the AC mains. The two wires after the bridge rectifier are generally not marked with L and N.

Yes, amateur, it's already DC

西里古1992

maychang posted on 2018-8-25 12:10 In the figure, a 24V voltage regulator is connected in parallel to the positive and negative output ends of the rectifier bridge, which is absolutely not allowed. If the voltage across the capacitor marked "electrolytic capacitor" in the figure...

Isn't the voltage regulator used to stabilize the voltage when it exceeds the instantaneous voltage? Why does it break? ! The original picture is this, a three-pin plug power supply circuit

西里古1992

maychang posted on 2018-8-25 11:33 In the figure, L and N usually refer to the live wire and the neutral wire of the AC mains. The two wires after the bridge rectifier are generally not marked with L and N.

I would like to ask about the capacitor step-down circuit. I chose a 1.8uF capacitor. After reading the relevant information, the power frequency is 50Hz, the half-wave rectifier 1uF output current is about 30mA, and the full-wave current output is 60mA. The rectifier is a bit amateurish. What kind of rectifier does this circuit (the complete one given later) belong to? Bridge rectifier?

maychang

Sirigu1992 posted on 2018-8-26 13:38 I want to ask about the capacitor step-down circuit. I choose 1.8uF capacitor. After reading the relevant information, the power frequency is 50Hz, the half-wave rectifier 1uF output current is about 30mA, and the whole wave...

What kind of rectification does this circuit (the full one given later) belong to? Bridge rectification? It is bridge rectification. The approximate calculation that each uF capacitor outputs about 60mA DC is applicable to this.

maychang

Sirigu1992 posted on 2018-8-26 13:32 Isn't the voltage regulator used to stabilize the voltage when the voltage exceeds the instantaneous voltage? Why does it break? ! The original picture is this, a three-pin plug power supply circuit

AC mains, capacitor voltage reduction, it is best not to take a voltage regulator and connect it directly in parallel with the positive and negative output ends of the rectifier bridge. For this capacitor step-down circuit, the steady-state current is not large after the power supply is connected (60mA per uF as you said on the 8th floor), but the instantaneous current when the power is turned on may be very large (related to the AC mains phase at the time of closing the switch), up to 200A or even larger, and the duration is very short, less than 1ms. If there is no parallel capacitor at both ends of the Zener diode, 1ms is enough to burn the Zener diode. To avoid burning the Zener diode, a resistor should be connected in series before the Zener diode on the AC or DC side of the rectifier bridge to limit the current spike that may occur at the moment of closing the switch.

maychang

Sirigu1992 posted on 2018-8-26 13:32 Isn't the voltage regulator used to stabilize the voltage when the voltage exceeds the limit? Why does it break? ! The original picture is this, a three-pin power supply circuit

It is best not to use this capacitor voltage reduction circuit, especially for beginners. This circuit is too dangerous and can easily cause electric shock and personal injury.

廖颂文

It is a π-type filter circuit

西里古1992

Sirigu1992 posted on 2018-8-26 13:32 Isn't the voltage regulator used to stabilize the voltage when the voltage exceeds the limit? Why does it break? ! The original picture is this, a three-pin power supply circuit

There is a varistor in front of my rectifier bridge, which should be used to prevent instantaneous overvoltage and protect the subsequent circuit! Also, my company has produced this circuit board, and my boss asked me to do unit testing. Is it dangerous?

qwqwqw2088

Sirigu1992 posted on 2018-8-26 13:32 Isn't the voltage regulator used to stabilize the voltage when the voltage exceeds the limit? Why does it break? ! The original picture is this, a three-pin power supply circuit

It is necessary for the technical department of an electronics company to have an isolation transformer. There is no danger when using an isolation transformer for testing.

maychang

Sirigu1992 posted on 2018-8-27 09:38 There is a varistor in front of my rectifier bridge, which should be used to prevent instantaneous overvoltage and protect the subsequent circuit! In addition, my circuit board company made it...

The circuit board has been made? In other words, the circuit has been determined? Of course, it is dangerous to connect it directly to the mains. According to what qwqwqw2088 on the 14th floor said, add an isolation transformer and experiment again.

topwon

And don't do a no-load test? The RC step-down power supply is actually a constant current source. If there is no load, the current 1.8x60mA=108mA calculated according to the capacitance value can only flow through the voltage regulator tube. In this way, the power consumption of the voltage regulator tube should be very large, so pay attention to its parameters and temperature.

topwon

Therefore, when using this kind of power supply, you must pay attention to the changes in the load and consider whether the component parameters are suitable under the minimum and maximum load limits. The set current should not be too large or too small. Therefore, you need to pay more attention when using this kind of power supply. The more stable the load, the better.

maychang

topwon posted on 2018-8-27 10:21 Also, don't do the no-load test? The RC step-down power supply is actually a constant current source. If there is no load, the current calculated according to the capacitance value is 1.8x60mA=108 ...

The first circuit, the original design uses a 5W voltage regulator tube, and the power dissipation of the tube is 0.1A*24V=2.4W when no-load. The original design fully considers the steady-state power dissipation.

西里古1992

maychang posted on 2018-8-27 10:10 Has the circuit board been made? In other words, the circuit has been determined? It is expected that when you turn on the AC power for the experiment, nothing may happen once or twice...

It is also possible to use an AC voltage regulator

西里古1992

maychang posted on 2018-8-27 10:10 Has the circuit board been made? In other words, the circuit has been determined? It is expected that when you turn on the AC power for the experiment, nothing may happen once or twice...

It is also possible to use an AC voltage regulator