Simple Theoretical question, 3 things in parallel

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
146
Lets say we have the following 3 things in parallel

A current source that is outputting 15 amps max

A battery that is at a low state of charge (for the sake of it, lets call it a lithium) that is at 26.6V

A constant load that requires 400W. (@ 26.6 Volts the load draws ~15A)

If I_total = 15A
and if I1 represents the current into the battery and I2 represents the current into the load
we know that I_total = I1 + I2 obviously.

What dictates how the current flows?

Think of this as an ideal type question ignoring as many nuances as possible.

Does the current flow into the battery and try to charge or does the current flow into the load?
At first glance, i would of said the the impedance of each is what dictates it. But now I'm not so sure?

Thoughts?
 
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Ian0

Joined Aug 7, 2020
9,817
It depends on what exactly you mean by a "constant load". Is the resistance of the load constant? Or is it an active load that keeps the power constant, regardless of the voltage?
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
146
It depends on what exactly you mean by a "constant load". Is the resistance of the load constant? Or is it an active load that keeps the power constant, regardless of the voltage?
Good quesiton

For the sake of the argument, lets say its an electronic load set at 400 watts
for constant power
 

LowQCab

Joined Nov 6, 2012
4,072
1)
You can't use a "Current-Source" as a Battery-Charger.
You need a "Current-Limited", Voltage-Regulator.

2)
The Current-Limit must be less than that which would over-heat,
and damage the Battery, or catch it on fire.

3)
The "Fixed-Load" plus the Load created by a discharged Battery are additive.

4)
Can the Fixed-Load operate on a wide range of Voltages ?

The best solution is to have a stand-alone-Power-Supply for the Load,
plus a dedicated Battery-Charger for the Battery.
The add some sort of "Oring" Circuit to insure no Power loss during lack of Mains Power.
.
.
.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
146
1)
You can't use a "Current-Source" as a Battery-Charger.
You need a "Current-Limited", Voltage-Regulator.

2)
The Current-Limit must be less than that which would over-heat,
and damage the Battery, or catch it on fire.

3)
The "Fixed-Load" plus the Load created by a discharged Battery are additive.

4)
Can the Fixed-Load operate on a wide range of Voltages ?

The best solution is to have a stand-alone-Power-Supply for the Load,
plus a dedicated Battery-Charger for the Battery.
The add some sort of "Oring" Circuit to insure no Power loss during lack of Mains Power.
.
.
.
Apparently you missed the part where i said ideal and your talking about battery temps?
(shaking my head)

And you are wrong on many points here.

"you cant charge a battery with a current source" ??

Really? Ever heard of a PV panel?
 

Ian0

Joined Aug 7, 2020
9,817
So what's going to happen?
The 15A constant current source will supply 15A. The battery is at 26.6V. The electronic load will take 400W which is 26.6V @ 15.03A. The extra 30mA will be supplied by the battery, which already is flat.
The battery voltage will reduce, let's say it reduces to 26.5V.
The electronics load will try to take 400W, which is now 15.09A @ 26.5V. The battery will now supply 90mA, and it's voltage will further reduce.
This will continue until the battery is ruined, because the load will take ever more current to maintain 400W, and the constant current source can only supply 15A.

It should be noted that the dynamic impedance of the electronic load is (26.6V-26.5V)/(15.03A-15.09A) which is -1.67Ω. Negavite impedances tend to result in instability or oscillation.
 

WBahn

Joined Mar 31, 2012
30,062
Good quesiton

For the sake of the argument, lets say its an electronic load set at 400 watts
for constant power
Just take it one piece at a time.

You've already stated what the total current is and that the current in the load plus the current in the battery must equal that?

What is the voltage across the load (at least initially)?

What is the current in the load need to be in order to achieve the 400 W that it's controller is going to adjust it's effective resistance in order to achieve?

What is the resulting current in the battery need to be?

What is the battery voltage (assuming it's achievable) at which no current flows in the battery)?

What happens if the battery voltage happens to be above this voltage? Does the battery charge, or discharge?

What happens if the battery voltage happens to be below this voltage? Does the battery charge, or discharge?

What does that tell you about the stability of the circuit?
 

MisterBill2

Joined Jan 23, 2018
18,518
The current will flow into each load in proportion to it's conductance, equal to the inverse of it's resistance. The voltage across both will be the same, determined by the instant load resistance presented by the parallel loads. That will vary as the battery charges.
 

Janis59

Joined Aug 21, 2017
1,849
Short words - the current path and value is dictated by Matrix, resulting on equations system on basis of Kirchoff current Law and Kirchoff Voltage Law. Less universal but mathematically simpler it may be analyzed also by means of Norton-Mayer or alternatively Thevenin theorems combined with Millman`s theorem.
 

Thread Starter

mike _Jacobs

Joined Jun 9, 2021
146
Short words - the current path and value is dictated by Matrix, resulting on equations system on basis of Kirchoff current Law and Kirchoff Voltage Law. Less universal but mathematically simpler it may be analyzed also by means of Norton-Mayer or alternatively Thevenin theorems combined with Millman`s theorem.
These answers crack me up. For every one decent answer on this forum usually given by a select few that are 10 X the answers that literally serve no purpose
 

MisterBill2

Joined Jan 23, 2018
18,518
We have an excess of unknowns here. " A current source that is outputting 15 amps max " does not adequately describe the compliance voltage of the current source. And without knowing what will happen due to the definition of current source it is not possible to provide a definite answer.
IF we assume that the " A current source that is outputting 15 amps max "is limited by it's compliance voltage, then the current split will be decided by a guy named FRED, who jsy walked into that same bar room.
 

BobTPH

Joined Jun 5, 2013
8,967
A current source that is outputting 15 amps max

A battery that is at a low state of charge (for the sake of it, lets call it a lithium) that is at 26.6V

A constant load that requires 400W. (@ 26.6 Volts the load draws ~15A)
The solution is actually quite obvious.

The current source adjusts to 26.6V. 15A is drawn by the load. Nothing is drawn by the battery.
 

Ian0

Joined Aug 7, 2020
9,817
The solution is actually quite obvious.

The current source adjusts to 26.6V. 15A is drawn by the load. Nothing is drawn by the battery.
Some tolerances would be useful as well. If all figures are taken as absolute, then 26.6V*15A is only 399W. The load needs another Watt from somewhere, so it has to come from the battery, which will reduces the battery voltage as it is already quite discharged.
 

sparky 1

Joined Nov 3, 2018
757
In a simple case, the purpose of explaining being to remind what is probably going and nothing new. Corrections are welcome.

A potential difference across a conductor produces an electric field that influences an excitement of free electrons
that transfers by charge carrier in a direction of polarity and distributes equally across the surface so that the potential inside the conductor is constant every place. The mobility corresponding to that portion having higher resistance (slower mobility) has a measured equivalence in calories of heat.
 
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k1ng 1337

Joined Sep 11, 2020
960
Here is a simulation of my interpretation. Assumptions:

1) V1 is an ideal voltage source set to 26.6V because no information about the battery was given other than it being "discharged".

2) I1 is an ideal current source set to 15A because no other information was given.

3) I2 is an active load set to 15.0A. It is 15.0A and not 15.04A etc. because 400.W / 26.6V = 15.0A rounded to 3 significant figures.

Therefore, I1 sources 15A, I2 sinks 15A and V1 sources / sinks 0A. Obviously the problem with this simulation is the lack of information and limited precision.

Untitled.png

This doesn't tell us anything useful because precision was lost on the element we are most interested in. However:

4) If the precision of all initial parameters are given to a standard of 4 significant figures, I2 becomes 15.04A. It is 15.04A because 400.0W / 26.60V = 15.04A rounded to 4 significant figures. If this is the case, I1 sources 15.00A, I2 sinks 15.04A and V1 sources the difference of 0.04A.

Untitled2.png
 
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