Reconciling Paper DC Network with BreadBoard Observations

[watchmaker]

Before moving onto opamps and AC analysis I wanted to verify my understanding of DC networks.

I was using a homework assignment in the Real Analog course and did not have the homework answers,  So I wanted to verify my answer.  Simple two source (opposed) series parallel. (Schematic below).

Was up to 4:30 AM Friday morning working to reconcile this.

The assignment (paper exercise) used 10 Vdc and 20 Vdc with resistors all under 40 ohms.  So when I first hooked it up I smoked a couple, 1/2 W then 1 W resistors.  So I dropped the sources down to 2.5 and 5 with all 1 W resistors. 

I then measured I(t), and all the I(Rs) as well as voltage drops using resistors in the same config as the assignment but with values I had that were near enough.  Redid the equations and the results did not reconcile.

FINALLY at 4AM I realized my 2.5 CV source was NOT CV; it was putting out 4V.  When I dropped this into my equations, it all reconciled.

So, apparently I do not know how CV works on a real world PSU.

Any recommended reading? 

THANKS!

[ledtester]

FINALLY at 4AM I realized my 2.5 CV source was NOT CV; it was putting out 4V.  When I dropped this into my equations, it all reconciled.

So, apparently I do not know how CV works on a real world PSU.

Perhaps the problem in this case (if you were using the schematic you posted) is that the power supply you used for the 2.5V source does not like to sink current.

Here is a discussion about how linear regulators respond to sinking current:

https://electronics.stackexchange.com/questions/262396/power-supply-response-to-sinking-current-overvoltage

If V1 is a typical series linear regulator (such as an LM7805 or LM317) it will not sink current and remain in regulation. Instead the output voltage will rise. ...

[Ian.M]

To breadboard this, I'd have scaled the resistors by a factor of 1000 (i.e. R1 1 ohm becomes 1K), and used 9V batteries for the sources, two in series for the nom. 20V one.   The resulting currents will be 9/10 * 1/1000 of the theory problem ones.

However if you don't need the pain of burning your fingers, its generally easier to put this sort of problem into LTspice as-is, but do remember SPICE programs need one node of the circuit to be grounded, and all parts of the circuit to have a DC path to ground.

[TimFox]

Yes, Spice requires a ground node and DC continuity from each node to ground to set the initial conditions.
However, you can add an absurdly high resistor (like 1000 G\$\Omega\$) from each floating node to ground to get past the initial error checking.

[watchmaker]

FINALLY at 4AM I realized my 2.5 CV source was NOT CV; it was putting out 4V.  When I dropped this into my equations, it all reconciled.

So, apparently I do not know how CV works on a real world PSU.

Perhaps the problem in this case (if you were using the schematic you posted) is that the power supply you used for the 2.5V source does not like to sink current.

Here is a discussion about how linear regulators respond to sinking current:

https://electronics.stackexchange.com/questions/262396/power-supply-response-to-sinking-current-overvoltage

If V1 is a typical series linear regulator (such as an LM7805 or LM317) it will not sink current and remain in regulation. Instead the output voltage will rise. ...

Well that sounds like what happened here.  Thanks.  I keep learning stuff!

[watchmaker]

To breadboard this, I'd have scaled the resistors by a factor of 1000 (i.e. R1 1 ohm becomes 1K), and used 9V batteries for the sources, two in series for the nom. 20V one.   The resulting currents will be 9/10 * 1/1000 of the theory problem ones.

However if you don't need the pain of burning your fingers, its generally easier to put this sort of problem into LTspice as-is, but do remember SPICE programs need one node of the circuit to be grounded, and all parts of the circuit to have a DC path to ground.

Well now I know better!  I thought the ground I put into Multisim was correct.  Did I not do it correctly?

Thanks!

[SteveThackery]

Is it worth us agreeing what the right results are by calculation, at this point?

In the attached diagram, the results in red are for the original voltages of 20V and 10V.  The results in green are for the reduced voltages of 5V and 2.5V.  Results are to three decimal places, and there will be rounding errors in the final digit or two.  As is customary in resistor network calculations, voltage sources are assumed to have zero internal resistance.

EDIT:  I show R1 to be 10 ohms in my diagram, when it should be 1 ohm. I used the correct value - 1 ohm - in the calculations.

Does anyone agree (or disagree) with my figures?

[Ian.M]

Your circuit has a small problem!^*, however the numerical results looks OK.

As an easy check, simplify it by deleting R2 and combining R3 and R4 to reduce it to a single series loop.   You can now rearrange the component order to combine the sources and remaining resistors to calculate I1.    For I1, I get  0.52A for your circuit (with 20V and 10V sources), and 0.9774 A for the original.

* You made a mistake transcribing the original, changing R1 from 1 ohm to 10 ohms in your diagram, but seem to be doing the math with the original value . . .

[watchmaker]

Is it worth us agreeing what the right results are by calculation, at this point?

In the attached diagram, the results in red are for the original voltages of 20V and 10V.  The results in green are for the reduced voltages of 5V and 2.5V.  Results are to three decimal places, and there will be rounding errors in the final digit or two.  As is customary in resistor network calculations, voltage sources are assumed to have zero internal resistance.

Does anyone agree (or disagree) with my figures?

Thank you all for taking the time with me.  I KNEW someone would ask me to show my work.

Steve, we agree on results for the assumed behavior of the network.  Below is my work for the assumed behavior and the actual behavior.  Also I show the observed measurements taken from the breadboard.  These agree with my calculations for the actual behavior and are internally consistent.

Again:  THANKS!

Oh, I also used the resistor values as measured with my Tweezers.

NUTS.  I DID rotate the pics in the MS Photo Editor; did not stick.  Sorry.

[SteveThackery]

Your circuit has a small problem!^*, however the numerical results looks OK.

As an easy check, simplify it by deleting R2 and combining R3 and R4 to reduce it to a single series loop.   You can now rearrange the component order to combine the sources and remaining resistors to calculate I1.    For I1, I get  0.52A for your circuit (with 20V and 10V sources), and 0.9774 A for the original.

* You made a mistake transcribing the original, changing R1 from 1 ohm to 10 ohms in your diagram, but seem to be doing the math with the original value . . .

Thanks, @Ian.M.  The 10 ohm value for R1 is a simple transcription error when I was drawing the schematic. I used the correct 1 ohm value in the calculations. I've added a note to my post. Thanks again for picking it up.

[ledtester]

Just wanted to demonstrate the Thevenin approach to this circuit...

The voltage at point A is the sum of the voltages when V1 is shorted + when V2 is shorted.

We will find:

V(A) = 0.9775. when V1 is shorted
V(A) = 18.045 when V2 is shorted

Thus, V(A) = 18.045 + 0.9775 = 19.0225.

We will also find that R2 never enters into this calculation.


We first start with the full circuit:



We will use R34 to designate the parallel combination of R3 and R4 and its value is 9.23.

We first short V2:



If we rearrange it we see that we just have a voltage divider with R1 and R34 which leads to a calculation of V(A) = 18.04.



We then short V1:



and when we re-draw it we see that we have a voltage divider with R1 and R34 but with the ground point in the middle. This just means that 0 volts is at the divider middle:



Note that R2 never enters into the calculation. Of course, knowing this in advance would greatly simplify the analysis.

Once you know the voltage at A, just subtract 10 to get the voltage at B and then all the currents are easily computed.

[SteveThackery]

This is the simplified circuit. R3 and R4 are combined into one. The voltage sources are on the left, the load resistors on the right.

As others have said, the presence of R2 does not affect the current in R1, R3 and R4. I've left it in the drawing because it does affect the current in V2.

[watchmaker]

Damn, you guys love to teach!  Thanks.

[watchmaker]

Just wanted to demonstrate the Thevenin approach to this circuit...

The voltage at point A is the sum of the voltages when V1 is shorted + when V2 is shorted.

We will find:

V(A) = 0.9775. when V1 is shorted
V(A) = 18.045 when V2 is shorted

Thus, V(A) = 18.045 + 0.9775 = 19.0225.

We will also find that R2 never enters into this calculation.


We first start with the full circuit:

(Attachment Link)

We will use R34 to designate the parallel combination of R3 and R4 and its value is 9.23.

We first short V2:

(Attachment Link)

If we rearrange it we see that we just have a voltage divider with R1 and R34 which leads to a calculation of V(A) = 18.04.

(Attachment Link)

We then short V1:

(Attachment Link)

and when we re-draw it we see that we have a voltage divider with R1 and R34 but with the ground point in the middle. This just means that 0 volts is at the divider middle:

(Attachment Link)

Note that R2 never enters into the calculation. Of course, knowing this in advance would greatly simplify the analysis.

Once you know the voltage at A, just subtract 10 to get the voltage at B and then all the currents are easily computed.

I did not think of Thevenin, but I did use V₂ to simplify as well.

Remember, I am showing for my breadboard (actual) ckt and its behavior (where the presumed to be a 2.5V CV source jumped to over 4 v.

That reduces it to VR₁+VR₃=.89 (V₁-V₂)

=I_T*R₁+I₃*R₃

=.89=.092+.7967