Input impedance of a battery?? - Page 1
Forum section: electronics / projects

author: Siddhat
author_rank: Regular Contributor
timestamp: February 27, 2020, 06:46:32 am
content: Hello everyone!I know that, the impedance as seen from the input terminals of a particular circuit (that resists the flow of current into that circuit) is known as its input impedance.But things got confusing while I was trying to make a buck converter to make a 12 Volt lead-acid battery charger.For certain calculations of the buck converter, I required the load to be connected to the output terminals of the buck converter and since I will be charging a 12 Volt lead-acid battery (which acts as load to the buck) from the buck converter, the impedance as seen from the output terminals of the buck converter into the battery (which equals the input impedance of the battery) acts as a load to the buck converter.So, is there anything like input impedance of a battery?. If there is, how do I calculate that? Also, does the internal resistance of the battery act as the input impedance??Anyone here an expert on these stuff?? Need help!Thank you!

author: Berni
author_rank: Super Contributor
timestamp: February 27, 2020, 07:57:54 am
content: Yes a battery can be imagined as a voltage source with a resistor in series (internal resistance)But due to the battery chemistry stuff this internal resistance is not a fixed value. It depends on the battery charge state, temperature, age, current in the last few seconds. But in general for a battery in working order this internal resistance is very low (Between a about an Ohm to a few miliOhm) as its required for the battery to provide a lot of power.Because of this you have to make a battery charger that can regulate both the output voltage and current. The battery just takes however much current its internal resistance allows. So connecting a empty car battery directly to a 14V supply with no current limiting will cause it to draw easily 500A provided that the power supply is capable of providing so much power and the cables don't melt (Tho after half a minute of this boiling acid would likely be spewing out of the battery). But since most power supplies are not capable of providing the 7kW of power required to do this they will get overloaded and drop in voltage so the current drops too, constantly running a power supply in such overload can make it overheat and die. This is the difference between a power supply and a battery charger

author: Siwastaja
author_rank: Super Contributor
timestamp: February 27, 2020, 08:44:48 am
content: Very low, milliohms. It shouldn't be a part of your basic buck charger equations, you are clearly misunderstanding something here. Show your math.

author: Lesolee
author_rank: Regular Contributor
timestamp: February 27, 2020, 09:30:08 am
content: Quote from: Siddhat on February 27, 2020, 06:46:32 amBerni is quite right. You would model the lead acid battery as a voltage source in series with a resistance. A flat 12V battery might measure 11V on its terminals, but the series resistance would be "quite high". That's very qualitative because I don't know what that value would be. If you were to draw one amp from this flat battery and the terminal voltage dropped by 2 volts you would calculate the series resistance as 2V/1A = 2 ohms.You typically charge lead-acid batteries from a constant voltage supply, but subject to a current limit. A 10 A charger, for example, might put out 14V, or less if the current reached 10A at a lower voltage. 13.8V would be a lower limit for the maximum output voltage, and maybe 14.5V as an upper limit.On a fully charged lead-acid battery (not being charged) it might sit at 12.6 V. Put a 1 amp load on it and the terminal voltage wouldn’t drop much at all, maybe 0.1 V. That means the internal resistance is now 0.1 V/1 A = 0.1 ohms.It means that the circuit model of a resistor and a voltage source has both elements which are variables. As the battery discharges, the series resistance increases and the voltage source decreases.

author: SiliconWizard
author_rank: Super Contributor
timestamp: February 27, 2020, 04:23:07 pm
content: Yep, Berni told you the basics.Thing is, not only the internal resistance is not fixed, but will vary depending on remaining capacity and current draw (when being loaded) - it will also be different when you are charging the battery from what you get when you are drawing current from it...One of the usual ways of not bothering with the battery's internal resistance when charging it is to charge it with a constant current - with a voltage compliance which is set at the maximum voltage allowed for the battery.

author: Siwastaja
author_rank: Super Contributor
timestamp: February 27, 2020, 06:08:57 pm
content: To design the buck for battery charging, you should choose your charging current (which will be the average inductor current), then choose switching frequency, amount of ripple current in the inductor (for example, 30-40% of average current), and finally, inductor L so that you get your desired current triangle ramp, from V(t) = L * di/dt.Control the switching to maintain the current you want - peak current mode control is fine, for example. Use an outer, slower control loop to maintain voltage regulation, by controlling the current setpoint. Or in the simplest case, just turn the control off once a certain voltage is reached, thus no voltage control loop necessary."Impedance of a battery" is never part of the equation, and only becomes relevant if you want to model the losses in the battery.

author: SiliconWizard
author_rank: Super Contributor
timestamp: February 27, 2020, 06:20:18 pm
content: Yep. For anyone interested in simulating batteries, I also suggest reading papers about battery models - there are many.