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I'm trying to draw a supply scheme for a class D integrated amplifier.

It requires 21V to 24V power supply and about 6/7A of current at 24V. The input is a car battery: 10V to 14.4V.

I found this on the web:

enter image description here

I've a few questions:

  1. Is the NE555 usable for this scope?
  2. How much output current this circuit have?
  3. If the answers to the first two are: "no" and "too low" can you suggest a circuit scheme that it's my case?

EDIT: If you did not understand well, the scheme of this DC boost must be added to the PCB of the amplifier. Also, the output current can also be 3A, i will put two of this circuit in parallel with diodes.

EDIT_2: Ok then, I can't use a NE555, this is a scheme of a DC DC boost circuit that I already have and works very well for my application.

enter image description here

What resistance with what value I've to change to get a 21-24V output? Because it is adjustable, and it want fixed.

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  • \$\begingroup\$ "The output is a car battery ...". I think you mean input. With Q1 and Q2 in that orientation you will never be able to pull C3 to V+ or to V-. \$\endgroup\$ – Transistor Jul 27 '17 at 9:16
  • \$\begingroup\$ Yes, the input, sorry @Transistor \$\endgroup\$ – Northumber Jul 27 '17 at 9:17
  • \$\begingroup\$ R1 is too low, and that's before we've got anywhere near to power. Visit Amadong or fleaBay, and buy a DC to DC boost converter module, 12v to 24v is a very common application. \$\endgroup\$ – Neil_UK Jul 27 '17 at 9:19
  • \$\begingroup\$ I don't want to buy one, I'm trying to draw a scheme to add it to the same PCB of the amplifier @Neil_UK \$\endgroup\$ – Northumber Jul 27 '17 at 9:20
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    \$\begingroup\$ @Northumber - "I don't want to buy one, I'm trying to draw a scheme to add it to the same PCB" Then you're out of luck, unless you want to do a great deal more work and make a much bigger pcb. Your current requirements simply cannot be met with a practical capacitively-driven circuit. C3 will need to be in the Farad range and the transistors are grossly inadequate, as is the drive for them. And the diodes, too. You need a standard DC-DC boost converter, and if you roll your own, at the very least you'll get horrendous noise injection unless you really know what you're doing. \$\endgroup\$ – WhatRoughBeast Jul 27 '17 at 9:28
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It requires 21V to 24V power supply and about 6/7A of current at 24V. The input is a car battery: 10V to 14.4V.

So the output power is about 22.5 volts x 6.5 amps = 146 watts.

Forget the circuit you have shown because it might deliver a couple of watts at best. Go for a proper boost regulator to step up 12 volts to 24 volts. Something like this: -

enter image description here

enter image description here

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  • \$\begingroup\$ And only expect it to work if you use the exact inductor specified - not just one with the same value that looks about the same size - unless you're willing to spend a lot of time learning the subtleties of inductor design. By the time you've built it and got it working, buying one will look like a much better option. \$\endgroup\$ – user_1818839 Jul 27 '17 at 13:46
  • \$\begingroup\$ @BrianDrummond LOL \$\endgroup\$ – Andy aka Jul 27 '17 at 13:52
  • \$\begingroup\$ And expect at least one iteration of the pcb layout before you get that right. \$\endgroup\$ – WhatRoughBeast Jul 27 '17 at 14:00
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Consider the LTC1270/1270A as a more highly integrated solution. With a 10 A switch you should be able to get about 4 A output. The data sheet is thin but the LT1070 datasheet and app note are great tutorials.

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've a few questions

Your switches will deliver 2x of the average current of 7a. With derating, you are likely talking about a pair of to247 devices for each side.

Check that against the datasheet specified there.

The topology here is a charge pump. They are simple but hard the champs at delivering loads of current.

Your power requirement is at the higher end of a single ended design. I would think about a fly back converter instead.

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