I ordered this test jig to securely hold my cells so I can measure them accurately.
I made some progress assembling it tonight, but I don’t understand how the leads are supposed to be attached.
It came with crimp ring attachments which have matching nuts. However, there is no electrical connection between the pin that attaches to the battery and the nut that attaches to the ring. So, how are the wire leads supposed to connect to the battery?
The center pin that you solder to is a voltage sense. Think of it as an extension of your dmm leads. The large outer contacts are for current handling and connect with ring terminals and larger gauge wire.
What do you plan on using it with. Few devices can utilize the 4point connection and take advantage of it.
But they don’t connect with the battery, so how do they help?
I need to buy a big 1-ohm resistor and then use my Fluke, right? Something like that. I saw a video that uses a resistor and uses some math to find out the internal resistance.
I could also just use them for slowly discharging the cell for discharge tests, right? Wouldn’t need all four wires.
I’m a bit confused also. I watched part of the video. Got bored watching the guy assemble the thing. Never saw him actually use it.
Are the ring terminals not connected to the same contact as the soldered on wire?
If so, other than providing a second connection point for a meter - what’s the deal?
fogus, a 1 ohm resistor is going to produce some serious heat if connected + to - across the battery. You sure you want to use that big a load?
How were you planning to use it?
All the Best,
Jeff
They do, the center pin is independently sprung load. I have many of these fixtures and use them often. Fully assemble and install a cell to see. I’ll take a picture of my setup for you to see.
[quote=fogus]
1ohm should be okay… 4.2V and 1ohm equals a maximum of 4.2Amps. Only the best cells will even crest 4A. IR would be determined by the delta V of the cell at rest to underload measured from the center pin divided by the V drop across the resistor (assuming a perfect 1 ohm and therefore 1:1 with amps)
I did fully assemble it and put a cell in and got connectivity neither between the ring connector and the interior pin nor between the ring and the battery electrode. Only the pin has an electrical connection to the battery electrode. The pin is electrically isolated from the chrome-steel-looking metal which makes contact with the ring crimp connector.
However, I think I’m still being stupid because this doesn’t make sense to me: why make someone solder onto the pin? It’s awkward compared to the ring connector. Why not just use two rings? There must be a reason.
Am I supposed to open the whole thing up and solder something inside where the spring-loaded pin holder is? Sounds crazy, I know.
Ah, wait, maybe I didn’t have the “car” tight enough. Only the pin was contacting the cell. When I moved it a notch or two tighter, the chrome-steel-looking-thing-where-the-pin-comes-out-of made contact with the cell electrode also.
I'll borrow funtastic's picture, and placed some labels.
When inserting the battery, the battery has to touch *both* the "center pin" (voltage sense) and the bigger cone (current), which means you'll need to slide the battery tight enough that it holds into place without falling.
I doubt it fits 21700 battery though.
(I'm looking for something similar that fits 21700 battery (if possible, protected 21700, but if not possible, fits unprotected 21700 is fine; preferably supports up to 10 Amp load, better is maybe up to 15 Amps). Haven't found one yet though)..
You may want to refer to this message thread for some related info.
The pin is spring loaded itself and had enough of a hold on the cell to suspend it, leading to my confusion. One notch tighter and it would have all made sense.
Yes, the inner pin and the outer cone both have to touch to utilize 4-wire sensing, primarily internal resistance. I’ve gotten confused when it wasn’t working right because I didn’t set it tight enough.
I understand the necessity of using "Four Wire versus Two Wire" techniques to reduce the effect of test lead resistance when measuring low resistances, but I do not understand why the "pins and outer cones" on Four-Wire battery holders (and YR1030 style Battery Internal Resistance Testers) need to be electrically separated where they contact the battery being tested.
QUESTION: What issues occur if there is only one robust connection to each end of the battery and the two test leads for each end of the battery are connected to only one connection at each end of the battery instead of having two separate connections at each end of the battery?
It doesn't seem like the two leads at each end of the battery need to be electrically isolated since they are electrically shorted when touching the battery and it seems the added resistance from having only one connector at each end of the battery instead to two would be negligible compared to the resistance of the test leads reading the voltage, but there must be some reason I am not understanding since it is always done this way with separated test leads.
I’m not an expert on electronics, but from what I understand, at low current load, the voltage sag may not be that noticeable.
But with higher current load, voltage sag happens when load is passing through the cable. (As to how much current the voltage sag becomes significant, I’m not very sure, but there should be a way to compute how much voltage sag versus the size of the wire we can live with, for the specific testing scenario)
Maybe if the 2-wire is really thick and short enough that the resistance becomes very low, then voltage sag may be much less noticeable, but it’s still better to separate the voltage sense (which can be a thinner wire) vs the current wire (somewhat thicker cable).
For instance, this video shows the voltage sag bcomes significant in his testing scenario (he’s using the earlier DL24P with 2-wire ; probably the problem will have been fixed if using the newer DL24P with 4-wire)…
The spring loaded pin allows measurement of cell voltage all the time in both the Open circuit and Loaded conditions, and this allows you to see the pull-in sag and monitor if a recovery bounce occurs when the load is removed.
So separate isolated leads are required—otherwise the Load is always connected and no Open circuit measurement is possible.
