battery voltage drops with high currents

[Integrated]

battery voltage drops as current goes up

Hi all,

my question is about use of battery in a robot, i used a battery pack , with serial and paralleled cells and 7v output, and it meant to produce 10000mha.

but when all the motors of robot (all toghether up to 25A) start sinking current , the voltage of the battery (input of the regulator) drops continoulsy and in less than five minutes, my regulator goes off. and the robot stops.

Is such thing normal , that when the current increases ,the voltage drops? i mean , when its said that the battery is 10AH , does it mean that with 20A in 30minutes the voltage is the same during all this time?

thanks and best regards.

 

[Davood Amerion]

voltage drop for high load

Actually when capacity of battery is K AH that means it produce K/10 Ampere in 10 hours.
when our circuit (system) consume K Ampere it can not working 1 hour (say 45 minutes or less).
and for 2K ampere condition is worse.

 

[hermin]

battery voltage drops after short time under load

qwll, theoretically if it says 10AH, it could supply a constant 10A for one hour, but this usually doesnt happen, its the maximum it could give so give it a 70% window of operation, let it operate at about 7A or lower to make it work properly, if you can, try to make use of two power supplies so the the battery will not die out so easily, also it preserves its life, the larger the current it delivers the more the battery loses its quality with time,

 

[xxargs]

unloaded battery voltage

Hi all,

my question is about use of battery in a robot, i used a battery pack , with serial and paralleled cells and 7v output, and it meant to produce 10000mha.

but when all the motors of robot (all toghether up to 25A) start sinking current , the voltage of the battery (input of the regulator) drops continoulsy and in less than five minutes, my regulator goes off. and the robot stops.

Is such thing normal , that when the current increases ,the voltage drops? i mean , when its said that the battery is 10AH , does it mean that with 20A in 30minutes the voltage is the same during all this time?




thanks and best regards.

No, battery have varying internal resistance depend of charge state and how much current drawing out.

Depend of lowest acceptable voltage, you can only take out 50% - 70% of Ah-capacivity for < 30 minutes discharge time (and watch battery temperature).

read discharge charts from battery manufacture to se how deep you can discharge and actual voltage for specefic load time.

remember, battey is driven on chemical process and take time for substance to moving place inside battery and give very unlinear properties ...

 

[Tohu]

2.5 volt russian battery

Hi,
all posts before is exactly right, but 5 min is too small time, what type of batteryes you use,may be you not use the right type.

 

[xxargs]

7.2 volt battery pack voltage drop under load

Hi,
all posts before is exactly right, but 5 min is too small time, what type of batteryes you use,may be you not use the right type.

hmm. 7 Volt -> 2 x 3.6 Volt LiIon ???

LiIon have much higher internal impedance compare to NiMh and old NiCd-battery still best for very high current delivery.

check type of battreies using in electrical model fligth and RC-car...

 

[pisoiu]

Try to take a look at Ultracaps, from www.epcos.com . Their internal resistance is much lower than accu's. These caps handle very well huge loads.

/pisoiu

 

[throwaway18]

Perhaps you could change your software so the robot only uses some of it's motors at once.

I agree with the points about checking the battery capability.

You should also check the thickness of the conductors from the battery.
Even for short distance you need thick cable to carry 25A without significant voltage drop.

A ultracap stores much less energy than a battery of the same volume.
You could not replace the battery with ultracaps but it might be
usfull to provide extra current for a few seconds.

With a physically-huge 1200F ultracap in theory I=C.dv/dt means you
could get 15A for 120seconds with the capacitor voltage dropping
by 2Volt. **broken link removed**
I looked at random is 2.8Kg and probably very expensive.

 

[artem]

For sealed lead acid battery (example 6V 7Ah https://lib.store.yahoo.net/lib/zbattery/bp7-6.pdf) maximum discharge current can be up to 105 A for 5 seconds . Internal resistance for fully charged battery is less than 16 mOhm. That is for your case means less than a volt drop if loaded with 25 A.

The question is - what battery type was used in robot ?
Lead acid batteries provide higher discharge current than most of other types, so worth to consider them also because of their price - due to wide ups usage those are very cheap.

 

[xxargs]

For sealed lead acid battery (example 6V 7Ah h**p://lib.store.yahoo.net/lib/zbattery/bp7-6.pdf) maximum discharge current can be up to 105 A for 5 seconds . Internal resistance for fully charged battery is less than 16 mOhm. That is for your case means less than a volt drop if loaded with 25 A.

The question is - what battery type was used in robot ?
Lead acid batteries provide higher discharge current than most of other types, so worth to consider them also because of their price - due to wide ups usage those are very cheap.

And possible need adjust shutdown voltage depend of current delevery...

For example Lead acid batteries not normay recomended going down lower than 10.8 Volt (12V-battery) in >1 h load, but for rapid disharge in short time can accept low voltage as 7.5 Volt (and rule is voltage going higher than 10.8 Volt after load removed in short time). Very fresh charged lead acid battery have 13.8 -15 Volt initial voltage and shutdown around 7.5 Volt (fast discharge) - you Power supply need handle around 7.5 Volt vary input without fault here in this case.

I thinking simular for LiIon-batteries needs using adjustable lowest voltage treshold depend current of load. (and possible need remove battery secury-circurits and replace with more advanced discharge control). Electrical model fligth people using now LiIon and this motors draw huge of current - study how this guys solve battery problem, electrical model fligth needs pretty optimizing on energy storage, weight, size and still needs very high current delevery capacity...

and remember - if battery voltage going down to half of unload voltage in hevay load sitiation, you draw maximum power battery can delevery (and heaten battery itself same as load power - and usable energi is 50% of battery energy storage)

(ie is case load have samme resistance as batterys internal resistance -> maximal power match)

... this situation can see on 3-5 Watt LED-flaslight and 2x CR123 Litium battery voltage of 6 Volt unloaded, going to 3.5 Volt loaded... and whole flashligt going hot in short time...

Added after 1 hours 54 minutes:

Hi all,

my question is about use of battery in a robot, i used a battery pack , with serial and paralleled cells and 7v output, and it meant to produce 10000mha.

but when all the motors of robot (all toghether up to 25A) start sinking current , the voltage of the battery (input of the regulator) drops continoulsy and in less than five minutes, my regulator goes off. and the robot stops.

Is such thing normal , that when the current increases ,the voltage drops? i mean , when its said that the battery is 10AH , does it mean that with 20A in 30minutes the voltage is the same during all this time?

thanks and best regards.

Hmm, if you using simple 7805-regulator to feed 5 volt logic and only 7 Volt battery - i thinking you have problem in high load situation...

even if use low-drop regulator your need 6.5 Volt minimum input for 5 Volt out in 1 A or more.

If case using LiIon-batter as 2 in series (7.2 Volt nominal):

typical nominal 3.6 Volt LiIon have 4.2 Volt/cell (8.4volt) in full charge state, and 3.0 Volt in discharge state (most LiIon battery protect circurit break battery circurit lower than 2.5 -2.8 Volt independ of current load - ie. cut circurit if battery voltage going under 2.5 Volt and result break power in shorter time and higher left charge state on high load situation)

your '7-Volt' battery have in real range of 6.0 - 8.4 Volt unloaded depend of charge state. If battery have 20 mOhm internal resistans, voltage going down 0.5 Volt in 25 A current load situation (ideal - in practical more voltage loss).

ie. you need better than 7 Volt unloaded charge state if you still want 5 Volt to logic under 25 A load. - and if logic stop < 4.75 Volt - reprecent 6.75 Volt unloaded charge state and you have around 30-50 % capacy left on battery...

If your problem is describen above - you first step try using separate battery (and 1.5 volt higher voltage) to feed logic and see if you can use higher yeld of big battery capacity. If so, you need reconstruct power supply to logic to handle 4 Volt to 10 Volt input for out @ 5 Volt [1], or use 3.3 Volt logic or use big batteries of 10.8 Volt nominell (3 x 3.6 Volt LiIon-cell in series) and possible reconstruct whole drive line of robots - pick easiest method ;-)



[1] needs switching regulator can work in both buck and boost-mode depend of input voltage or using boost regulator to make around 8-10 Volt and using linear regulator to take down to 5 Volt. if using big capacitans of 10 Volt you have nice spare power for logic in cases of high load (and low voltage) situation on robot work