Help me make a line power supply 50 and how to make a power supply

This one.. Do you just have one transformer? I can't do it complicated, I just need to use 4 rectifier diodes and 1 capacitor.

First of all, the transformer may have to be dismantled and rewound, the output voltage is related to the number of turns of the coil, the more turns the voltage is larger, the smaller the number of turns the voltage is smaller, the current is larger to use a thicker wire to wind, the number of turns of the primary and secondary is proportional to the voltage of the primary and secondary, so calculate how many turns you need.

The transformer comes out of AC, to become DC, to use a full-bridge rectifier circuit, you find it by yourself is very easy to find, now out of DC is DC, but the current waveform is too large, you need to connect a capacitor in parallel, the size of the capacitor is best above 1000uf. That's it.

TL431 is used as a reference voltage, your circuit is unreasonable, he can regulate the voltage, but the load capacity is extremely poor, mainly limited by the resistance value of R4. Because its resistance value is too large, it will divide a lot of voltage, and the remaining voltage is very low, so the load voltage is **, the heavier the load, the lower the voltage. You can try it, if you connect a 100k resistor as a load, the output voltage is still stable at 4V, but if you reduce the load resistance to a certain extent, the output voltage will start to **.

The power supply should not only have an output voltage, but also the ability to carry a load, that is, how much current it can output.

The maximum output current of your circuit is the supply voltage divided by the resistance of R4. That is, assuming that the load resistance value is zero (the heaviest load), how much current can your power supply provide, here R4 limits this current value, because the load current must pass through R4, and this circuit 5V can only provide 5mA at most after R4! R5R6 and U3 have to be divided into parts, how much is left for the load?

Think about it, don't you?

No!

When the output current changes, the voltage at R5 and R6 changes, and the control voltage of TL431 also changes. The output voltage also changes.

You can connect a resistor in parallel at the output and change the resistance of the resistor, and you can see that the output voltage will change accordingly.

The correct thing to do is to move R4 to the right of R5 and R6 and to the left of U3.

Fine-tuning R6 will do the trick. It should be noted here that the load capacity of R4 is too poor, and if you need to stabilize 4V, you can change R4 to 13 ohms. In this way, the current of U3 can be controlled within 100 mA even when there is no load. After changing it, you can readjust the R6.

The minimum current of TL431 cannot be less than 1mA, and your R4 is 1K, which can only provide 1mA of current. The correct way is to select R4 according to the negative current, e.g. a few hundred ohms to tens of ohms.

Professionals know at a glance that it can't be done, it can only be regarded as a voltage-limiting circuit, and it can't be loaded.

Yes, but the load should not be too heavy.

The first tube amplifies the signal and controls the second tube to oscillate, and the oscillation signal generates sound for the piezoelectric ceramic sheet, while controlling a thyristor (consisting of two transistors) to turn on the lamp.

The diagram above is a rough idea and has not been tested.

Depending on the selected supply voltage:

R1 is taken in the range of 10K to 50K, the larger it is, the weaker the light required, but the more unstable it is, and it is easy to malfunction.

R2 is the upper bias resistance of BG2, and the voltage is divided with R5 to obtain a base voltage of about 3V, and the value is about 20K 50K.

r3 r4 r6, which is the resistor that controls the frequency, is calculated according to the formula according to the selected capacitance.

R5 calculates the partial pressure of about 3V based on R2.

R7 is the emitter resistor, which controls the quiescent current of the oscillating tube, and the quiescent current is calculated to be 3 6 mA.

R8 is an analog thyristor ruggedized resistor, the smaller it is, the less likely it is to mislead it, but the stronger the trigger signal required.

C1 C2 C3 is the capacitor that controls the frequency, the frequency is about 1 (2 R3C3 6), and the R value is generally much larger than the RBEs of the transistor

C4 is the audio bypass capacitor, take 10 20 F.

C5 is the audio coupling capacitor, take 10 20 F.

In parallel with the R6 are piezoelectric ceramic sheets, which are used to produce sound.

Connected to the BG4 emitter is a light bulb. If you switch to LED lights, you need to connect the appropriate resistors in series.

The resistor capacitance values provided above are approximate reference values, and the specific values need to be determined by testing.

The sound stops when the light is on (BG3 on), and if the time is too short, you can find a way to delay it with an integral circuit instead of C5.