Proper load for zb2l3 module

vanaedium

New member
Joined
Dec 18, 2017
Messages
4
Hi Guys
I'm going to receive 15 zb2l3 modules with two 7,5 ohm 5wresistors. What is the proper load for this module to get 1.0A discharge rate?
I guess 2 R in parallel to get 3,75 ohm? Am I correct?
Thanx
 
U=I*R

You have voltage of 3.7 and want I to be 1. => 3.7/1 = 3.7Ohm So yes you need to parallel those to so you will get the 1A load.
 
Check them as the ones I got were above rated Ohms by quite a bit. I ended up changing them out for 2x 7 Ohm measured and i get 1.1A at 4.2 down to .9 at the end
 
I have to agree with jdeadman: check the resistors for their values. Mine have ranged from 7.1 - 7.9 ohms. If you have a bunch of mixed values, try to match them up so that each Zb2l3 has basically the same load.

It is also good to calibrate the ZB2L3's before you start using them. They can give fairly different voltage readings straight from factory.
 
Hi guys I built 10 discharge modules and I was very happy by them, especially the test speed, but after checking the same cells with my Foxnovo (to check the accuracy) I found zb2l3 are missing an average of 150mAh and in some cases 350mAh. So I had to recharge and restart the slow test with Foxnovo. I checked the Resistors and they are quite accurate to 3.7-3.75. The Voltmeter seems legit and all fine, but the weird thing is that the results lack consistency. I start all tests with full voltage 4.21 4.22. Should I start a calibration? That looks complicated. All modules behavior are the same and all readings are the same. Should I drop and keep going with foxnovo (and next week with opus) or these modules need just tuning and then they are accurate?
 
It's not about tuning. They don't do a CC discharge, current will change together with the dropping voltage while discharging. No matter what you do, you should stick to one method and one device to get consistent results. Ideally it should be a device that does CC, like the Opus. And the Foxnovo possibly as well, don't know. But don't use them both, use either and stick with it.
 
As DarkRaven said they just test differently. Stick to one and use that as the number you want.


There is really no cheap unit like those that fully get accurate numbers in my world. You get what you pay for :p
 
DAmn it! Well they are cheap individually but to get those I had to buy 15 and with shipping and customs more than 100 in total. I'm still waiting the 2 opus. Live and learn.
Thanx Guys
G
 
And your mAh ratings are just base lines. If they are consistently 150mAh off, then I think that's pretty darned good. Unless you mean 150mAh off from each other, then not so good.
If it's the former, that means that the cell is just being discharged under a different curve, as mentioned my DarkRaven, by not being a CC discharge. Personally I think this is ok as you aren't going to be constantly discharging your packs at the same amperage draw until they are flat line anyways. So the non-CC may be closer to a real world discharge.
However, it does not show a worse case scenario, which CC discharge does show. So take that to mind as well.

If the zb chargers are staying consistent, and they work as expected, use them as a guideline just like using the opus as guideline. We know the opus isn't 100% accurate and is actually slightly generous in its readings. So if the zb charger is lower than the Opus, maybe it's closer to the actual mAh rating. But still, this is a baseline. I would always round down when taking measurements to get safer readings.
 
The only problem when it is not CC is that you have to watch out for some special, worn out cells. I've found cells with 2000mAh at CV and 500mAh or less at CC with 1A.
CV will tell you the capacity of the cell, CC will tell you the usable capacity at a certain current. With a real world load you will probably haven't either, but with CC you know at least that the cell will be able to deal with a certain current. With CV you don't know, unless you constantly monitor the process.
 
Hi daromer (and forum members too : )

I realizethat this is an old thread. If you would prefer that I start a new thread, just let me know and I will delete this post and re-post (or go away, ha!) as you advise. So here goes my first post on SecondLifeStorage . . .

First I'll mention I'm not an electronics or storedenergypower expert. I'mgood enough with a soldering iron,basic multi-meter use, and basic electronics repairs. My "challenge" :

I just received a ZB2L3 v2.3g. I'm hoping I can use it to test 12v sealed lead acid batteries, typically 12v 7Ah or 12v 9Ah.I've salvaged somepos and neg F2 leadsfrom an old unit. I trimmed off maybe 30% of the wire diameter, then tinned with solder, then spun the tinned end on abench grinder to make a nice pin/plug that fits into theZB2L3 screw down terminals. I put some shrinkwrap on the "pin" ends and the positive F2 connector.Here's a pic if interested:

image_zxsnkf.jpg


If I'm not already completely wrongwith this project:If I attach the two 5w 7.5 ohm resistors in parallel, I'll get a ~1A load. But I have afeeling thaton these12v 7Ah or 12v 9Ah batteries,those resistors are going to heat up really fast and I supposefail in some way internally. A couple months ago when I was researching this I saw a Youtube with someone using one of theseZB2L3 and a 50w resistorattached to a PC CPU heatsink, with the heatsinkfan running:
(Oops, can't add links on 1st post: will add later)

(that example is a 6 ohm)
Here's the full selection of those:
(Oops, can't add links on 1st post: will add later)

Will theZB2L3 work for my needsif I use a resistor like those RH-50'swith a powered CPU heatsink? Or is this just not going to work, or would work just fine with the two 5w 7.5 ohm resisters that came with theZB2L3?

Sorry for the elongated post. I look forward to any comments or suggestions you care to offer.

Regards . . .


Brcobrem said:
 
Brcobrem said:
Hi daromer (and forum members too : )

I realizethat this is an old thread. If you would prefer that I start a new thread, just let me know and I will delete this post and re-post (or go away, ha!) as you advise. So here goes my first post on SecondLifeStorage . . .

First I'll mention I'm not an electronics or storedenergypower expert. I'mgood enough with a soldering iron,basic multi-meter use, and basic electronics repairs. My "challenge" :

I just received a ZB2L3 v2.3g. I'm hoping I can use it to test 12v sealed lead acid batteries, typically 12v 7Ah or 12v 9Ah.I've salvaged somepos and neg F2 leadsfrom an old unit. I trimmed off maybe 30% of the wire diameter, then tinned with solder, then spun the tinned end on abench grinder to make a nice pin/plug that fits into theZB2L3 screw down terminals. I put some shrinkwrap on the "pin" ends and the positive F2 connector.Here's a pic if interested:

image_zxsnkf.jpg


If I'm not already completely wrongwith this project:If I attach the two 5w 7.5 ohm resistors in parallel, I'll get a ~1A load. But I have afeeling thaton these12v 7Ah or 12v 9Ah batteries,those resistors are going to heat up really fast and I supposefail in some way internally. A couple months ago when I was researching this I saw a Youtube with someone using one of theseZB2L3 and a 50w resistorattached to a PC CPU heatsink, with the heatsinkfan running:
(Oops, can't add links on 1st post: will add later)

(that example is a 6 ohm)
Here's the full selection of those:
(Oops, can't add links on 1st post: will add later)

Will theZB2L3 work for my needsif I use a resistor like those RH-50'swith a powered CPU heatsink? Or is this just not going to work, or would work just fine with the two 5w 7.5 ohm resisters that came with theZB2L3?

Sorry for the elongated post. I look forward to any comments or suggestions you care to offer.

Regards . . .


Brcobrem said:
heatsinkfan running:
Vishay Dale RH0506R000FE02
(that example is a 6 ohm)
Here's the full selection of those:
https://www.mouser.com/Passive-Components/Resistors/Wirewound-Resistors/Wirewound-Resistors-Chassis-Mount/_/N-7fx9f?keyword=Dale%20RH-50


Can anyone recommend a forum where theremightbe experiencethat would be able to help with my project?
Thanks.


 
Keep in mind doing a "full capacity test" & flattening any LA battery will significantly shorten it's life every time.
LA batteries don't like being below typically about 70% & being flattened really damages them.
Google for "life vs depth of discharge of lead acid battery".
They should be recharged to full every time & regularly "top-up charged" due to self discharge.

Maybe have a think about the current you want to use to test your SLA batteries with.
Then use Ohms law(s) to work out the resistance.
V= I x R or
I = V/R or
R = V/I
So eg at 12V (V) you want say 0.75 A (I) so it's R = V/I = 12/0.75 = 16 Ohms

Once you know the resistance, you can find the power the resistor(s) will dissipate by
P = Vsquared/R ie V x V/R or
P = V x I
So for the eg above V = 12V, I = 0.75A, so P = V X I = 12 x 0.75 = 9 watts
Similarly, for a 1A load, it would be 12 ohms & dissipation would be 12W
Yes, you'll need a decent heatsink & robust resistors (or many smaller ones) like your link.

The ZB2L3 can only do up to 3A (maybe stick to max 2A or less) & 15V.
 
Redpacket said:
Keep in mind doing a "full capacity test" & flattening any LA battery will significantly shorten it's life every time.
LA batteries don't like being below typically about 70% & being flattened really damages them.
Google for "life vs depth of discharge of lead acid battery".
They should be recharged to full every time & regularly "top-up charged" due to self discharge.

Maybe have a think about the current you want to use to test your SLA batteries with.
Then use Ohms law(s) to work out the resistance.
V= I x R or
I = V/R or
R = V/I
So eg at 12V (V) you want say 0.75 A (I) so it's R = V/I = 12/0.75 = 16 Ohms

Once you know the resistance, you can find the power the resistor(s) will dissipate by
P = Vsquared/R ie V x V/R or
P = V x I
So for the eg above V = 12V, I = 0.75A, so P = V X I = 12 x 0.75 = 9 watts
Similarly, for a 1A load, it would be 12 ohms & dissipation would be 12W
Yes, you'll need a decent heatsink & robust resistors (or many smaller ones) like your link.

The ZB2L3 can only do up to 3A (maybe stick to max 2A or less) & 15V.

HiRedpacket,

Thank you for that wonderful explanation and google researchreference. I'm one of thosenumskulls who can technicallydo the math, but someonewho's brain works better with a real-worldexample.Theexamples were veryhelpful. I "get it" now! Whoopee!

Question: When you say "flattened" do you mean draining the battery until it's dead? If yes, and I understand the ZB2L3 manual/directionscorrectly,theZB2L3 will run the test until the battery drains down to a certain voltage, then the test stops. I think the low voltage "cutoff" voltage is determined by theZB2L3 based on the voltage it senses when the test begins. This meansthat the battery has to be in relatively good condition and fully charged to specification.

I'm really busy for the next couple weeks but I will return to this little project. When I do I'll reply back and let you know how I made out (or what I fried, ha!) and how theZB2L3 held up to the task.

Regards . . .
 
Try repeated tests of the same cell and see if you get similar results each time.
I gave up on testers like that after getting different results each time, even though it was the same charger, same cell and same capacity tester each time.
 
Brcobrem said:
Question: When you say "flattened" do you mean draining the battery until it's dead? If yes, and I understand the ZB2L3 manual/directionscorrectly,theZB2L3 will run the test until the battery drains down to a certain voltage, then the test stops. I think the low voltage "cutoff" voltage is determined by theZB2L3 based on the voltage it senses when the test begins. This meansthat the battery has to be in relatively good condition and fully charged to specification.

Flattened is when the state of charge is close to zero, the cell might have some voltage on it but little useful charge left & the remaining voltage quickly drops under load.
0V is a (bad!)giveaway....

Not sure how the ZB2L3 determines low voltage cutoff - I would have thought is was programmable?
This link suggests it'll automatically set that limit but you ccan vary it with the buttons:
https://www.instructables.com/id/ZB2L3-BATTERY-CAPACITY-TESTER/
 
Redpacket said:
Keep in mind doing a "full capacity test" & flattening any LA battery will significantly shorten it's life every time.
LA batteries don't like being below typically about 70% & being flattened really damages them.
Google for "life vs depth of discharge of lead acid battery".
They should be recharged to full every time & regularly "top-up charged" due to self discharge.

Maybe have a think about the current you want to use to test your SLA batteries with.
Then use Ohms law(s) to work out the resistance.
V= I x R or
I = V/R or
R = V/I
So eg at 12V (V) you want say 0.75 A (I) so it's R = V/I = 12/0.75 = 16 Ohms

Once you know the resistance, you can find the power the resistor(s) will dissipate by
P = Vsquared/R ie V x V/R or
P = V x I
So for the eg above V = 12V, I = 0.75A, so P = V X I = 12 x 0.75 = 9 watts
Similarly, for a 1A load, it would be 12 ohms & dissipation would be 12W
Yes, you'll need a decent heatsink & robust resistors (or many smaller ones) like your link.

The ZB2L3 can only do up to 3A (maybe stick to max 2A or less) & 15V.
Redpacket said:
Keep in mind doing a "full capacity test" & flattening any LA battery will significantly shorten it's life every time.
LA batteries don't like being below typically about 70% & being flattened really damages them.
Google for "life vs depth of discharge of lead acid battery".
They should be recharged to full every time & regularly "top-up charged" due to self discharge.

Maybe have a think about the current you want to use to test your SLA batteries with.
Then use Ohms law(s) to work out the resistance.
V= I x R or
I = V/R or
R = V/I
So eg at 12V (V) you want say 0.75 A (I) so it's R = V/I = 12/0.75 = 16 Ohms

Once you know the resistance, you can find the power the resistor(s) will dissipate by
P = Vsquared/R ie V x V/R or
P = V x I
So for the eg above V = 12V, I = 0.75A, so P = V X I = 12 x 0.75 = 9 watts
Similarly, for a 1A load, it would be 12 ohms & dissipation would be 12W
Yes, you'll need a decent heatsink & robust resistors (or many smaller ones) like your link.

The ZB2L3 can only do up to 3A (maybe stick to max 2A or less) & 15V.
Hi Redpacket,

Three months later . . .I'm finally able to returnto this thread.So sorryfor the delay. If I may:

Above you said,"Similarly, for a 1A load, it would be 12 ohms & dissipation would be 12W".

I understand the math, but have a newbie electronics hardwarequestion please: If I could only find a single12ohm 15W resistor, thatwould work because I've at least met the 12W requirement of this scenario. Is thatthinking right?

Similarly, If I could onlyfind 6 pieces of 12 ohm2 watt resistors, I could put them inparallelandthat would work?

Is the key herein thisscenario,thatthe resistorsneed to be 12ohm resistorsif inparallel? Or if the resistors arein series, thecombination of resistors would need to total12 ohms total,and each resistor would need to berated for at least 12 watts (andhigherwattagerated12 ohmresistorswouldbe too)?

I hope I am not just muddying the watersfurther with these questions. I've just never had anyone explain to me how to combine or substitute components to meet the minimum requirements of the circuit.

Thanks for any clarification/training you care to provide (or anynewbiebutt kickingyou feel is necessary :>

Regards . . .
 
Distributing the power you want to dissipate across several resistors is a good idea - resistors can get very hot (>100degC) dissipating say 10W without heatsinks.

Per the math above if you wanted to share the power dissipated between two resistors with the same voltage you could either:
a) use two resistors in series (using the original single resistor value)
b) use two resistors in parallel (using double the original single resistor ohms value for each one)

Brcobrem said:
If I could only find a single12ohm 15W resistor, thatwould work because I've at least met the 12W requirement of this scenario. Is thatthinking right?
A: Yes, but you'll need a heatsink.

Brcobrem said:
Similarly, If I could onlyfind 6 pieces of 12 ohm2 watt resistors, I could put them inparallelandthat would work?
A: No. If you just paralleled them on the same voltage, they would all try & dissipate 12W each & burn out.
If you put some of them in series, then yes they would each dissipate less. You'd have to calculate the current & dissipation
eg 3x 12 ohms in series across 12V = 12/(12+12+12) = 0.333A. Volts per resistor = 12/3V = 4V. So Pwr = 4x 0.333 = 1.333W each.

Brcobrem said:
Is the key herein thisscenario,thatthe resistorsneed to be 12ohm resistorsif inparallel? Or if the resistors arein series, thecombination of resistors would need to total12 ohms total,and each resistor would need to berated for at least 12 watts (andhigherwattagerated12 ohmresistorswouldbe too)?
Key is the combination of resistors across the load needs to be = the resistance you need to draw the desired load current.
And each resistor you've chosen needs to be rated for the power it will be dissipating in that combination.
 
Back
Top