Timeline for What is the formula for how long a capacitor can power a circuit
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 3, 2019 at 9:29 | vote | accept | Jake quin | ||
| Aug 2, 2019 at 18:25 | comment | added | Hitek | @DKNguyen - I agree... | |
| Aug 2, 2019 at 9:05 | answer | added | Russell McMahon♦ | timeline score: 0 | |
| Aug 2, 2019 at 3:03 | comment | added | DKNguyen | @Hitek Dead voltage at no-load and dead voltage while loaded are different things. Of course, there is also the fact that discharging a battery too deeply damages it, so you don't want to discharge it below that even if it can still provide current. | |
| Aug 2, 2019 at 0:47 | comment | added | Mattman944 | Elliot gave the right answer for your original question. It is reasonable to assume that the current is about constant as the voltage drops from nominal to the minimum voltage that the circuit will operate. As the voltage drops further down, the current will rarely remain constant and the equation is not applicable anymore. | |
| Aug 2, 2019 at 0:35 | comment | added | Hitek | @DKNguyen - The lead-acid battery in my truck starts the engine in two revolutions every time, even though the battery's voltage drops to 9.5 Volts while starting, so it is apparently still providing current at that voltage... | |
| Aug 2, 2019 at 0:22 | comment | added | Hitek | @MarkU - It's the voltage delta, or difference in voltage... | |
| Aug 1, 2019 at 23:38 | answer | added | D.A.S. | timeline score: 1 | |
| Aug 1, 2019 at 23:03 | comment | added | Jake quin | @MarkU a supercapacitor might do, but in the future when i might create a fairly low power capacitor and not need to go that far. I am still waiting for Elliot to answer me what does ΔV means. so i can definitively compute. Do you know what ΔV means? | |
| Aug 1, 2019 at 22:56 | comment | added | MarkU | You might look into something called a “Super Capacitor”, made for this kind of application. | |
| Aug 1, 2019 at 22:52 | comment | added | DKNguyen | @Jakequin You sure can. Might be best to look at your electrical specifications in the datasheet for your MCU to find the actual maximum loads though. You might find current you can trim like clock speed or peripehrals that you know you aren't using. | |
| Aug 1, 2019 at 22:50 | comment | added | Jake quin | @MarkU cant i just assume maximum load? for a typical micro controller board that powers from usb 5v at 0.5A. if it consuming less power then the better since the capacitor can last as long. Or is there something else? | |
| Aug 1, 2019 at 22:44 | comment | added | MarkU | Microcontroller is one of the most complicated kinds of loads, it’s not easy to model. | |
| Aug 1, 2019 at 22:32 | comment | added | Jake quin | @MarkU also i forgot to ask in the question what type of capacitor is best suited for this applications? Electrolytic , ceramic or something else? | |
| Aug 1, 2019 at 22:30 | comment | added | Jake quin | @MarkU i am not very well versed with electronics, I am still trying my best to learn. My circuit involves a microcontroller some leds, diodes, and some other modules. With the leds turning on and off that makes it not a constant current circuit, so a constant resistance i guess?? | |
| Aug 1, 2019 at 22:24 | comment | added | MarkU | Elliott’s answer explains the physics, but to answer the “applications” question about how much time your circuit will run, more information is required. The minimum voltage that your circuit can still run? The load characteristics, is it constant current or constant resistance, or something nonlinear like a diode or a silicon chip? The capacitor’s voltage decreases as charge is removed, how fast it discharges depends on how much current the load draws depending on voltage. | |
| Aug 1, 2019 at 22:05 | answer | added | Elliot Alderson | timeline score: 5 | |
| Aug 1, 2019 at 22:02 | comment | added | DKNguyen | A battery doesn't fall to 0V when it's dead. A capacitor does. A battery is dead long before it drops to 0V. For example, a lead-acid battery charges up to a maximum of 13.8V and is considered dead (can't provide current anymore) when it's 11.4V. If you are using a capacitor to power something, then you must treat it similarly: It doesn't matter if your capacitor is truly dead when it's 0V if whatever you're powering requires at least 3V. | |
| Aug 1, 2019 at 21:57 | history | asked | Jake quin | CC BY-SA 4.0 |