Battery Circuit Incomplete?

[Stains]

This is battery only, no recharging.
It seems that the neg of the two batteries in series should go to ground.
What do the zener diodes in the circuit accomplish?

https://obrazki.elektroda.pl/88_1307038148.jpg

 

[KerimF]

You are right. "The neg of the two batteries in series should go to ground."
Here the two zeners seem to act like ordinary diodes perhaps just to lower the battery 6V to become closer to 5V.

 

[srizbf]

the zener acts as ordinary diode as said.
and the two diodes also prevent the 2-6v lithium pack prevent from inter battery back current circulation.

 

[Stains]

I WIRED IN THE NEGATIVE TO GROUND AND I HAVE GOOD VOLTAGE.
WHEN I INSERTED THE BATTERY PACK IN THE INTENDED DEVICE IT SAID LOW BATTERY. THESE ARE BRAND NEW LiSO4 CELLS STORED ABOUT 1 YEAR.
HOW WOULD I MEASURE THE CURRENT OUTPUT FROM THESE BATTERY CELLS WITHOUT INSERTING THEM INTO THE INTENDED DEVICE.

 

[KerimF]

HOW WOULD I MEASURE THE CURRENT OUTPUT FROM THESE BATTERY CELLS WITHOUT INSERTING THEM INTO THE INTENDED DEVICE.

We usually measure two things on a battery:
Its open circuit voltage (Voc).
Its internal resistance (R_int).

I think you already measured its voltage and got about 6V.
Now we need to measure its internal resistance.

We can do it while charging or discharging it.
The idea is to pass a suitable current in one direction for a very short time (AC current can also do the job but doing the measurement using DC is rather easier).
In your case, I think a current about 1A is a good choice for a reasonable accuracy since your fuse is 10A (The less the testing current is, the less the accuracy of the measured resistance R_int will be).

First let me assume that your battery is well charged so that it can provide 1A (if necessary we will make it around 500mA) for a few seconds.
To calculate the value of the discharging resistor R_dis, we use the simple Ohm law:
R_dis = V_bat / I_dis
if V_bat=6V and I_dis=1A then R_dis= 6Ω
In this short time, the power dissipation is:
P_dis = V_bat * I_bat = 6 * 1 = 6 Watt
Maybe you can find a 10W power resistor of 6.8 Ω (standard value).
Or you can use 20 resistors connected in parallel; each 500mW 120R.

The test steps:
Connect your voltmeter to the two terminals of the battery under test.
Write the exact reading of its voltage. This is its open circuit voltage (Voc).
While your voltmeter is connected, connect the two terminals of R_dis (you have prepared) and read the exact new voltage on the battery. This is the battery voltage under load(Vload).
Remove R_dis and the test is over.

I_dis = Vload / R_dis
R_int = (Voc - Vload) / I_dis
Finally:

R_int = R_dis * (Voc - Vload) / Vload

Now after getting R_int, we can know in advance the battery voltage under any load (Rload) or discharging current (I_dis) :

Vload = Voc * Rload / (Rload + R_int)
Vload = Voc - I_dis * R_int

In case the battery is under charge:
V_chg = Voc + I_chg * R_int
Note: Voc increases with time during the charge so to measure its instant value, one may remove the charger momentarily (to let I_chg=0).

Kerim

Edited:
If 1A is rather high for the test you can try using 10 resistors only of 120Ω / 500mW.
In this case R_dis = 12Ω and I_dis≈500mA. (P_dis ≈ 3 Watt)