0
\$\begingroup\$

I have a load which can also be set to "Battery test". I can easily discharge a Li-ion battery using the CC (contant-current) mode, which is also a condition recommended by the manufacturers. For example, a typical recommendation would be:

Discharge CC (18A) to 2.0V @ 20degC

I have worked with CC before and it works as expected. But what about CV (contant-voltage)? One can never see that on the specs/recommendations. If I set one knob to CV and use the "Battery Test" setting, it says that I can't use it. So I assumed that the load doesn't really like the idea of CV and I was wondering why?

I could switch to another type of load (not "Battery Testing"), where I can actually set the CV, but I'm afraid that this would destroy the battery. Can anyone help me to figure out whether or not I could force CV into the battery and why is there no CV into the battery-testing option? Thank you

\$\endgroup\$
2
  • 1
    \$\begingroup\$ How do you suppose a constant voltage load would work on a battery where voltage is defined by the chemistry? It would draw a very high current at first which might be too much for the battery to handle - so it is an unsafe operation. No time for a full answer right now - sorry. \$\endgroup\$
    – Arsenal
    Commented Jan 30, 2018 at 13:14
  • \$\begingroup\$ I always thought that the load can sink that current, but it makes sense that the current is too high. Thank you for the information. \$\endgroup\$
    – Physther
    Commented Jan 30, 2018 at 13:21

3 Answers 3

2
\$\begingroup\$

If you use constant voltage less than the battery voltage the only thing limiting the current is the battery resistance. That, being small, means the initial current is very high and uncontrolled. This will result in overheating and potential fire/explosion.

The whole point of CC discharge is to limit that heating effect.

\$\endgroup\$
2
  • \$\begingroup\$ Thank you. As I wrote above, I thought that the current can actually be sinked by the load, but I guess I was wrong \$\endgroup\$
    – Physther
    Commented Jan 30, 2018 at 13:22
  • 1
    \$\begingroup\$ @Physther that's the point though, in order to present a constant voltage the circuit needs to adjust the load it presents to the battery. When the battery voltage is high that presented resistance is small. \$\endgroup\$
    – Trevor_G
    Commented Jan 30, 2018 at 13:25
1
\$\begingroup\$

The datasheet has specified a CC discharge at 19A to 2V.

This suggests something like a LiFePO4 battery which has max 3.65V and anywhere from 1.8 to 2.5V low voltage cutoff.

Let's say capacity is 20Ah, so you're discharging at the C-rate.

Typical internal impedance is somewhere between 0.5 mohm and 5mohm, let's call it high at 5mohm.

A step change of just 5mV difference to the terminal voltage would produce an instantaneous current of 1A, and 500mV would cause an instantaneous current of 100A to flow.

The very low source impedance produces large variations in current for small variations in voltage - it is better to control the current until the voltages have stabilised.

\$\endgroup\$
0
1
\$\begingroup\$

I think you misread or mis-interpretted the datasheet.

If it read...

Discharge CC (18k) to 2.0V @ 20degC

.... then I would recognize this as a standard (one time) capacity test for 3V Lithium coin cells which start at 3V and never ( ever !) for LiPo as this causes permanent damage. But may ok for max current for Li-Fe.

Li-Fe example

70Ah (ESR 2mΩ)
Charge 3.6V Discharge 2.0V
Max Charge <=3CA Max Discharge <4CA or <12CA pulse
Recommended Charge 18A(CC), 3.6V (CV) , 10%A cutoff

Since the model for a battery is similar to a capacitor

Ic=ΔV/ESR+CΔV/dt
For ΔV=Vinit-Vload = 1.6V and applying a CV load of 2.0V to a cell at 3.6V
ESR = 2mΩ, the 1st part becomes 1.6V/2mΩ ~ 800 Amps
or Pd=I²ESR= 800²*2mΩ = 1280 Watts ( Pfft or kaboom)

\$\endgroup\$
2
  • \$\begingroup\$ Thank you for the answer. If you would actually look at this liionbms.com/pdf/aleees/SE70AHA.pdf, you will see what I wrote. I randomly found that one now, but I saw it also in other batteries specifications \$\endgroup\$
    – Physther
    Commented Jan 30, 2018 at 13:23
  • \$\begingroup\$ That spec also has errors Nominal Capacity / impedence 70Ah / 2 ohms (??) Operating Voltage Charge 3.6V Discharge 2.0V \$\endgroup\$
    – D.A.S.
    Commented Jan 30, 2018 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.