Timeline for Are linear voltage regulators more efficient than switching regulators at very light loads?
Current License: CC BY-SA 4.0
18 events
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| Aug 8, 2022 at 0:54 | comment | added | CaveScientist | First of all I cant choose the linear regulator it is built in with the sensor but thanks anyway. As I promised, the full project can be found here: electronics.stackexchange.com/questions/630343/… | |
| Aug 8, 2022 at 0:15 | answer | added | Jens | timeline score: 0 | |
| Aug 7, 2022 at 19:55 | comment | added | Neil_UK | If you want a really low power buck, then you need one that will skip / shut down at very low currents, and have very low Iq when it does. Take care with choosing linear regulators, some have surprisingly high Iq. The only solution is plenty of time, a spreadsheet, a good supply of coffee and data sheets. | |
| Aug 7, 2022 at 19:01 | answer | added | TimWescott | timeline score: 3 | |
| Aug 7, 2022 at 18:16 | comment | added | user57037 | Yes usually at very low load currents, linear regulators are more efficient. The buck will need some type of internal linear regulator to start up, so it is difficult for it to beat a linear regulator. However, there is also a wide range of quiescent currents for bucks, and often you can find one that is low enough for your application, even if it is a micro-power application. | |
| Aug 7, 2022 at 18:08 | answer | added | tobalt | timeline score: 0 | |
| Aug 7, 2022 at 18:08 | comment | added | CaveScientist | Oh I am glad my thinking is right. My whole application is an alarm system with bunch of delay circuits and many transistors. There will be a buck converter but it will only activate when a motion is detected. I tried to do the project with Arduino before but burned it with a boost converter somehow. Also the interrupts just confuse me, sometimes buttons don't work. This time I will try to go purely analog. I'll create a separate post to acknowledge my design (forum format I guess) and I'll post a link here. | |
| Aug 7, 2022 at 17:43 | comment | added | bobflux | Your whole application is a PIR sensor? Or will the buck power something else? | |
| Aug 7, 2022 at 17:28 | comment | added | Andrew Morton | The calculations say yes. Why do you doubt them? Supporting material: When is an LDO More Efficient Than a Switching Power Converter? | |
| Aug 7, 2022 at 16:54 | comment | added | CaveScientist | The linear regulator will always be there (I can desolder it from the sensor if battery span will greatly increase otherwise I dont want to touch it :D), I can add a buck converter if I want to beforehand or I can let the sensor take 9V. The regulator is, let me google... It is HT7533 | |
| Aug 7, 2022 at 16:33 | history | edited | ocrdu | CC BY-SA 4.0 |
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| Aug 7, 2022 at 16:28 | history | edited | JRE | CC BY-SA 4.0 |
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| Aug 7, 2022 at 16:27 | review | Close votes | |||
| Aug 8, 2022 at 13:26 | |||||
| Aug 7, 2022 at 16:18 | comment | added | Unimportant | "and let the sensor's linear regulator obtain 3.3V". Please clarify. This linear regulator is in the sensor so it's quiescent current draw will be present in either scenario? Or does the switching regulator replace the linear regulator completely ? | |
| Aug 7, 2022 at 16:11 | history | edited | CaveScientist | CC BY-SA 4.0 |
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| Aug 7, 2022 at 16:10 | comment | added | Elliot Alderson | Please provide a link to the datasheet of your linear regulator. I think you have overlooked the regulator's quiescent current in your calculations. | |
| S Aug 7, 2022 at 16:04 | review | First questions | |||
| Aug 7, 2022 at 16:33 | |||||
| S Aug 7, 2022 at 16:04 | history | asked | CaveScientist | CC BY-SA 4.0 |