Isolating variable with constant power load

Blomqvist

Joined Jun 10, 2015
12
Hey all!

Disclaimer: This is NOT a homework question.

Situation: This is a personal project where I am trying to model a split rail configured circuit with two constant power loads, Pp and Pn, with a "shared" common return to ground. Reistance, R, is equal across all lines. Pp does not necessarily equal the magnitude of Pn.

Issue: The fact there are two constant power loads makes it difficult to extract an expression for I+ or I- based on the change in power consumed by the loads. I have so far developed an expression for the current as a function of the current in the opposite leg (ie I+ as a function of I- and vice versa), but I need to isolate purely I+ without any coupled dependancy.

Work done so far...

Note:
I+ and I- are the currents sourced by the positive rail and negative rail respectively. These quantities are henceforth referenced as I1 and I2 respectively.
Knowns: V, R, Pp, Pn

1. I began by developing an expression for I1 and I2 as a fucntion of the voltage drop across the load.
$$I_1 = \frac{1}{R}\left(V - \frac{2}{3}V_{pos} - \frac{1}{3}|V_{neg}|\right)$$

$$I_2 = -\frac{1}{R}\left(V - \frac{2}{3}|V_{neg}| - \frac{1}{3}V_{pos}\right)$$
2. Next I put Vpos and Vneg in terms of the power and their respective current through the load.
$$P_p := V_{pos}I_1$$

$$P_n := |V_{neg}|I_2=NP_p=N V_{pos}I_1$$

where
$$N = \frac{P_n}{P_p}$$
3. Substituting into the expressions for I1 and I2 reduces to
$$I_1 = -\frac{P_p}{3R}\left(\frac{I_2}{(I_2)^2 - \frac{V}{R}I_2 + \frac{2N}{3R}P_p}\right)$$

$$I_2 = \frac{NP_p}{3R}\left(\frac{I_1}{(I_1)^2 - \frac{V}{R}I_1 + \frac{2}{3R}P_p}\right)$$​

4. From here on, I'm attempting to isolate I1 by considering the voltage drop across one mesh.
$$V=I_1R + \frac{P_p}{I_1} + (I_1 - I_2)$$
5. Which, when I1 from step 3 is substituted into the mesh equation in step 4, the following behemoth is found:
$$I_1 = \frac{V(I_1)^2\left(3R(I_1)^2 - 3I_1V+2P_p\right)}{6(I_1)^4R^2 - 6RV(I_1)^3 +(I_1)^2(7P_pR+NP_pR)-3P_pVI_1 + 2(P_p)^2}$$​

Comments moving forward: Provided step 5, I have clearly hit a snag. I am not sure how best to proceed to isolate the current I1. I considered a Taylor expansion (or some variant) in the denominator and bound the expression, but this may be unnecessarily tedious. Unfortunately, any other manual approach seems equally tedious.

Any help would be greatly appreciated!

WBahn

Joined Mar 31, 2012
26,398
You start off with a contradiction. You emphasize that they are constant power loads, and then state they you are looking to solve for something based on the change in the power consumed by the loads. If the power consumed by the loads is changing, how are they constant power loads?

Why the absolute value operators in your initial equations? Surely the sign of Vneg and Vpos has an effect. And most definitely the sign of Vneg has a HUGE effect on the power consumed by Pn -- if it is positive Pn is consuming power but if it is negative it is producing power. Big difference!

Then all of those absolute value operators magically disappear.

Don't go throwing inappropriate operators at equations due to inconvenient polarity choices on defined variables. If you've defined Is as being sourced by the negative rail of the bottom supply, then (assuming the two loads ARE loads) I2 will be negative (look at your equation for I2). But now look at your equation for Pn -- the absolute value operator insures that you will be multiplying a positive value by a negative value yielding a negative power.

Your expression for Pn should simply be

Pn = -Is · Vneg

since Is is entering the negative terminal of the load (as defined by Vneg).

Blomqvist

Joined Jun 10, 2015
12
Thanks for the response, WBahn!

If the power consumed by the loads is changing, how are they constant power loads?
The system poses only a moderate contradiction. I'm attempting to develop a relation for one's leg's current purely on how the power changes across each load. The key here is to note the power is not changing variably as a function of any condition within the circuit. Replace the constant power load model with an equivalent ideal switch-mode power supply and you can see how the "load of the load" might change variably over time while remaining isolated from the circuit shown above. The takeaway here is, at any instant in time, the power is constant and can be calculated.

Why the absolute value operators in your initial equations?
Ultimately it came down to relative value vs absolute value. By convention, I could reference everything to ground, or I could proceed with absolute reference. The latter allowed me to discard negatives purely so I didn't need to deal with them. You're absolutely correct, though. The sign is important and are accounted for from step 3 and below.

Blomqvist

Joined Jun 10, 2015
12

Per WBahn:

The absolute values have been removed and signs reinstantiated. However, since power is consumed at the load, the power remains positive.

$$P_p := V_{pos}I_1$$

$$P_n := NP_p = (-)V_{neg}(-)I_2=N V_{pos}I_1$$

where $$N = \frac{P_n}{P_p}$$

Consequently, this only affects expressions in Step 1. See corrections below:

$$I_1 = \frac{1}{R}(V - \frac{2}{3}V_{pos} + \frac{1}{3}V_{neg})$$

$$I_2 = \frac{1}{R}(V - \frac{2}{3}V_{neg} + \frac{1}{3}V_{pos})$$

where Vneg is a negative value measured referencing ground.

Additionally, for the sake of readability, the following expression is modified from Step 5.

$$V=2R\left(\frac{I_1^4 - \frac{V}{R}I_1^3 + \frac{P_p(N+7)}{6R}I_1^2 - \frac{P_pV}{2R^2}I_1 + \frac{{P_p}^2}{3R^2}}{I_1\left(I_1^2 - \frac{V}{R}I_1 +\frac{2P_p}{3R}\right)}\right)$$

WBahn

Joined Mar 31, 2012
26,398
Thanks for the response, WBahn!

The system poses only a moderate contradiction. I'm attempting to develop a relation for one's leg's current purely on how the power changes across each load. The key here is to note the power is not changing variably as a function of any condition within the circuit. Replace the constant power load model with an equivalent ideal switch-mode power supply and you can see how the "load of the load" might change variably over time while remaining isolated from the circuit shown above. The takeaway here is, at any instant in time, the power is constant and can be calculated.

Ultimately it came down to relative value vs absolute value. By convention, I could reference everything to ground, or I could proceed with absolute reference. The latter allowed me to discard negatives purely so I didn't need to deal with them. You're absolutely correct, though. The sign is important and are accounted for from step 3 and below.
And yet, in doing so (discarding negatives so that you didn't need to deal with them) you ended up with equations that are simply wrong in Step 2. If the results of Step 2 are used in Step 3 or elsewhere, by what magic are they not going to make those equations wrong, too?

As a rule, I don't see much point in wading through equations when the equations they are based on are already known to be wrong. But, looking at your equation for Step #4, I can tell at a glance that is it wrong. You have that a voltage is equal to the sum of two voltages and a current -- guaranteed to be wrong.

Assuming that your Step 5 equation is correct, I don't understand the problem. I thought the problem was that you couldn't find a way to isolate I1 from I2 -- but I don't see I2 in that equation at all. So haven't you separated them?

What I see is a fourth order polynomial. My gut feel is that this is what you would expect since you have two powers (one for each load) and power is related to the square of the current flowing in it. You should also expect multiple solutions -- for instance, setting the power to zero for one of the loads could equate to short circuiting the load or to open-circuiting it. Since you could do either for either load, there would need to be four different solutions corresponding to each possibility.

As a check on your equation, you might consider determining I1 for all four cases and see if setting power powers to zero yields the expected four roots. Since you only have the equation for I1, two of those roots should be zero (if you open-circuit Pp, then I1 must be zero), but if you short circuit Pp, then you should get a different I1 depending on whether Pn is open or short.

I don't follow how throwing absolute value operators around has anything to do with whether something is referenced to ground or not -- absolute reference is not the same as absolute value.

You don't need to reference anything to ground -- you can remove your ground reference symbol and never refer to it in any way -- in order to set up your equations correctly.

You should also annotate your diagram with any quantity you define, such as I+ or I2, showing its location and polarity. Not only does that save you from having to write a bunch of text describing what you mean by each of the terms, but it also drastically reduces the likelihood that the reader will miss important distinctions like what polarity you mean for I2 when you say that the negative rail is sourcing it.

Note that if you redefine I2 to go the other way (the current sunk by the negative rail), then all of the problems that led you to get really sloppy and throw in a bunch of inappropriate absolute value operators simply go away. It also increases the symmetry of your equations and means that you can think in terms of both I1 and I2 being positive, which is more natural for humans and further reduces the chance of making silly mistakes.

WBahn

Joined Mar 31, 2012
26,398

Per WBahn:

The absolute values have been removed and signs reinstantiated. However, since power is consumed at the load, the power remains positive.

$$P_p := V_{pos}I_1$$

$$P_n := NP_p = (-)V_{neg}(-)I_2=N V_{pos}I_1$$

where $$N = \frac{P_n}{P_p}$$
But your I2 is going to have a negative value while your Vneg is going to have a positive value -- so that's three minus signs making Pn negative!

If you would annotate your diagram with your definitions for I1 and I2, you will see this. You have defined your I2 to be coming OUT of the negative terminal of the bottom supply, so it is going to be negative. That is consistent with your equation in Step 1.

Blomqvist

Joined Jun 10, 2015
12
looking at your equation for Step #4, I can tell at a glance that is it wrong.
I realized I Ieft off the R at the end of the expression in Step 4 10min after submitting the topic. It should read:
$$(I_1 - I_2)R$$

I didn't want to create a new reply and when I responded to you, I forgot to include that correction.

I don't understand the problem. I thought the problem was that you couldn't find a way to isolate I1 from I2 -- but I don't see I2 in that equation at all.
I want to isolate I1 from the expression. Meaning I want an equation for I1 as a function of V, R, P.

But your I2 is going to have a negative value while your Vneg is going to have a positive value -- so that's three minus signs making Pn negative!
I apologize. I made another error while attempting to correct for a mistake. I do understand how the current flows through the circuit. The err here is that I pulled out the negative sign for both I2 and Vneg, which broke the rule I was establishing for Vneg in the first place! That is, the sign of the value is negative and not the quantity Vneg. By placing the negatives (-) out in front of each quantity, I was demonstrating how their signs cancel. This convention isn't obeyed anywhere else and I apologize.

I have verified every expression is correct by placing dummy values in, but I now understand it wasn't clear what convention (if any) I was following. Truth be told, the expression in Step 1 and 2 were developed on a different day. Step 3 course corrected and when I shared my work, I forgot there was an inconsistency across steps. The project began as an attempt to create an excel calculator, but I didn't want to worry how the user interpreted Vneg (negative value or positive value of negative quantity). So I forced the value to always be positive and I handled the sign in the expressions. When I began substituting I1 and I2 into expressions, the sign became an issue again and I corrected those manually.

Vneg is now a negative value and should reflect as such moving forward.

MrAl

Joined Jun 17, 2014
8,996
Do you know for sure a solution is possible?

Blomqvist

Joined Jun 10, 2015
12
Do you know for sure a solution is possible?
I'm not sure. If a solution does exist, however, it would most likely be terribly unwieldly. This leads me to suspect an approximation might be a better avenue to pursue. What are your thoughts?

MrAl

Joined Jun 17, 2014
8,996
I'm not sure. If a solution does exist, however, it would most likely be terribly unwieldly. This leads me to suspect an approximation might be a better avenue to pursue. What are your thoughts?
Hi,

id have to think about it, but if you only had one branch one load could you solve that?

WBahn

Joined Mar 31, 2012
26,398

Per WBahn:

The absolute values have been removed and signs reinstantiated. However, since power is consumed at the load, the power remains positive.

$$P_p := V_{pos}I_1$$

$$P_n := NP_p = (-)V_{neg}(-)I_2=N V_{pos}I_1$$

where $$N = \frac{P_n}{P_p}$$

Consequently, this only affects expressions in Step 1. See corrections below:

$$I_1 = \frac{1}{R}(V - \frac{2}{3}V_{pos} + \frac{1}{3}V_{neg})$$

$$I_2 = \frac{1}{R}(V - \frac{2}{3}V_{neg} + \frac{1}{3}V_{pos})$$

where Vneg is a negative value measured referencing ground.

Additionally, for the sake of readability, the following expression is modified from Step 5.

$$V=2R\left(\frac{I_1^4 - \frac{V}{R}I_1^3 + \frac{P_p(N+7)}{6R}I_1^2 - \frac{P_pV}{2R^2}I_1 + \frac{{P_p}^2}{3R^2}}{I_1\left(I_1^2 - \frac{V}{R}I_1 +\frac{2P_p}{3R}\right)}\right)$$
Are you changing the definition of Vneg? If so, PLEASE provide an updated diagram in which you define ALL of the quantities you use. Otherwise it just becomes a confusing mess.

As it stands, your Vneg is defined at the voltage difference across the load irrespective of where your ground is. I see no compelling reason to change it. If you make Vneg a measurement relative to ground, then what node is Vneg referring to? The power dissipated in the lower load is no longer related to Vneg since Vneg will now include a voltage drop across the center resistor.

I just don't see what is so hard about providing an annotated schematic. Here:

I've defined both mesh currents so that they circulate clockwise.

Now your mesh equations are simply

$$2R \cdot I_1 \; - \; R \cdot I_2 \; = V \; - \; V_{pos}$$

$$-R \cdot I_1 \; + \; 2R \cdot I_2 \; = V \; - \; V_{neg}$$

Assuming your equation #5 was correct in the original post:

$$I_1 = \frac{V(I_1)^2\left(3R(I_1)^2 - 3I_1V+2P_p\right)}{6(I_1)^4R^2 - 6RV(I_1)^3 +(I_1)^2(7P_pR+NP_pR)-3P_pVI_1 + 2(P_p)^2}$$
Why not just write it the way polynomials are usually written (assuming I didn't mess it up):

$${6R^2 \cdot I_1^4 \; - \; 9RV \cdot I_1^3 \; + \; \( \(7+N$$RP_p \; + \; 3V^2\) \cdot I_1^2 \; - \; 5VP_p \cdot I_1 \; + \; 2(P_p^2)} \; = 0\)

MrAl

Joined Jun 17, 2014
8,996
Hello,

I briefly tested the function for I1 in step 3 of the first post and it seems to work.
What i am thinking is that if this is a 4th degree equation then maybe a numerical solution would be better because a direct symbolic solution would be too huge to work with unless you can program it into a home made program yourself. Ive done this before with very very large thousands of character equations, but you have to ask if it is worth it.

Blomqvist

Joined Jun 10, 2015
12
but if you only had one branch one load could you solve that?
Assuming only one branch with one source and one load, there are two solutions that can be derived from the following resulting expression (assuming my math is right):
$$I_1^2 + \frac{V}{2R}I_1 - \frac{P}{2R} =0$$

Why not just write it the way polynomials are usually written (assuming I didn't mess it up):
I'll study this step tomorrow and get back to you!

a 4th degree equation then maybe a numerical solution would be better because a direct symbolic solution would be too huge to work with unless you can program it into a home made program yourself. Ive done this before with very very large thousands of character equations, but you have to ask if it is worth it.
I did just that and it wasn't pretty to say the least. Perhaps I can consider an approximation?

MrAl

Joined Jun 17, 2014
8,996
Assuming only one branch with one source and one load, there are two solutions that can be derived from the following resulting expression (assuming my math is right):
$$I_1^2 + \frac{V}{2R}I_1 - \frac{P}{2R} =0$$

[reference to numerical solution]
I did just that and it wasn't pretty to say the least. Perhaps I can consider an approximation?
Not sure what you mean by "it wasn't pretty".
Are we talking about the same thing?
I was talking about a numerical solver that solves equations numerically such as:
a4*x^4+a3*x^3+a2*x^2+a1*x+a0=0

The numerical solver would find out the value(s) of x.
There will be 4 solutions, some may be complex.

With a numerical solver you just hand it the equation and it will find the solutions and for a 4th degree equation it will work pretty fast.
Alternately you could provide your own algorithm such as Lin-Bairstow. There are of course other algorithms.
The aforementioned method is pretty fast too. It will spit out the four solutions and you can test the real ones to see which one is correct.

There is an article in Wikipedia for "Bairstow" if you look that up.

The reason i suggested a numerical method was because the symbolic solution looks so complicated and with many characters in the equation.
I'll try it again though later today and see if i can get it down to earth.

There also may be a trial and error numerical method because the circuit itself is so simple.
There could be other interesting possibilities too. Let's not rush this one it's too interesting to just pass by when one solutions is found.

MrAl

Joined Jun 17, 2014
8,996
0.361423 0.364791 0.456824 0.450000
0.363073 0.363921 0.455113 0.453412
0.363495 0.363707 0.454687 0.454262
0.363601 0.363654 0.454581 0.454475
0.363627 0.363641 0.454554 0.454528
0.363634 0.363637 0.454548 0.454541
0.363636 0.363637 0.454546 0.454544
0.363636 0.363636 0.454546 0.454545
0.363636 0.363636 0.454545 0.454545
0.363636 0.363636 0.454545 0.454545
0.363636 0.363636 0.454545 0.454545

I dont know if i got lucky here or not but the last entries in that table are correct to that number of digits for I1 and I2.
I1 is first column, I2 is third column, the others are just previous values and can be ignored.
So i start off with a guess for I2 (0.45) and after 7 iterations it converges.
It would take more iterations though if the guess for I2 was more incorrect.
Interesting though, when i set I2=16 amps it comes out close after 10 iterations.
16 amps is way off from 0.45454545 amps the correct solution.

It's a very simple algorithm where i use the guess for I2 to predict I1, then use that I1 to check I2. If not the same as the guess, a correction is made.
What i did not do yet is test for convergence for any situation.
[Note: However, see Bisection Method]

Last edited:

MrAl

Joined Jun 17, 2014
8,996
Hello again,

After converging on an unexpected result, i came to the realization that there could be up to four real solutions, and this follows from the fourth degree equation too.

Let's look at a couple simple examples.

1. V1=2, I1=3, V2=4, I2=5
let's say that is one solution.
Now because if we are given a number:
K=A*B

and we know that K is 6 (and let's say for simplicity A and B have to be integers not equal to 1) then there are two solutions:
A=2, B=3
B=3, A=2

Therefore if the power K was 6 and then all we know is that V1*I1=6
and we can make either one 2 and the other 3 (integers for now again) and so we dont know if I1 is 3 or 2.

Since we have two such power values, we can at least see three solutions because we can either:
1. Swap the first pair
2. Swap the second pair
3. Swap both pairs.

Now we might be able to say that a 3rd degree poly would solve this, or there could be another solution and it would also have to be real.
The problem is that if V1 changes, then we cant calculate I1 can we without knowing V1 also.
To state this another way, the equations in part 3 of the first post may have three or four solutions. It is possible that the fourth solution is zero but we could test that.

Solving with some power values and V=3 and R=2, i get three real solutions:
I1=0.79235551110658, I1=0.92397950443343, I1=0
and two imaginary solutions. That's because it ended up being a 5th degree equation the way i did it in the final solution.
Those two non zero values actually work in the circuit also so they must be correct.

Last edited:

WBahn

Joined Mar 31, 2012
26,398
Hello again,

After converging on an unexpected result, i came to the realization that there could be up to four real solutions, and this follows from the fourth degree equation too.
Why is this unexpected? I said back in Post #5:

What I see is a fourth order polynomial. My gut feel is that this is what you would expect since you have two powers (one for each load) and power is related to the square of the current flowing in it. You should also expect multiple solutions -- for instance, setting the power to zero for one of the loads could equate to short circuiting the load or to open-circuiting it. Since you could do either for either load, there would need to be four different solutions corresponding to each possibility.
In general, for most power levels achievable in each load it can be achieved with either a lower voltage and higher current or a higher voltage and lower current.

MrAl

Joined Jun 17, 2014
8,996
Hi,

Second, i guess i must have missed that part of the post. Sorry about that.
When i got the alternate solution with the algorithm i set up, i realized this. But yeah of course there could be four solutions.
But also interesting maybe is if there is one real solution, then there must be at least one more real solution because that would be required to factor the poly.
With the test i did i got two real and two complex. I cant say if that will always happen though.

The algorithm is pretty cool, but for now i can only get it to converge on one of the two solutions and zero. I guess zero comes with open circuits.

Blomqvist

Joined Jun 10, 2015
12
Not sure what you mean by "it wasn't pretty".
Are we talking about the same thing?
My bad, what I was referring to was witnessing symbolic gobbledygook of the resulting solutions solved by wolfram. When I say "it wasn't pretty", I meant completely unusable in its present form.

To state this another way, the equations in part 3 of the first post may have three or four solutions. It is possible that the fourth solution is zero but we could test that.
A fourth solution for I1 could be zero only if I2 became zero, which is a true statement only if the power across the load is zero or the input voltage is zero. When I first ventured into this problem, I played with Excel's "Solver" function to see how well it could converge on a zero. It did very well to identify one of the zeros, but not the zero that may actually meet the source voltage requirement. Zero was an actual zero for the solution, but only under artificial limits. I didn't spend much time tuning the boundaries.

The algorithm is pretty cool, but for now i can only get it to converge on one of the two solutions and zero. I guess zero comes with open circuits.
I had a lot of fun playing with this idea! Thank you!!

Why not just write it the way polynomials are usually written (assuming I didn't mess it up):
Today I confirmed your math and you were correct. However, it was a real "doh" moment for me when I realized how easily the function could be expressed as a basic quartic polynomial. Since I'm using excel to develop the calculator, I decided to breakdown the general formula for the roots. While I could have brute forced the solution, I found "hard coding" solutions into excel proved to ultimately have greater efficiency and precision that I couldn't ignore. While I could have converged on a given zero for the function, I wouldn't be able to determine which solution I was going to receive.

Consequently, by finding the four solutions, I determined the Solver function did a great job identifying a solution. Unfortunately, it picked the "wrong" solution 3 out of 4 times. I suppose this is more me not yet being able to identify which solution is "more" correct. For instance, see the image below:

Rows 24 thru 27 provide the four solutions to the quartic. Solution #1, that is X1, turns out to be the correct solution for the system. However, X2 is also a solution that fulfils the requirements (to my knowledge) of the system. So, I need to figure out how to distinguish these roots more thoroughly. On the other hand, While X3 and X4 are solutions, they violate the source input voltage requirement.

The Delta value in Row 17 is a depressed quartic. In this example, it is positive. This means either all four roots are real or none are. However, to narrow in on the nature of these roots, Row 15 has a P value that is negative, which when considered in conjunction with the sign of the Delta value, ensures the function has four real and distinct roots. Exciting!

In closing, I think this just about wraps up the work for the calculator. While I still need to consider how the simulated system converges on only one solution (X1) when there are two valid solutions (X1 and X2), the calculator is working as it should. I really appreciate the help I have recieved from both you, WBahn and MrAI! If you have any additional comments, I look forward to reading them as well.

MrAl

Joined Jun 17, 2014
8,996
Hi,

I see what might be a couple problems.

First, your equation in line 8 has coefficients a,b,c,d yet your list shows a,b,c,d,e.

When i use a,b,c,d,e for powers of x from x^4 down to x^0 (5 co's) i get different results:
I1=0.099870913157
I1=0.150685362896
I1=6.624003783663
I1=13.37543994028

I used a different approach to get the coefficients but they are the same as yours.