Hi guys I want to ask sth
I am working on a dc dc buck-boost converter which feed a battery and a load in parallel .
the batery can be modeled as an internal voltage sourse in series with its internal resistance which is 0.1Ohm.
the output of the converter is in two stages: In the first stage when the converter charges the battery the converter feed the battery with an 20 A current and its internal voltage is 59.7V so the voltage on the battery (and the output voltage of the converter) is Vb1= 59.7+20*0.1=61.7V
During the second stage when the battery is charged the converter feed the battery with 1A current and the internal voltage of the battery is 61.5V so the voltage on the battery (and the output voltage of the converter) is Vb2=61.5 +0.1 * 1 =61.6V.
The question is would in real world this slight difference between the voltages Vb1 and Vb2 of the two stages create problems in the operation of the converter or on the circuit?

 

In each stage, is the charger maintaining a constant current or a constant voltage?
What determines the end of stage 1?

Assuming that the switch from stage 1 to stage 2 is a detectable transition of some sort, then that event can switch the voltage and current behavior of the supply.

 

In each stage the voltage and the current are constant and I suppose that we can detect each stage by measuring the volrage on the battery but is it posdible to design a so accurate converter/system with only 0.1V voltage difference of these stages?

 

The voltage and the current CAN'T both be constant. As the loading conditions change - as the battery charges - so either the voltage or the current MUST change.
What kind of battery is it, lead acid, NiMh, Li-ion...?

 

Actually I dont have exact characteristcs of the battery but only this simple model..
Can the feeding current and the voltage of the battery stay constant and only the internal voltage of the battery changes?

 

This is possible if, and only if, the internal resistance of the battery falls in the precise way required to make this possible.
i = the current
Vt = battery terminal voltage
Vi = internal battery voltage
r = resistance of the battery
Vt = i * r + Vi
Rearranging gives: r = (Vt - Vi) / i
It is unlikely the internal resistance will follow this exact pattern.
For instance with a lead acid battery, the charger holds the voltage constant and as the battery charges the current drawn falls. When the current has fallen to a low value, this indicates that the battery is charged
With a NiMh battery, the charger keeps the current constant and as the battery charges, the terminal voltage increases. When the voltage reaches a particular value this indicates that the battery is charged (it is more complicated than that for fast chargers but we'll keep it simple).

You might have a read here: http://batteryuniversity.com/learn/

 

Yes I understood thanks! Probably my exercise(as a theoritical exercise) refers to the steady states and not in the mibble transient stages.But what about this voltage margin? Can this create problems and faults buring the operation of the system?

 

As I said before if there is a clear definition of what should be happening in each stage, and a clear means of deciding when to switch between states then, no, there is no problem. But you will need tie down those details to get anywhere near a design.

 

Yes I have clear definition of what the converter has to do i these two stages so there is no problem..Thanks!