Lightning Bolt / General Electronic Question
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Simple question hopefully. I've found partial answers here and there but not exactly what I am looking for.
When the lightning bolt indicator comes on, I was under the impression that is a voltage drop issue not a current issue. Recently people said it could be both. Things tend to draw as much current as they need, right? So even though something draws more current to keep the voltage up, this wouldn't really be a current issue, just a voltage issue right?
My main question in regards to this is how is this affecting the Rpi power supply? What is actually going on? I am trying to learn more about circuits and power (the Rpi projects were the gateway :0 ) and can not seem to work out with paper mathematics how drawing more current drops the voltage in the circuit. Is there a simple equation for this?
Thank you in advance
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@trailjacker Here is the simplest way I can think of to help you. Imagine taking a 9V battery that is no longer fully charged. If you measure the voltage across it with a meter, you would still measure close to 9V because a voltmeter does not draw current from the device or circuit under test, in this case the battery. Voltmeters have high input resistances so that they do not affect the circuits that you are testing them with. Now if my battery is bad and I put it in a device, it's job is to deliver a certain current at a certain voltage but because the battery is bad and has no charge, when the circuit demands a current, the voltage across the battery collapses. This is what is happening with the lightning bolt. The setup is demanding power which the supply cannot deliver so the voltage supplied is falling slightly below 5V as the current demand is increased. The power pack is failing to deliver a regulated stable voltage when the circuit demands a current that is close to its rated value. This is why it is a good idea to get a supply that can deliver up to 3A because at half load you can pretty much be certain the pack can maintain its voltage.
If you load up a motor with a physical force the motor starts to decrease speed as the torque from the physical load ramps up. Eventually the motor slows down so much it stalls and you are left with zero speed and maximum torque.
It's the same for a battery. Without a load it runs at full speed (open circuit voltage) and as you load it up the terminal voltage lowers as the current taken increases. Eventually, with a shorted out battery the current taken is at maximum but the terminal voltage is zero.
The internal resistance of the cell causes this to happen. If a cell didn't have internal resistance it could supply any amount of current without the terminal voltage falling (an impossibility of course).
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@rbaker That was a very good explanation and definitely cleared things up! Thank you very much for the help.
My next question now is if the Pi limits itself to 2.5 amps regardless of the power supply potential, would it be better to have a charger with = or >5.25v rather than a charger with 3A if the voltage is where the issue is?
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@trailjacker Well now you must understand that if you power the Pi with anything greater than 5V you stand a good chance of destroying it. 5V is the maximum, actually it's about 5.1V or something with the official power pack. With regard to current, the Pi doesn't limit itself to 2.5A. The Pi will draw current depending on what you want it to do. So at boot up, it might just draw 800mA or so. When you decode 1080p video it will draw more. If you start loading your USB ports with accessories then you could be looking at 2A or more. So what I am saying is that I run a 3A 5V fully regulated supply because I know my setup will have keyboard encoders, controls, leds etc all controlled by the Pi. I need to know that my supply can deliver what is asked of it. 3A is sufficient for me. 5A would be overkill. The misconception is that if a power supply is rated at 5A it then delivers 5A to whatever's connected to it. This is not the case. It means that you can be sure of a stable 5V out at a current draw up to 5A. Any more and that 5V would start to collapse and the thing would probably catch fire!
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Also, cheap USB cables are bad news even for good power supplies. They can often fail to deliver current properly or even worse create voltage drops leading to the lightning bolt.
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Ok last question on this issue I think...
So in the case that "not all power supplies are created equal" and undervoltage may occur, I keep running into this statement that makes me think i can use my phone's two port quick charge usb adapter to power my pi to ensure the pi can draw everything it needs:
Many users worry that that they may "supply too much power" by using a higher rated supply. The Pi will only draw as much current as it requires and can not use more then 2.5A (Pi3) or 2A (Pi2/B+) as this is limited by a fuse, so there is no benefit in a higher rated supply. (Earlier models had a smaller polyfuse - probably 1.1A.)
But you are saying this is incorrect? I apologize if I am beating a dead horse, and am just making sure I understand well enough so I know what max PSU to buy/use as i tax the pi cpu pretty hard.
Edit : Obviously the the pi fuse would probably blow if I hooked it straight up to a high power source, but in theory using just the pi with its on-board cpu, blutooth, usb, etc, it should never draw a ridiculous amount from any power output device, correct? Assuming I am using low powered usb devices like a keyboard, mouse, and a couple of controllers. I would also assume that worst case scenario with this setup would just be power supply inefficiency?
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@trailjacker said in Lightning Bolt / General Electronic Question:
Many users worry that that they may "supply too much power" by using a higher rated supply.
Yes, because they do not understand that in electronics, a device will draw the current that it demands. For example, I buy a green LED. It requires 25mA to illuminate. I decide to power it with a 3A power supply. The power supply laughs at it and delivers the 25mA with 2.975A spare capacity to power other things like a Pi.
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