Battery Voltage

[martin1]

I read here on the Fieldlines that as soon as you reach battery voltage you can assume the internal resistance in the battery to be zero and the battery will clamp the voltage.

But why is the measured charge voltage higher than the battery voltage? In my experience, at any given moment a higher charge voltage will result in a higher current and that doesn't make sense if the voltage it's really clamped.

[electronbaby]

For all practical purposes, the battery does clamp the voltage...UNTIL it starts to become charged.

At this point the voltage could be allowed to rise, and current will rise with it. This is a good way to damage a battery (and a wind turbine), so any type of "charge controller" whether it be a conventional charge controller (for PV)or a DIVERSION LOAD CONTROLLER (for wind) will help to keep the battery from getting overcharged. There are also other very creative ways to do this as well.

[Flux]

I am not sure what is bothering you.

Yes the internal resistance is negligible so forget it. The battery will clamp the voltage, but what voltage? is that confusing you?

The emf of a battery ( virtually the same as volts as the internal resistance is low) depends on electrochemical reactions . A battery on charge will have a higher voltage than one on discharge and the actual voltage depends on the current and state of charge.

During charge the voltage will rise and also at any instant forcing more current will also result in a rise of voltage due to gas formation on the plates ( surface charge). This has nothing to do with internal resistance ( an order of magnitude smaller and you won't even see it at normal currents).

"But why is the measured charge voltage higher than the battery voltage? In my experience, at any given moment a higher charge voltage will result in a higher current and that doesn't make sense if the voltage it's really clamped."

I think this is where you are confused, What is charge voltage? your idea of it seems strange. Surely the charge voltage is that between the battery terminals on charge.

I have a feeling you are considering constant voltage charging circuits and are considering the charge voltage to be that of a constant voltage source that you connect to a battery. This mode of charging is not the same as current limited charging from wind or solar.

In this case you can use a constant voltage over a limited range and the battery will adapt but you can't cover a wide range of charging conditions without having current limit to cater for a low battery. With constant voltage charging you can only deal with a battery whose terminal voltage is kept within close limits and the battery surface charge mechanism will regulate the current to suit ( basically a battery on float or very near it). A low battery will draw excessive current and you will have to limit it in some way. As soon as this happens the battery terminal voltage will not be that of the source emf and you must consider the charging voltage to be that measured on the battery not the source.

When we talk about a battery clamping the voltage we are normally looking at wind or solar or some other current limited source. In the case of wind the open circuit source voltage can be very high ( 100's of volts) but as soon as you connect the battery it will drop to battery voltage and the mill will dictate the current ( not the battery).

The battery voltage will still change with state of charge ( not internal resistance but surface charge). The battery will be an effective clamp until fully charged then things get out of hand and you need an external voltage limit to prevent excessive gassing.

Flux

[Flux]

Perhaps your basic understanding is wrong when I read your question again.

"But why is the measured charge voltage higher than the battery voltage? In my experience, at any given moment a higher charge voltage will result in a higher current and that doesn't make sense if the voltage it's really clamped."

Going back to this. If a battery acted like a perfect clamp your argument is wrong.

Any attempt to exceed the clamp voltage would result in an increase in current only limited by the external circuit resistance. For example if the battery clamped at 12v then if you connected a 24v source with no internal resistance the current would be infinite. If the source resistance was one ohm then you would have 12v ( 24-12) forcing current through one ohm and 12A would flow.

You can't clamp a voltage source to any lower value and maintain that voltage. If the source has no internal resistance there will be infinite current and a bang.

I think you have confused battery internal resistance with that of the source. If the battery had internal resistance ( significant) then it can't act as a voltage clamp.

Flux

[martin1]

Thanks Flux. I have already accepted that there is no internal resistance but I'm still confused. I will give a real life example.

Right now I see 7 amps going to my battery and I measure 13.1 Volts over the battery. If I remove all the wires going to the battery I measure 12.8 Volts.

I assume the 13.1 Volts is the charge voltage and the 12.8 Volts is the battery voltage. Both measured on the battery.

Assume an ideal charger with no resistance was connected to the very same battery. Then as soon as it hits 12.8 Volts the current will go infinite high, is this right?

If so, why is it that the battery needs 13.1 Volt at only 7 amps when there is resistance in the charger?

The obvious answer would be that the difference, in this case 0.3 volt, is the voltage drop in the windings but I can't see how that ends up in the battery.

There's a reason why I'm asking all these questions.

[Flux]

The difference between 13.1v on charge and 12.8v off charge is this issue of the surface charge.

You need to look at the curves of voltage for a battery on charge and discharge to see this properly.

Fully charged a battery will read typically 12.8 when standing idle, it will always be above this on charge and below this on discharge.

"Assume an ideal charger with no resistance was connected to the very same battery. Then as soon as it hits 12.8 Volts the current will go infinite high, is this right?"

No you just produce more surface charge and the terminal voltage rises to meet the charger voltage ( At least within the battery current handling capability and below gassing volts).The current will rise rapidly but fall quickly again as the battery charges and at gassing volts it will be fully charged.

If the battery is not fully charged and you push the charger emf above the surface charge voltage range then the current will rocket, Similarly once you discharge with heavy loads to below the discharge voltage curve the current goes right up and if you short the terminals then you get a feel of the true internal resistance, a typical small battery will reach 1000A so you are looking at something near 12mOhms internal resistance or lower.

Don't confuse this with the voltage needed to fully charge a battery. You will not maintain 7a at 13.1v and it will never charge fully.

Flux

[martin1]

What I really wanted to know but didn't quite know how to put it was, if you design a charger circuit with no output impedance (theoretical) then within a small range the current will be proportional to the voltage. Judging from your answers I think that's true.

It's not that I want to design a constant voltage circuit. It has more to do with stability and stress issues. The buck topology differs a lot from the boost in this regard. In a boost converter the inductor and generator is always connected in series and the coil resistance will help limiting the current. In a buck converter, during the OFF state, the generator is disconnected and the inductor, the free wheeling diode and the battery become a circuit of its own.

[Flux]

 "if you design a charger circuit with no output impedance (theoretical) then within a small range the current will be proportional to the voltage. Judging from your answers I think that's true."

I suppose instantaneously that is more or less true but the current is also going to be dependent on the state of charge and with zero impedance the thing is going to get charged pretty quickly.

I suppose this is of some minor consequence if you have voltage feedback of battery volts on the converter but I doubt that you will get an impedance low enough to bother you with normal wiring the connecting leads to the battery alone will be very significant ( even though you may try for fractions of an ohm. The freewheel diode also has some sort of square law voltage current characteristic.

My converters don't have voltage feedback and are really energy converters rather than zero impedance sources.

If you are using any form of output filter capacitor then it will swamp any effects of a battery and if you keep the filter capacitor tiny then the connecting leads and battery length will look like an inductor at any reasonably fast switching frequency.

You would have an interesting time trying to model a battery, it will have internal resistance, a far greater resistive like component from the gas potential surface charge thing. It will look at low frequency as a capacitor to charging current and at high frequency it will look like an inductor proportional to the distance between its terminals. It probably looks like other things as well so just stop worrying and carry on.

Flux