Ideal current sources in series

Thread Starter

MrChips

Joined Oct 2, 2009
21,052
Here is an interesting problem posted by @MrAl in another thread.

Ideal current sources in series.jpg

Two ideal current sources set for different current values are in series in a circuit.
What is the total current in the circuit?

I will provide an answer after you have had a chance to think about it.
 

crutschow

Joined Mar 14, 2008
24,914
It's the same answer as if you have two ideal different voltage, voltage-sources in parallel.
It's indeterminant.

So which is larger infinity or two times infinity?
 

LvW

Joined Jun 13, 2013
1,016
However, it is a pure theoretical question without any practical relevance because there are no "ideal" current sources.
A series connection of two REAL current sources can be found in each diff. amplifier with an active (current mirror) load...and it works!
 

MrAl

Joined Jun 17, 2014
7,581
Hi,

Yes this is where we see that if there is no wiggle room it cant be done.
So many circuits depend on the tiny differences we usually dont look at but are ever present.
For example, put a 1 yottaohm resistor in parallel with one of the current sources and suddenly it becomes more practical.

The simulator MicroCap will not allow things like this. It will complain before it does any analysis. There are various element connections that cause this too besides the two current source paradox. It does not accept an ideal DC voltage source driving only an ideal inductor for example.
 

WBahn

Joined Mar 31, 2012
25,738
Here is an interesting problem posted by @MrAl in another thread.

View attachment 172577

Two ideal current sources set for different current values are in series in a circuit.
What is the total current in the circuit?

I will provide an answer after you have had a chance to think about it.
This is like asking what the voltage is across two ideal voltage sources in parallel -- the solution is indeterminate.

If you make just one of them non-ideal, no matter how high the resistance, then there's a solution. But limits on the max voltage either can deliver and you can also arrive at a solution.

But let me take a guess at the "answer" you are going to provide.

There are 6 A flowing through the overall circuit with the other 6 A flowing in the infinite resistance of the 12 A source due to an infinite voltage across it. This means there are 6 A flowing in the resistor.

The problem with this is that there is an equally valid solution in which that other 6 A is flowing through the resistor (so 12 A total) and then through the infinite resistance of the 6 A source.

And of course there's every possible splitting of that 6 A between the two source resistances.

Hence, indeterminate.
 

Thread Starter

MrChips

Joined Oct 2, 2009
21,052
All good answers.

Let us see if we can make the problem more manageable.
We will model a constant current source as capable of having an infinite voltage source and infinite series resistance.
Now let us impose some real-world limitations. We will set a limit on the voltage to 1TV.
The series resistance can be any positive value from 0 to ∞ ohm.

Have a go again.
 

WBahn

Joined Mar 31, 2012
25,738
All good answers.
So where is the answer that you said you would provide?

Let us see if we can make the problem more manageable.
But if you have an answer to the original question, what need is there to make it more manageable? In particular, what is it that is in need of managing?

We will model a constant current source as capable of having an infinite voltage source and infinite series resistance.
Modeling a Norton source (an ideal current source in parallel with an equivalent resistance) as a Thevinen source (an ideal voltage source in series with an equivalent resistance) only works for finite-valued parameters.

So your basic model breaks down even before you get started. Placing limitations on it after the breakdown may or may not repair the break -- you'd have to show that it does before any results you get could be considered valid.

Now let us impose some real-world limitations. We will set a limit on the voltage to 1TV.
The series resistance can be any positive value from 0 to ∞ ohm.
If you are going to do this, you don't need to go through the break in the first place. Simply place a 1 TV limit on the voltage of your current sources and restrict the output resistance to be any nonnegative finite value. You can always then examine the behavior in the limit that the resistance becomes arbitrarily large.

Have a go again.
To what end? What is the point of the exercise?
 

mvas

Joined Jun 19, 2017
537
I will guess ... 6 Amps in the real world.
My reasoning is ...
The 6 Amp Current Source can keep raising its internal resistance, as high as needed, to limit to 6 Amps - near infinity is possible
The 12 Amp Current Source will eventually run out of voltage, when it hits its real world finite Voltage Limit, 6 amps will be flowing
 
Last edited:

MrAl

Joined Jun 17, 2014
7,581
Hi,

I thought i would look at the theoretical solution and see what becomes of it.

First, i started with MrChips circuit in post #1, but added a center tap to the resistor and grounded that. I then solve for the voltage at the top of the two current sources Vout.
So it's the same, except now there are two resistors going to ground and i assume they have the same value without specifying that value yet.
I also model both current sources as having a resistance in parallel, R1 for I1 and R2 for I2. That makes them non ideal.
To make either one ideal again, all i have to do is let their respective resistances go toward infinity.
Ok, so writing the equations for this setup and making both parallel resistances the same value (Rp) i get this:
Vout=(I1*Rp)/2-(I2*Rp)/2

That would represent the non ideal case.
To find the ideal solution, i factor:
Vout=(I1-I2)*Rp/2

and then let Rp go to infinity.
It's plain to see that the voltage Vout goes to minus infinity now when I2>I1 which in this case really says there is no solution.

Taking a quick second approach, if we leave the circuit the same as post #1, we cant even write an equation because there is no rule for two ideal current sources in series. Recall that with just one current source we would write I1 times something or I2 times something for that one loop, but since the current in the loop is paradoxically two different values, we cant write a reasonable equation. Even if one current was zero, we still could not do it. The only way i think would be if both currents were the same and in the same direction but then it would quickly reduce to just one current source and then the paradox would be eliminated.

With these 'proofs' if you will, the only answer is that there is no solution because the circuit is impossible using ideal current sources.
 

Thread Starter

MrChips

Joined Oct 2, 2009
21,052
Here are my answers I wish to offer.

1) As many members have said, the current is indeterminate. This is a theoretical problem with an obvious paradox.

2) If we were to model an ideal current source as an infinite voltage source with infinite impedance, we can still apply Ohm's Law,

I = V / R

where we apply limits to infinity

I = (V → ∞) / (R → ∞)

For every large value of V, there is a value of R which will result in a finite value of I, and vice versa. This is still solvable.

There are two possible cases, (a) where the current source is bipolar, i.e. the direction of the current can be reversed, and (b) where the current source is unipolar, i.e. the direction of the current cannot be reversed.

Let us consider the limitation of (b) where the current source cannot change polarity.
The second limitation I wish to impose is that the maximum voltage is limited.

Ideal current sources in series_2.jpg

Using the above model of infinite voltage V and infinite resistance R for each current source, I will assume that source A is set for a higher source current than B.

As source A attempts to increase the current through the loop, VA will reach its maximum value and RA will be forced to zero.
Simultaneously, source B will attempt to reduce the loop current by setting VB to zero and RB in response to VA.

Thus the current I in the loop is VA / RB.
The lower current setting wins.

According to my analysis, @mvas takes home the prize.
 

MrAl

Joined Jun 17, 2014
7,581
Here are my answers I wish to offer.

1) As many members have said, the current is indeterminate. This is a theoretical problem with an obvious paradox.

2) If we were to model an ideal current source as an infinite voltage source with infinite impedance, we can still apply Ohm's Law,

I = V / R

where we apply limits to infinity

I = (V → ∞) / (R → ∞)

For every large value of V, there is a value of R which will result in a finite value of I, and vice versa. This is still solvable.

There are two possible cases, (a) where the current source is bipolar, i.e. the direction of the current can be reversed, and (b) where the current source is unipolar, i.e. the direction of the current cannot be reversed.

Let us consider the limitation of (b) where the current source cannot change polarity.
The second limitation I wish to impose is that the maximum voltage is limited.

View attachment 172655

Using the above model of infinite voltage V and infinite resistance R for each current source, I will assume that source A is set for a higher source current than B.

As source A attempts to increase the current through the loop, VA will reach its maximum value and RA will be forced to zero.
Simultaneously, source B will attempt to reduce the loop current by setting VB to zero and RB in response to VA.

Thus the current I in the loop is VA / RB.
The lower current setting wins.

According to my analysis, @mvas takes home the prize.
Hi,

Well just one question then, are you solving for ideal current sources or non ideal? I sounds like non ideal because for one you side with "mvas" which states that it is a real life case.

Also, not sure if this matters here but you could look at it.
The form of a limit that results in infinity/infinity is an indeterminant form so to find the limit you have to differentiate.
 

Thread Starter

MrChips

Joined Oct 2, 2009
21,052
I am saying that there is no solution for "ideal" current sources.
If you impose "real-world" limitations then there is a solution.

P.S. This is the difference between science and engineering.
In science, you work with ideal hypothetical concepts.
In engineering, you have to look outside the box, be pragmatic and ask "what if" questions. What if this is not an ideal case? What if we impose real world limitations?
 

MrAl

Joined Jun 17, 2014
7,581
I am saying that there is no solution for "ideal" current sources.
If you impose "real-world" limitations then there is a solution.

P.S. This is the difference between science and engineering.
In science, you work with ideal hypothetical concepts.
In engineering, you have to look outside the box, be pragmatic and ask "what if" questions. What if this is not an ideal case? What if we impose real world limitations?
Hi,

Oh yes ok.

Normally a fixed current source tells us that there is a certain fixed current in a single wire so the current source is like a specification of the current in that one wire, so when we go to write an equation we use that current as the spec for the current in the wire, but when we have two different current specifications for the same wire it just does not make sense at all because in a single homogeneous wire we can only have one current. So the question itself is paradoxical to begin with.
 
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