Hello
I made the following circuit but it didn't work, is the schematic correct?
No. No base drive.
What voltage and frequency is "AC"? If household AC, expect a small explosion and/or fire.
What is Q1 supposed to do? Its base isn't connected so it won't conduct. If you did connect it to pin 1 of U3 which is what you may have intended then it would switch on when the load current exceeded about 0.7A and apply the full rectified voltage to the output, bypassing the regulator completely! Quite unclear why Q1 is there at all, the LM7824 should do everything you want.
Should we assume that J1 gets the output of a 24V step-down transformer? If it's direct from the mains then jremington's warning should be noted!
and bad smell
Q1 is a bootstrap high current driver for more output current. (As noted, the base must be connected.)
Oh yes, it's an application circuit in the datasheet.
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How should the circuit be? can you draw me
The transistor base needs to be connected to pin 1 of the LM7824. In the schematic, it is not. I suspect R3 is too small, and you are missing the required capacitor on the output of the LM7824.
Don't try to build this on a breadboard as the tracks will burn.
You're missing two things. A connection dot below the base of the transistor and an ISOLATION transformer. Your schematic shows the components connected directly to the mains with only 50V compoonents. Everywhere on the planet the mains is 2x to 5x 50V so you can be SURE of flames when first connect. Not to mention that there will be FATAL voltages EVERYWHERE in your circuit. Instead of trying to "roll your own", just BUY a certified module.
Why are you asking US to draw a schematic when the part you referenced INCLUDES one?
50v 2200uF at 7824 output, how should the correct drawing be?
Too big, 10 uF is all that is needed. See the LM7824 data sheet for example application circuits.
Does it work properly?
On a schematic, it is best to clearly indicate connections by dots on crossings, as you see along the bottom (negative) rail in the schematic above. As it is, connections above that rail are ambiguous.
See the clear difference in the snapshot below. The crossing on the right is assumed not to connect.
What I don't like about this circuit is that the current at which the bypass transistor starts to take some load is determined by the value of R3 and the base-emitter voltage of Q1. BJT Veb is about 0.7 volts but temperature dependent, rising with temperature. So if Q1 starts to conduct a significant amount of current it gets hotter which tries to reduce the current so more of the load current has to be supplied by the regulator which we were trying to help!
Is it worth building a linear regulator when you can buy 5 switched-mode ones for ~$10?
So forward loss on those modules is 1.5V times 3A max current, or 4.5W! Better have a heat sink!
But yeah, this is a much better solution than rolling your own. You can't buy the piece parts for what these modules sell for assembled and delivered.
BTW, the link says "OUT OF STOCK" Available date UNKNOWN
100 Ohms is WAY too big for R1, in both the upper (wrong) and lower (otherwise correct) circuits.
Trial and error is not the best approach to circuit construction. There are design principles that can be applied to this problem.
Why are you bothering with this, when you can buy a 24V power supply this cheaply?
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i will use 10R