Hi, I've got a cheap 5v (I don't know how many amps) power bank that consists of a single cell battery. I want to measure if it is capable of powering up my arduino project (and also a raspberry pi) but I don't know how to measure anything. Actually I know it's capable because I've tried powering it via usb and worked, I left it on for about 5hs, but I want to know how much longer can it be powered that's why I want to know if it is possible to measure the Amps output?
At first I tried measuring with a 100ohms resistor the volts being delivered and it says 3.92... Shouldn't I get 5v? so that means that the battery is not good?
I need to understand a bit more of how this works, here I attach an image of what I did in order to measure it.
Take out the resistor, it's doing nothing except wasting power. And it looks like it's just measuring the battery voltage, not the output of the 5V boost converter.
Get a USB power meter. The majority of them will measure voltage, amps, mAh, and some even keep track of Wh (watt-hours). For discharge testing the capacity of the bank you can pair it with a USB load with varying levels of sophistication. My favorite are the ones with a built-in screen.
Oh I see, I thought there was a way of measuring it without buying stuff.. I'll see what I can do, Amazon doesn't ship to my country 
In order to measure current you need to break the connection in half where you want to do the measurement, and put the current meter in at that point.
Take a cheap USB cable, chop it in half, and solder the black wires together. Set your multimeter to a current setting and use it to bridge the broken red wire, so that all the current flowing through it goes through the meter to be measured.
I will note that the capacity of battery banks is usually listed as the sum of the listed capacity of the cells inside. For a single 18650 like that it's probably somewhere between 1,800 and 2,600 mAh depending on how much they cheaped out. Because of the 5V boost converter, the actual output capacity needs to be derated by the boost ratio (3.7/5) and the efficiency (probably at least 90%). The actual capacity available to power your Arduino is probably only about 1,200 to 1,800 mAh.
Jiggy-Ninja:
In order to measure current you need to break the connection in half where you want to do the measurement, and put the current meter in at that point.Take a cheap USB cable, chop it in half, and solder the black wires together. Set your multimeter to a current setting and use it to bridge the broken red wire, so that all the current flowing through it goes through the meter to be measured.
I will note that the capacity of battery banks is usually listed as the sum of the listed capacity of the cells inside. For a single 18650 like that it's probably somewhere between 1,800 and 2,600 mAh depending on how much they cheaped out. Because of the 5V boost converter, the actual output capacity needs to be derated by the boost ratio (3.7/5) and the efficiency (probably at least 90%). The actual capacity available to power your Arduino is probably only about 1,200 to 1,800 mAh.
and then divide the answer by any number between o and 100 because such banks are notoriously over-rated by the manufacturer/seller. I once bought one of fleabay that was of quite a high rating but failed to give its rated AH output. Although it seemed heavy enough all was revealed when I pulled it apart to discover two large steel plates making up around 80% of its weight.
jackrae:
and then divide the answer by any number between o and 100 because such banks are notoriously over-rated by the manufacturer/seller. I once bought one of fleabay that was of quite a high rating but failed to give its rated AH output. Although it seemed heavy enough all was revealed when I pulled it apart to discover two large steel plates making up around 80% of its weight.
Not to mention the lies that are just outright absurd, like claiming 9,990 mAh capacity in an 18650.
bigclivedotcom's Youtube channel is almost nothing but buying cheap ebay (or dollar store) crap and taking it apart for examination. Almost every "10W" LED he buys barely pulls 3W from the wall.
Thanks so much to everyone, I measured about (directly powered by power supply) 273mA on my raspberry pi and about 286mA with the battery, under a specific CPU stress test. So that means that the battery pherhaps isn't providing 5v maybe a little bit less or the power supply provides more, I could measure tension but I don't want to break my cable any more haha..
So the answer is yes, it's powerful enough to power those devices. Now I would like to see how much energy is remaining on the battery is that possible? (again without buying stuff).
EDIT> I'm going to do it the stupid way, I'm going to fully charge the battery and then power the raspberry pi with it running a bash script every 10 minutes that writes the time on a file, and whenever it finishes I will know how much energy it can provide. I would like to have another way so that I can ask for how much battery is remaining but anyways this way I have an estimative value.
You can measure the voltage on the battery (not the 5V output, directly on the battery) to get an approximation of the amount of charge left in it. Each battery has a certain relationship between the charge left and the terminal voltage.
How long do you want this to last? At 250 mA a single cell battery bank will be drained in hours. Power management is a significant undertaking in its own right, you can't just "bolt it on" after the fact if you don't want to be lugging around car batteries.
Your idea of logging duration to estimate 'life' is a good one if you can keep the current fairly constant
Can I suggest that you tap into the battery directly and log its voltage at the same time (you will have to determine if the 3.7 to 5 conversion is common -ve or common +ve as your measurement will be affected by whichever it is.
Now you should end up with a file system containing the two variables (time and voltage) From this you can create a look-up table which estimates capacity against voltage
Jiggy-Ninja:
You can measure the voltage on the battery (not the 5V output, directly on the battery) to get an approximation of the amount of charge left in it. Each battery has a certain relationship between the charge left and the terminal voltage.How long do you want this to last? At 250 mA a single cell battery bank will be drained in hours. Power management is a significant undertaking in its own right, you can't just "bolt it on" after the fact if you don't want to be lugging around car batteries.
I've measured the battery after my raspberry pi powered off (some energy might still be remaining, not enough to power the approximately 1.5W required by it but still some) and got 3.5V, what does this number mean? I thought that it should have been almost zero. The log showed that it lasted for about 2h10min consuming approx. 300mA.
jackrae:
Your idea of logging duration to estimate 'life' is a good one if you can keep the current fairly constant
Can I suggest that you tap into the battery directly and log its voltage at the same time (you will have to determine if the 3.7 to 5 conversion is common -ve or common +ve as your measurement will be affected by whichever it is.Now you should end up with a file system containing the two variables (time and voltage) From this you can create a look-up table which estimates capacity against voltage
Yes the current was fairly constant, after doing the logging I tested with the synthetic cpu load test and the current value was always the same (it increases about 1mA for half a second but decreases again). I don't understand the part of +-ve sorry.
Your battery bank probably has a converter unit contained within it to convert the battery voltage up to 5 volts. The 5 volt output may have its negative terminal directly coupled to the battery negative terminal (I used -ve to denote negative). Alternatively it may have the 5volt positive terminal directly connected to the battery positive terminal. (I used +ve to denote positive) A third alternative (unlikely) is that neither 5 volt terminal is directly connected to a battery terminal; ie the converter is an isolated one.
If the negative terminals are common then your measurement will be easy as the micro measures positive voltages and you will be measuring a voltage ranged from 0 to +4.2 volts. If however the positive terminals are commoned then you won't be able to directly measure the battery voltage as it's actual value will be negative relative to the micro ground reference.
Fjallbacka:
I've measured the battery after my raspberry pi powered off (some energy might still be remaining, not enough to power the approximately 1.5W required by it but still some) and got 3.5V, what does this number mean? I thought that it should have been almost zero. The log showed that it lasted for about 2h10min consuming approx. 300mA.
Certainly not! Batteries have a fairly characteristic discharge curve that depends on their chemistry. Most of them will maintain a relatively constant voltage for most of their capacity, and then the voltage basically falls off a cliff towards the end. It is definitely not linear. Here's a generic discharge curve for rechargeable lithium batteries:
The legend with "C" numbers refers to different discharge currents. C ratings are used to scale the currents to the battery's capacity. 1C is the current necessary to dischage the battery's rated capacity in one hour. 2C is twice that current, 18C is 18x the 1C current, etc. So for a 2300 mAh battery, a current of 1C is 2300 mA.
As you can see from the graph, for most of the discharge levels most of the battery's energy is gone by the time the terminal voltage hits 3.5V
The different chemistries will also have their own unique characteristics. Lithium batteries are the delicate little snowflakes of the battery world; overcharging or overdischarging can make them burst into flame. The Note 7 was recalled (twice!) because of battery issues. Any responsible use of a lithium battery will have some kind of protection circuit (either built into the battery or in the circuit) that will prevent the battery from being damaged.
jackrae:
Your battery bank probably has a converter unit contained within it to convert the battery voltage up to 5 volts. The 5 volt output may have its negative terminal directly coupled to the battery negative terminal (I used -ve to denote negative). Alternatively it may have the 5volt positive terminal directly connected to the battery positive terminal. (I used +ve to denote positive) A third alternative (unlikely) is that neither 5 volt terminal is directly connected to a battery terminal; ie the converter is an isolated one.If the negative terminals are common then your measurement will be easy as the micro measures positive voltages and you will be measuring a voltage ranged from 0 to +4.2 volts. If however the positive terminals are commoned then you won't be able to directly measure the battery voltage as it's actual value will be negative relative to the micro ground reference.
It's a common +ve situation, if I measure the positive terminal of the tester with positive terminal of the circuit I get 4.12V (fully charged)
Jiggy-Ninja:
Certainly not! Batteries have a fairly characteristic discharge curve that depends on their chemistry. Most of them will maintain a relatively constant voltage for most of their capacity, and then the voltage basically falls off a cliff towards the end. It is definitely not linear. Here's a generic discharge curve for rechargeable lithium batteries:
The legend with "C" numbers refers to different discharge currents. C ratings are used to scale the currents to the battery's capacity. 1C is the current necessary to dischage the battery's rated capacity in one hour. 2C is twice that current, 18C is 18x the 1C current, etc. So for a 2300 mAh battery, a current of 1C is 2300 mA.As you can see from the graph, for most of the discharge levels most of the battery's energy is gone by the time the terminal voltage hits 3.5V
The different chemistries will also have their own unique characteristics. Lithium batteries are the delicate little snowflakes of the battery world; overcharging or overdischarging can make them burst into flame. The Note 7 was recalled (twice!) because of battery issues. Any responsible use of a lithium battery will have some kind of protection circuit (either built into the battery or in the circuit) that will prevent the battery from being damaged.
Oh, I see I thought battery discharging cycle was linear, thanks for the info.. Now I've got somewhere at home (I just have to search) another power bank that has 2 cells but it's not working properly it discharges really fast (I don't know if it's a battery issue or a converter one). It was used to charge an iPhone no more than 5 times I think. I wanted to know if I could have those batteries connected in series to the battery of this power bank in order to increase it's capacity? Or do I need to measure something in order to see if they are compatible, before attempting on doing this.
Fjallbacka:
It's a common +ve situation, if I measure the positive terminal of the tester with positive terminal of the circuit I get 4.12V (fully charged)
That's not really what that means. You've probably got a common -ve, but here's how to test it. Put your multimeter into continuity mode, and press one lead to the shell of the USB connector, which is probably connected to the 0V line. Then try touching the other lead to the two battery terminals and see which one gives you the continuity beep.
Oh, I see I thought battery discharging cycle was linear, thanks for the info.. Now I've got somewhere at home (I just have to search) another power bank that has 2 cells but it's not working properly it discharges really fast (I don't know if it's a battery issue or a converter one). It was used to charge an iPhone no more than 5 times I think. I wanted to know if I could have those batteries connected in series to the battery of this power bank in order to increase it's capacity? Or do I need to measure something in order to see if they are compatible, before attempting on doing this.
It's possible that that battery bank has crap cells in it from some Chinese counterfeit manufacturer that are only partially filled with material.
Jiggy-Ninja:
That's not really what that means. You've probably got a common -ve, but here's how to test it. Put your multimeter into continuity mode, and press one lead to the shell of the USB connector, which is probably connected to the 0V line. Then try touching the other lead to the two battery terminals and see which one gives you the continuity beep.
It's possible that that battery bank has crap cells in it from some Chinese counterfeit manufacturer that are only partially filled with material.
Forgive me for this late response, but I've just got the other power bank that I was talking about. This one is rated 5600 mAh, I measured the output tension of the circuit and it's approximately 5V so I can assume that the circuit works fine. But I want to increase the capacity with the other one I've got, can I connect all 3 cells in parallel and expect them to work? or might there be some limitations with he circuit booster?
Fjallbacka:
Hi, I've got a cheap 5v (I don't know how many amps) power bank that consists of a single cell battery. I want to measure if it is capable of powering up my arduino project (and also a raspberry pi) but I don't know how to measure anything. Actually I know it's capable because I've tried powering it via usb and worked, I left it on for about 5hs, but I want to know how much longer can it be powered that's why I want to know if it is possible to measure the Amps output?
At first I tried measuring with a 100ohms resistor the volts being delivered and it says 3.92... Shouldn't I get 5v? so that means that the battery is not good?
I need to understand a bit more of how this works, here I attach an image of what I did in order to measure it.
Hello,
To test the current you need a meter of some sort. If you have a volt meter you can use that if it has a low voltage setting. Basically you use a low value resistor and then measure the voltage across the resistor and then use Ohm's Law to calculate the current.
Some meters have a current scale but you have to be careful when doing it that way.
To test the actual capacity however is a lot harder, but nonetheless doable.
To test the capacity in a given application (like running an Arduino) you can either connect the Arduino or a resistor that is approximately the same value load as the Arduino presents. If you use the Arduino you have to know the current, but with a resistor (more recommended) you dont.
You do have to measure the voltage of the battery pack, and make a log of the voltages you read versus the time you read them. You can then calculate the actual capacity of the pack.
The idea is to calculate the current for each reading, and use the time as a weighing factor. For example, say you are using a 10 ohm resistor and you take the following set of readings (hours:minutes, voltage) and your cut off voltage point is 3 volts:
01:00, 5v
02:00, 5v
03:00, 4v
04:00, 3v
In the above at 1 o'clock we had 5v, and at 2 o'clock we had 5v for example.
We then make a table that includes the current levels knowing the resistance:
01:00, 5v, 0.5 amps
02:00, 5v, 0.5
03:00, 4v, 0.4
04:00, 3v, 0.3
Now from 1:00 to 2:00 we had 0.5 amps and that was for one hour, so we have so far:
Total=1hr times 0.5 amps=1*0.5=0.5 ampere hours.
Next from 2:00 to 3:00 we went from 0.5 to 0.4, which is an average of 0.45, and that was for one hour so we have:
SubTotal=1hr times 0.45 amps = 1*0.45=0.45
and now we add that to the total from before:
Total=Total+0.45=0.5+0.45=0.95 ampere hours.
From 3:00 to 4:00 we went from 0.4 to 0.3 which is an average of 0.35 and for one hour that gives us a subtotal of 0.35, so we add that to the previous total and we get:
Total=0.95+0.35=1.30 ampere hours.
So in this simple example we got an estimate of 1.3 ampere hours for the capacity of the pack.
A couple notes:
- The resistance load should be close to the load of the application this is going to be used for. If not, the final result has to be adjusted and that's not easy to do.
- The readings should be closer together than the above, probably one reading per 30 minute interval, but i've used every minute before too. Using automatic logging software you can log every second if you wish.
- The results are pretty definitive, but you do have to make sure the pack is charged as it is normally when using it in the actual application.
- Minimum test equipment required is a meter, preferably a digital meter.
- To find the average (as in the example) just add the two respective readings and divide by two. Thus the average of 0.5 and 0.4 is:
(0.5+0.4)/2=(0.9)/2=0.45