Timeline for How to store and step 12V to 5V with capacitor and regulator? [closed]
Current License: CC BY-SA 3.0
34 events
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| Mar 23, 2016 at 4:47 | history | closed |
Scott Seidman PeterJ uint128_t Dave Tweed |
Needs details or clarity | |
| Mar 22, 2016 at 22:03 | review | Close votes | |||
| Mar 23, 2016 at 4:47 | |||||
| Mar 22, 2016 at 15:30 | comment | added | ftwv | @StainlessSteelRat agree. The above circuit looks way more complicated than it should be. | |
| Mar 22, 2016 at 15:27 | comment | added | StainlessSteelRat | If something does not work, simplify. My money is on 7805 pins. Remove caps and load. Add a resistor as load. Pin 1 to +12, 2 to gnd and 3 to +5V. Verify operation. Replace resistor with +5V load. Verify operation. 2nd possibility, possible overload or thermal shutdown. If you are creating a battery charger, then this is a definite possibility. 1.5A typical max for TO220 package. Finally, add caps, one at a time or all. | |
| Mar 22, 2016 at 13:35 | history | edited | ftwv | CC BY-SA 3.0 |
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| Mar 22, 2016 at 13:23 | history | edited | ftwv | CC BY-SA 3.0 |
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| Mar 22, 2016 at 13:01 | comment | added | brhans | The circuit diagram is fine. The problem is probably in the way you've constructed it (strongly hinted at by your statement "there is a significant drop in voltage (measuring with multimeter at the points before and after C1 on circuit)". Post a few good hi-res photos from different angles of what you've actually constructed and we might be able to see what you've done wrong. | |
| Mar 22, 2016 at 11:48 | comment | added | KilowattLaser | Maybe the cap is bad? | |
| Mar 22, 2016 at 11:35 | history | edited | ftwv | CC BY-SA 3.0 |
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| Mar 22, 2016 at 11:20 | history | edited | ftwv | CC BY-SA 3.0 |
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| Mar 22, 2016 at 11:17 | comment | added | ftwv | Hi Scott Seidman, no. Is a usb power pack. | |
| Mar 22, 2016 at 11:16 | comment | added | Scott Seidman | Are you trying to charge tell battery? Why not ask how to do that? | |
| Mar 22, 2016 at 10:57 | history | edited | ftwv | CC BY-SA 3.0 |
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| Mar 22, 2016 at 10:48 | comment | added | ftwv | Hi Asmyldof, yes it's a usb power pack, didn't know how to represent in a circuit. Do you recommend adding two 7805 components? | |
| Mar 22, 2016 at 10:43 | comment | added | Asmyldof | And also, what 5V battery? I cannot right now think of a single rechargeable topology that's exactly 5V. Unless you mean a USB power pack, which will draw current to its own maximum until the voltage drops to 4.6V or so, which means the 7805 is HEAVILY overloaded if the pack is limited at 2A or more. | |
| Mar 22, 2016 at 10:42 | answer | added | Andy aka | timeline score: 2 | |
| Mar 22, 2016 at 10:36 | history | edited | ftwv | CC BY-SA 3.0 |
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| Mar 22, 2016 at 10:32 | comment | added | CharlieHanson | You can replace the 5V battery with a resistor because it you are trying to get the circuit to work in principle. One step at a time. | |
| Mar 22, 2016 at 10:27 | comment | added | MarkU | Heat sinking will also be important; a 7805 is very inefficient at regulating 12V down to 5V. The heat loss will be (VIN-VOUT)*(whatever current the load draws). You may be surprised by how little load current it takes to make the bare 7805 really really hot. | |
| Mar 22, 2016 at 10:27 | comment | added | CharlieHanson | Also, check that (a) you have connected the pins correctly; and (b) if you are using a battery, check that it is charged, or consider a bench-type power supply. | |
| Mar 22, 2016 at 10:27 | comment | added | ftwv | Hi Neil_UK, thanks for the info. C1 is a 47uF / 50V e.cap, can you suggest a more appropriate capacitor to use? Can't replace the battery component with a resistor as I'm trying to step down to a 5V battery | |
| Mar 22, 2016 at 10:24 | comment | added | JRE | Check the datasheet. You need a 0.33µF from the 7805 input to ground, and a 0.1µF capacitor from the 7805 output to ground. These two must be connected to the 7805 right at the 7805. | |
| Mar 22, 2016 at 10:24 | comment | added | MarkU | Might also help if you could post a photo of the actual prototype, since this is a very common type of circuit and it could either be a construction error or a measurement error. | |
| Mar 22, 2016 at 10:22 | comment | added | MarkU | Unclear what "before and after C1" means; do you mean the multimeter (+) probe at different points along the +12V input wire to the 7805? | |
| Mar 22, 2016 at 10:21 | comment | added | CharlieHanson |
As in "to the left" and "to the right" of C1? They should all be the same voltage. Perhaps edit the picture to mark the exact points that you are measuring, and the voltages you get at each.
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| Mar 22, 2016 at 10:19 | comment | added | ftwv | Hi CharlieHanson, measuring with multimeter at the points before and after C1 on the circuit | |
| Mar 22, 2016 at 10:17 | history | edited | ftwv | CC BY-SA 3.0 |
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| Mar 22, 2016 at 10:17 | comment | added | Neil_UK | C1 is tiny for energy storage! Might be better to replace the 5v battery with a resistor that draws the amount of current you expect to want to source. It would make simulation, testing, understanding what's supposed to be happening a lot easier. | |
| Mar 22, 2016 at 10:15 | history | edited | ftwv | CC BY-SA 3.0 |
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| Mar 22, 2016 at 10:15 | comment | added | CharlieHanson | And are you testing the schematic in simulation or have you actually made the circuit? | |
| Mar 22, 2016 at 10:14 | history | edited | ftwv | CC BY-SA 3.0 |
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| Mar 22, 2016 at 10:13 | review | First posts | |||
| Mar 22, 2016 at 10:28 | |||||
| Mar 22, 2016 at 10:13 | comment | added | CharlieHanson | What do you mean by "measuring before and after C1" ? | |
| Mar 22, 2016 at 10:11 | history | asked | ftwv | CC BY-SA 3.0 |