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ripple voltage on the battery

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Derun93

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Hi, I have a balancing circuit as shown in figure. Cells are balancing each other by transporting enrgy from higher energy cell to lower energy cell. My problem is that; the duty of the capacitors is to smooth battery voltages but when I change the battery capacities, the ripple voltage changes. I do not know why. I thought, the value of the capacitors can be found as ILoad*the time swith is opened/ ripple voltage. Please help me to find the value of the capacitor and why the ripple voltage depends on the battery capacity.
THANK YOU.
balancing circuit.png
 

Ripple voltage does not depend on battery capacity, it depends on battery impedance. If you want to know how the impedance changes in your simulation, review the battery model parameters.
 

Higher battery capacity (within family) usually means lower
battery impedance (more plates = less R, more C). But also
you want to accommodate / tolerate changes in the cell
impedance as aging tends to drive that up (increasing ripple
in a switcher type supply).

First question is, do you care about ripple? Beyond the need
to make your control electronics tolerate it?
 

With control algorithm, ı can tolerate it but is there any limit for ripple voltage to not ruin battey chemistry? Also what is the formula of the ripple voltage for this circuit.
Thank you for answers by the way.
 

Hi, I have a balancing circuit as shown in figure. Cells are balancing each other by transporting enrgy from higher energy cell to lower energy cell. My problem is that; the duty of the capacitors is to smooth battery voltages but when I change the battery capacities, the ripple voltage changes. I do not know why. I thought, the value of the capacitors can be found as ILoad*the time swith is opened/ ripple voltage. Please help me to find the value of the capacitor and why the ripple voltage depends on the battery capacity.
THANK YOU.
View attachment 131952
do not use inductive voltage or current sense.
use low inductive wiring between switch to source and battery load (twisted pair)
use low inductive probe with ground lead <1cm to prevent measurement errors of ground loop and EMI.
If inductor is just for LPF , then measure Vripple across battery with AC coupler low impedance termination I.e. 50 Ohm with suitable C. If going into ADC , use precision rectifier with gain, otherwise coax with series R and 50 Ohm termination at scope for clean ripple measurement. Battery load ESR impedance ought to be <<<1 Ohm. If you wish to measure ESR of battery, drive a 50 Ohm pulse gen into battery with ultralow ESR series cap and measure ripple voltage and compute attenuation and load ESR with low ESL cable.
 
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Thanks Sunnyskyguy for your advices. I am going to use your instructions but right now I need exact formula of the ripple voltage. Because it is my thesis project and ı need to write how ı determine the value of the circuit components.
Thank you.
 

Thanks Sunnyskyguy for your advices. I am going to use your instructions but right now I need exact formula of the ripple voltage. Because it is my thesis project and ı need to write how ı determine the value of the circuit components.
Thank you.

try this . Using equivalent Capacitance of battery Ah

I=C ΔV/ΔT for lossless battery for ramped voltage

Rb = ESR of battery = ΔV/ΔI [Ω]

or in other words ripple ΔV=ΔI*Rb

But if there is series inductance from power source, this integrates the pulse voltage and attenuates ripple..

All capacitors and batteries alike in the same quality, size and voltage range may have different Ah ratings like capacitors have different uF ratings.
From my experience low ESR e-caps are have a C*ESR=T time constant <10 µs and smallest <1µs, while general purpose (G.P.) caps are > 100 µs to > 1000 µs in larger sizes.

You can research all battery chemistry and sizes to find the ESR and ESR*C or ESR*Ah product constants for quality and types.

Generally the RdsOn in the Switch must be << Load. Since Rb or ESR of battery is very low (mOhm), the use of SMPS topology raised effective impedance by duty cycle of switch and inductive impedance. But Switch loss is a tradeoff with cost and C/x charge rates and ripple voltage depends on Z=|ωL| which depends on cost budget, charge rate and pulse frequency

Higher f also adds cost for EMC reduction, but reduces cost of L due to high currents and saturation.

So performance specs and cost are biggest tradeoffs.
 
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All batteries can be approximated as a capacitor with a large series resistance; the series resistance is sufficiently large to give a large time constant for discharge but not large enough for the battery to appear as a constant current source. The "chemistry" within the cell decides the max current available (shorted current) but more importantly it modifies the discharge curve.
 

I am using lithium-ion battery cells. if ΔV=ΔI*Rb and we assume Rb is constant, do you mean ESR of the capacitor and load not effect the ripple voltage? Because ΔI depends on the battery voltage, inductor, the time that switch is on and resistive losses like Rdson and ESR of the inductor. If we keep simple and say ΔV=10mV, and ΔI=10A how could I find capacitor value ?
 

ΔV=ΔI*Rb
ΔV=9A*0.000688ohm
ΔV=0.0061V
but as you can see in picture1, ΔV is approximately 45mV. Picture2 shows current on the inductor. I want to get ripple voltage smaller and also to keep current 9A so do I need to get capacitor bigger?
batteryvoltage.pngcurrent.png
 

No technical battery balancing solution uses capacitors to reduce the ripple caused by pulsed balancer current. It seems to me as a purely theoretical concept with no practical relevance.

If you worry about the ripple in your design, use larger inductance for the switched mode circuit and possibly smoother current setpoint profiles or higher pwm frequency.

And find out if the impedance parameters assumed by your Matlab battery models are realistic at all.
 

45mV ripple is pretty low and would be acceptable for powering
most digital circuitry at 2.5V and above. Going lower for a
battery charger probably brings no benefit. You're talking
maybe 1-1.5% of charging voltage - why so fussy?
 

If the ripple voltage is low, control of the switches are gonna be easier. Also , when the capacity of the battery decreases, internal resistance increases thus ripple voltage is getting huge. That is why I am fussy:) Furthermore, I could not find a formula for capacitor value by keeping ripple voltage constant.
 

Furthermore, I could not find a formula for capacitor value by keeping ripple voltage constant.

Whether you will need a capacitor or an inductor depends on the phase of the ripple current (yes, the ripple voltage is the effect and not the cause, I think) and how does it change with load and battery state (health of the charge). I hope you will not find a Warburg impedance at work!
 

@c_mitra ... Disagree with high series R.

LiPo short term load current can exceed >> 10A thus ESR is in xx to xxx mΩ

Assume capacity is Vo=3.7 to 3.0V @3000mAh= 3Ah = 10,800 amp-seconds and we know Ic=CΔV/Δt

thus C=IcΔt/ΔV for IcΔt=10,800 As and ΔV=0.7V

then C=15.4 kilo-farads
if ESR was say 0.1Ω
then ESR*C=1,540 seconds =0.43h

This is not the recommended charge rate but the minimum charge time if self heating could be dissipated to prevent damage. Thus lower ESR the better.

Always follow Mfg specs for charge rate eg C/10 where here C means 20 hr std rate.

You can lower ESR of a battery to 1mΩ with an ultralow ESR Cap but then that only lasts for a few microseconds to ms depending on ESR of battery , cap and C.
 
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t

All capacitors and batteries alike in the same quality, size and voltage range may have different Ah ratings like capacitors have different uF ratings.
From my experience low ESR e-caps are have a C*ESR=T time constant <10 µs and smallest <1µs, while general purpose (G.P.) caps are > 100 µs to > 1000 µs in larger sizes.

.

This is an excellent rule of thumb -and easy to remember too- to make a quick assesment the figure of merit of a particular filter capacitor
 

then C=15.4 kilo-farads
if ESR was say 0.1Ω
then ESR*C=1,540 seconds =0.43h
... ...
You can lower ESR of a battery to 1mΩ with an ultralow ESR Cap but then that only lasts for a few microseconds to ms depending on ESR of battery , cap and C.

I understand your point but do not agree completely. Let me try to explain.

Batteries (cells) are capacitors only at a low discharge rate. When you try a high discharge rate, another time constant, due to the chemical charge transfer process with the cell reactions, appears. The electrochemical rate (charge transfer) process appears as an impedance (Warburg Impedance) that delays the charge delivery.

At high current levels, the effective AH capacity (boilerplate value) of most batteries drop miserably. The phase angle is around 45 deg (commonly) and is rather constant over a wide range of frequency. The impedance prevents perfect reduction of ripple using conventional capacitors.

At high discharge rates, cells cannot be modelled as capacitors because of this problem.
 

I understand. Warburg impedance is new to me.

The equivalent circuit from float charge voltage to resting full charge is a completely different model with high resistance and pulse load fall time below is a much shorter time constant than ESR*C. So the RC equiv. model is much more complex.

Normally a car battery of 55 Ah is never sold as a 1 million Farad capacitor with low ESR, they rate them based on CCA for a 5V drop from 12.5V so 750CCA rating is equivalent to 6.67 mΩ with a simple C value of 55Ah*3600s/0.5V =IΔt/ΔV or ~400,000 Farads would translate as a current pulse rise time of 400k*6.67m= of 2.6 kilo-seconds.

We can conclude the battery Model is a small RC time constant for pulse loads in parallel with a large RC constant for bulk storage and probably a large number of RC values in between in parallel. So it is not a perfect model.

It is more like the usual decoupling of a power supply and board with 100uF//1uF//0.1uF//100pF from AC to RF all lumped together except with bigger ESR values and larger C vales than small caps. each with different ESR*C values.
 
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