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Proper load for zb2l3 module
#11
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.
vanaedium and Korishan like this post
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#12
Hi daromer (and forum members too : )

I realize that 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 stored energy power expert. I'm good 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 some pos and neg F2 leads from an old unit. I trimmed off maybe 30% of the wire diameter, then tinned with solder, then spun the tinned end on a bench grinder to make a nice pin/plug that fits into the ZB2L3 screw down terminals. I put some shrink wrap on the "pin" ends and the positive F2 connector. Here's a pic if interested:


If I'm not already completely wrong with this project: If I attach the two 5w 7.5 ohm resistors in parallel, I'll get a ~1A load. But I have a feeling that on these 12v 7Ah or 12v 9Ah batteries, those resistors are going to heat up really fast and I suppose fail in some way internally. A couple months ago when I was researching this I saw a Youtube with someone using one of these ZB2L3 and a 50w resistor attached to a PC CPU heatsink, with the heatsink fan 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 the ZB2L3 work for my needs if I use a resistor like those RH-50's with 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 the ZB2L3?

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

Regards . . .

(05-08-2019, 08:25 PM)Brcobrem Wrote: heatsink fan 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
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#13
(05-08-2019, 08:25 PM)Brcobrem Wrote: Hi daromer (and forum members too : )

I realize that 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 stored energy power expert. I'm good 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 some pos and neg F2 leads from an old unit. I trimmed off maybe 30% of the wire diameter, then tinned with solder, then spun the tinned end on a bench grinder to make a nice pin/plug that fits into the ZB2L3 screw down terminals. I put some shrink wrap on the "pin" ends and the positive F2 connector. Here's a pic if interested:


If I'm not already completely wrong with this project: If I attach the two 5w 7.5 ohm resistors in parallel, I'll get a ~1A load. But I have a feeling that on these 12v 7Ah or 12v 9Ah batteries, those resistors are going to heat up really fast and I suppose fail in some way internally. A couple months ago when I was researching this I saw a Youtube with someone using one of these ZB2L3 and a 50w resistor attached to a PC CPU heatsink, with the heatsink fan 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 the ZB2L3 work for my needs if I use a resistor like those RH-50's with 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 the ZB2L3?

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

Regards . . .

(05-08-2019, 08:25 PM)Brcobrem Wrote: heatsink fan 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 there might be experience that would be able to help with my project?
Thanks.
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#14
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.
Running off solar, DIY & electronics fan :-)
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#15
(05-10-2019, 10:51 AM)Redpacket Wrote: 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,

Thank you for that wonderful explanation and google research reference. I'm one of those numskulls who can technically do the math, but someone who's brain works better with a real-world example. The examples were very helpful. 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/directions correctly, the ZB2L3 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 the ZB2L3 based on the voltage it senses when the test begins. This means that 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 the ZB2L3 held up to the task. 

Regards . . .
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#16
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.
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#17
(05-13-2019, 02:04 PM)Brcobrem Wrote: Question: When you say "flattened" do you mean draining the battery until it's dead? If yes, and I understand the ZB2L3 manual/directions correctly, the ZB2L3 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 the ZB2L3 based on the voltage it senses when the test begins. This means that 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-B...TY-TESTER/
Running off solar, DIY & electronics fan :-)
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#18
(05-10-2019, 10:51 AM)Redpacket Wrote: 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.
(05-10-2019, 10:51 AM)Redpacket Wrote: 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 return to this thread. So sorry for 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 hardware question please: If I could only find a single 12 ohm 15W resistor, that would work because I've at least met the 12W requirement of this scenario. Is that thinking right? 

Similarly, If I could only find 6 pieces of 12 ohm 2 watt resistors, I could put them in parallel and that would work?

Is the key here in this scenario, that the resistors need to be 12 ohm resistors if in parallel? Or if the resistors are in series, the combination of resistors would need to total 12 ohms total, and each resistor would need to be rated for at least 12 watts (and higher wattage rated 12 ohm resistors would be too)?

I hope I am not just muddying the waters further 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 any newbie butt kicking you feel is necessary : >

Regards . . .
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#19
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)

(08-11-2019, 08:08 PM)Brcobrem Wrote: If I could only find a single 12 ohm 15W resistor, that would work because I've at least met the 12W requirement of this scenario. Is that thinking right? 
A: Yes, but you'll need a heatsink.

(08-11-2019, 08:08 PM)Brcobrem Wrote: Similarly, If I could only find 6 pieces of 12 ohm 2 watt resistors, I could put them in parallel and that 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.

(08-11-2019, 08:08 PM)Brcobrem Wrote: Is the key here in this scenario, that the resistors need to be 12 ohm resistors if in parallel? Or if the resistors are in series, the combination of resistors would need to total 12 ohms total, and each resistor would need to be rated for at least 12 watts (and higher wattage rated 12 ohm resistors would be 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.
Running off solar, DIY & electronics fan :-)
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