Hello everyone,

I am having a problem using my Voltage regulator, I don't know why but it drains all the power of my 2 Li ion batteries really fast (only few second).
That is really zeird, the batteries are new and the voltage regulator also.
I am using two of this batteries in series : https://www.tinytronics.nl/shop/nl/...650-li-ion-batterij-2600mah-5.2a-icr18650-26j
And this voltage regulator : https://ph.rs-online.com/web/p/low-dropout-voltage-regulators/7313662/
I don't think it is because of the Quiescent current because as said in the specification it is very low, only 0.01mA.

If you have any idea of what is going on that would be really helpfull,
thanks in advance.

 

I think anything that drains a 7.2 volts 2.6Ah source (Your two batteries in series) is going the get VERY hot. You have quoted the specified quiescent current but have you measured the input current into the reglator. Also you have not stated if you have any load on the output of the regulator. My feeling is that one or both of the cells is faulty as the current required to discharge tose cells so quickly would be many tens of amps.

Les.

 

Hello,

Wich version of the regulator do you have?
The 10 μA given in the datasheet is for the 5 pin regulator in shut down mode.

Bertus

 

Thanks for your reply,

Hello,

Wich version of the regulator do you have?
The 10 μA given in the datasheet is for the 5 pin regulator in shut down mode.

Bertus

thanks for your reply.
This is the exact version that I have :
https://nl.farnell.com/rohm/ba33dd0...ey=https:nl-NL/Element14_Netherlands/w/search

 

I think anything that drains a 7.2 volts 2.6Ah source (Your two batteries in series) is going the get VERY hot. You have quoted the specified quiescent current but have you measured the input current into the reglator. Also you have not stated if you have any load on the output of the regulator. My feeling is that one or both of the cells is faulty as the current required to discharge tose cells so quickly would be many tens of amps.

Les.

Thanks for your reply.
There is no load on the output, I was just trying the voltage regulator and measured the output current

 

That seems to be a contraddiction. How can you measure the output curren when there is no load on the output ?

Les.

 

You really have only four possibilities, listed in descending order of likelihood:

1. One or both of your batteries is dead (i.e., very high internal resistance);
2. Your regulator is dead; or
3. You've got something wired wrong; or
4. Something I haven't thought of yet.

Frankly, I have a very hard time believing that you've actually drained a battery with a capacity of 2600 mAh in just a few seconds. 2600 mAh is 2.6 Ah, which is 2.6 * 3600 ampere-seconds, or just over 9000 ampere-seconds. To drain that battery in, say, 10 seconds, would require almost a thousand amps, and the battery cannot deliver that much current.

 

Hello,

Looking at the datasheet of the 3 pin regulator, the input current (Quiescent Current) is about 2 mA at no output current and 7 volts input voltage.
(see fig 1 in the datasheet).
The Quiescent Current seems to grow when the output current is getting larger. (see fig 8 in the datasheet)

How do you want to measure the output current?
You must have a load to be able to measure a current through it.

Bertus

 

That seems to be a contraddiction. How can you meadure the output curren when there is no load on the output ?

Les.

I just plug the batteries on the voltage regulator and then i put my multimeter just at the output of the voltage regulator, at the beginning everything seems to work but after a few second, maybe one minut,e the current surprisingly become very high and then nothing, and when i measure the voltage directly on my batteries after this the voltage is very low. and after some minutes the voltage of the batteries start to become higher until it reach the normal voltage, I don't understand how it is possible, I am probably doing something wrong, but I don't know what.

 

Hello,

Looking at the datasheet of the 3 pin regulator, the input current (Quiescent Current) is about 2 mA at no output current and 7 volts input voltage.
(see fig 1 in the datasheet).
The Quiescent Current seems to grow when the output current is getting larger. (see fig 8 in the datasheet)

How do you want to measure the output current?
You must have a load to be able to measure a current through it.

Bertus

What do you mean by a load to measure the current?

 

You really have only four possibilities, listed in descending order of likelihood:

1. One or both of your batteries is dead (i.e., very high internal resistance);
2. Your regulator is dead; or
3. You've got something wired wrong; or
4. Something I haven't thought of yet.

Frankly, I have a very hard time believing that you've actually drained a battery with a capacity of 2600 mAh in just a few seconds. 2600 mAh is 2.6 Ah, which is 2.6 * 3600 ampere-seconds, or just over 9000 ampere-seconds. To drain that battery in, say, 10 seconds, would require almost a thousand amps, and the battery cannot deliver that much current.

Thanks for your reply,
I am also really surprised, the batteries and the voltage regulator are all new so I don't think it is that,
And i don't know how I could plug it wrong because there is only One cable on plus and one cable on gnd and the output, and this is good.

 

Hello,

The way you say : one cable on gnd and the output, it sounds like the output is shorted.
This is the way it is done on the standard 78XX series:



Bertus

 

Hello,

The way you say : one cable on gnd and the output, it sounds like the output is shorted.
This is the way it is done on the standard 78XX series:

View attachment 165128

Bertus

This is exactky what I did, I think (in the picture)

 

Hello,

Do I see this correctly?
Is there a strand from the black wire shorting the input?



Bertus

 

It sounds like you have almost certainly destroyed you regulator by shorting the output with your meter. Some regulators have thermal shutdown. I have not studied the datasheet on your regulator so I don't know if that one does have thermal shutdown. I'm surprised that it did not go on fire.

Les.

 

It sounds like you have almost certainly destroyed you regulator by shorting the output with your meter. Some regulators have thermal shutdown. I have not studied the datasheet on your regulator so I don't know if that one does have thermal shutdown. I'm surprised that it did not go on fire.

Les.

Yep, the measurement method hasn't been described adequately, but I'm guessing it was meter set to measure current, then probes across regulator output, presenting almost a dead short to the regulator. It would deliver as much current as it could (or up to its limit.) Pure luck if the regulator isn't damaged.

This rapid discharge would temporarily get ahead of the battery chemistry, causing batteries to appear dead, but then recover with time. The spontaneous recovery of voltage described below makes it pretty clear that the batteries aren't dead (yet!) But they're clearly getting discharged really, really fast... like from a meter acting as a short across the regulator!

the current surprisingly become very high and then nothing, and when i measure the voltage directly on my batteries after this the voltage is very low. and after some minutes the voltage of the batteries start to become higher until it reach the normal voltage,

@benjamin Peer, the problem here is that you can't measure current in any meaningful way unless you have a compete circuit, including adequate impedance of some form, to measure it in. When you do have something to measure, like current through a a resistor or light bulb, you put the meter in SERIES with the load (resistor, light, etc.) not in PARALLEL with it like you would do for a voltage measurement. If there's no load, there's nothing resisting current flow through the meter, so you'll get dangerously high current flowing.

 

The appearance of the cells in the photo suggest that they have protection boards that may protect against excessive output current. Note the appearance of a small "step" in the heatshrink covering near the end of the cell - that is typical of protected cells. Cell manufacturers don't make protected cells - they all come from third-party vendors.

The cells themselves, if they came from and ebay vendor or some Chinese vendor, are probably actually a small fraction of the rated capacity. One ampere hour is not unusual for such cells. I've seen photos of tiny lithium polymer cells with a volume on the order of 2 mL packaged inside an 18650 tube.

As others have said, if the capacity of the cells is even a small fraction of what is claimed, anything that would discharge them in 10 seconds would get extremely hot. The insulation probably would have melted off the wires.

In any case, LDO regulators with PNP pass transistors are inefficient because the base drive for the transistor flows to "ground" instead of into the load. Based on the curves in the datasheet, the PNP in that device has a current gain of around 10 at 1 ampere.

 

Thank you all for your reply,
Before I break another voltage regulator can you tell me if this wil work?(picture 1)
I added a diode to secure the circuit, and as I don't have LED I have plug this bluetooth chip to see if it works, but as I don;t want to break everything I am waiting for your response to know if I can plug the batteries.
Also if I understand good what you said, to measure the voltage or the current (let's focus on the voltage for the moment) I should plug the multimeter this way? (see picture 2)

 

They setups for measuring voltage and current are totally different. The way you connect the probes to the circuit varies, and where you plug probes into your meter varies as well. Here are two images showing the basic concept:



Images are from this page, in the "Using a multimeter" section:
https://www.sciencebuddies.org/scie...nces/how-to-use-a-multimeter#usingamultimeter

Sparkfun also has good descriptions:
https://learn.sparkfun.com/tutorials/how-to-use-a-multimeter/measuring-voltage

https://learn.sparkfun.com/tutorials/how-to-use-a-multimeter/measuring-current

 

Thank you all for your reply,
Before I break another voltage regulator can you tell me if this wil work?(picture 1)
I added a diode to secure the circuit, and as I don't have LED I have plug this bluetooth chip to see if it works, but as I don;t want to break everything I am waiting for your response to know if I can plug the batteries.
Also if I understand good what you said, to measure the voltage or the current (let's focus on the voltage for the moment) I should plug the multimeter this way? (see picture 2)

Forgot to answer the other part of your question: using a Bluetooth board as a test load is probably fine. It may have spiky, inconsistent current draw which makes it hard to get solid readings from, but it will definitely prevent the short circuit scenario.

Note that you still need different wiring configurations for current vs. voltage measurements. I'd suggest you read all three links in my previous post, just to make sure it really makes sense, before proceeding.