# Triangular-wave to sinewave converter with symmetrical circuit

#### RitterTree

Joined Dec 18, 2022
23
I don't think I have a solid approach when analyzing complex circuits like these. I hope some of you can give me general directions and help me tackle this one.

Given that the diodes are ideal and $$V_s(t)$$ is a triangular waveform generator, I am asked to:

1. Derive the analytic expression of the transcharacteristic $$V_o \; = \; f(V_s)$$ and draw it.
2. Recognize eventual properties and characteristics of the circuit.

Whenever I analyze circuits with diodes I always start by supposing that all of them and the AC generator are off to check their actual bias. I also try to redraw the circuit to simplify it a bit and maybe find familiar configurations. No luck with this one.

I started with something like this:

and ended up with this:

I won't bother you with my first try, I ended up failing to find a clean solution so I asked my professor about it. He didn't have much time and the only thing he managed to tell me was that all the time I wasted on my calculations could have been saved if only I had noticed that this was a symmetric circuit and then he showed me the graph of the transcharacteristic which was, in fact, symmetric to the $$V_o = V_s$$ line.

I went back at it again and this is how I proceeded:
This time I considered only half the circuit:

If both $$D_1$$ and $$D_2$$ are off, then I can say that $$V_{A1} = V_{A2} = V_s(t) = 0V$$, $$V_{K1} = \frac{\left( R/4 + R/4 \right)}{\left( R/4 + R/4 \right) + R/4}9V = \frac{2}{3}9V = 6V$$ and $$V_{K2} = \frac{R/4}{R/4 + R/4 + R/4}9V = \frac{1}{3}9V = 3V$$.

Therefore, until $$V_s(t)$$ doesn't reach $$3V$$, $$V_o(t) = V_s(t)$$.

When $$V_s(t) = 3V$$, $$D_2$$ is forward biased and the circuit becomes:

Applying Millman two times I can derive $$V_o(t)$$:

First I simplified the branch like so: $$\frac{\frac{9V}{\frac{R}{4}+\frac{R}{4}}}{\frac{4}{R}+\frac{1}{\frac{4}{R}+\frac{4}{R}}} = \frac{2}{6}9V = 3V$$.

And then derived $$V_o(t) = \frac{4V_s(t)+3\left(3V\right)}{4+3} = \frac{4}{7}V_s(t)+\frac{9}{7}V$$.

My problem lies here: I know that the value of $$V_{K1}$$ has changed but I don't know how to calculate it..

If I'd be able to, I could say for what value of $$V_o(t)$$ $$D_1$$ would get forward biased and the circuit would become:

And therefore $$V_o(t) = \frac{4\frac{V_s(t)}{R}+4\frac{9V}{R}}{\frac{4}{R}+\frac{4}{R}+\frac{28}{11R}} = \frac{11}{29}V_s(t)+\frac{11}{29}9V$$.

But now I wouldn't know how to justify that the other half of the circuit is the opposite of the one I just analyzed..

#### MrAl

Joined Jun 17, 2014
11,466
I don't think I have a solid approach when analyzing complex circuits like these. I hope some of you can give me general directions and help me tackle this one.

View attachment 296123

Given that the diodes are ideal and $$V_s(t)$$ is a triangular waveform generator, I am asked to:

1. Derive the analytic expression of the transcharacteristic $$V_o \; = \; f(V_s)$$ and draw it.
2. Recognize eventual properties and characteristics of the circuit.

Whenever I analyze circuits with diodes I always start by supposing that all of them and the AC generator are off to check their actual bias. I also try to redraw the circuit to simplify it a bit and maybe find familiar configurations. No luck with this one.

I started with something like this:

View attachment 296124 and ended up with this: View attachment 296125

I won't bother you with my first try, I ended up failing to find a clean solution so I asked my professor about it. He didn't have much time and the only thing he managed to tell me was that all the time I wasted on my calculations could have been saved if only I had noticed that this was a symmetric circuit and then he showed me the graph of the transcharacteristic which was, in fact, symmetric to the $$V_o = V_s$$ line.

I went back at it again and this is how I proceeded:
This time I considered only half the circuit:

View attachment 296126View attachment 296127

If both $$D_1$$ and $$D_2$$ are off, then I can say that $$V_{A1} = V_{A2} = V_s(t) = 0V$$, $$V_{K1} = \frac{\left( R/4 + R/4 \right)}{\left( R/4 + R/4 \right) + R/4}9V = \frac{2}{3}9V = 6V$$ and $$V_{K2} = \frac{R/4}{R/4 + R/4 + R/4}9V = \frac{1}{3}9V = 3V$$.

Therefore, until $$V_s(t)$$ doesn't reach $$3V$$, $$V_o(t) = V_s(t)$$.

When $$V_s(t) = 3V$$, $$D_2$$ is forward biased and the circuit becomes:

View attachment 296128

Applying Millman two times I can derive $$V_o(t)$$:

First I simplified the branch like so: $$\frac{\frac{9V}{\frac{R}{4}+\frac{R}{4}}}{\frac{4}{R}+\frac{1}{\frac{4}{R}+\frac{4}{R}}} = \frac{2}{6}9V = 3V$$.

And then derived $$V_o(t) = \frac{4V_s(t)+3\left(3V\right)}{4+3} = \frac{4}{7}V_s(t)+\frac{9}{7}V$$.

My problem lies here: I know that the value of $$V_{K1}$$ has changed but I don't know how to calculate it..

If I'd be able to, I could say for what value of $$V_o(t)$$ $$D_1$$ would get forward biased and the circuit would become:

View attachment 296129

And therefore $$V_o(t) = \frac{4\frac{V_s(t)}{R}+4\frac{9V}{R}}{\frac{4}{R}+\frac{4}{R}+\frac{28}{11R}} = \frac{11}{29}V_s(t)+\frac{11}{29}9V$$.

But now I wouldn't know how to justify that the other half of the circuit is the opposite of the one I just analyzed..

Hello there,

I used to be interested in these circuits back in the 1980's and sometime later also when people on the internet began to ask questions about these types of wave shaping circuits. The real life circuits use transistors and there are actual IC's that were made using that idea, like, if i remember right, the 8038 IC from i think Intersil. That was a long time ago.

This diode circuit really isn't that hard to solve you might just be getting confused because there are several diodes and resistors and at first it looks hard to do. You do already know that because this circuit is symmetrical and time is not a difficult problem because there are no storage elements, that only two diodes are ever on at the same time, and because the input is a triangle wave that goes from 0 to some max (for one half cycle) that means only one diode will be on at first, then as the voltage climbs another diode turns on later. That's the only affect time has on this problem.

If the diodes are ideal then that means that they either a short when 'on' or they have a forward voltage Vd when 'on', and when 'off' they are open circuit. The interesting thing about diode circuits though is that they are often two variable problems, both voltage AND time, so we have to calculate some voltages and solve for some time values.

Given that, you can calculate when the first diode turns on by looking at just three resistors, two voltage sources, and of course that one diode voltage drop. Again, since they are ideal diodes you can replace that diode with a voltage source in order to figure out the voltages at the resistor nodes and the output. After that, it is just a matter of calculating when the other diode turns on, then replace that with a voltage source also equal to the forward voltage drop of the diode (usually either 0v or 0.7v).

Because at the start of the triangle, assuming you start it at 0v, there are no diodes conducting, the output becomes the input for that time until the first diode turns on, and then really you just have a three source, multiple resistor problem, where you can use superposition (if you like) to calculate the voltages as the triangle rises.

See if that helps.
Keep in mind this really just turns into a several source and multiple resistor problem, so if you know how to calculate the output of a resistor network with several voltage sources, you can solve this one too, simply by calculating when the circuit topology changes due to the changing input triangle. The only difference is you have to solve for some time values as well, namely, two time values for each half cycle as the triangle is rising, and two when it is falling but those two should be simpler due to the fact that there are no storage elements in this circuit.
If you still have trouble, try setting the triangle to one single constant value and calculating the output with just one diode turned on (which means it becomes a voltage source, and if the forward voltage is to be zero then that voltage source values is zero, and if it is 0.7v as is often the case, then it becomes 0.7v). You can then move to other triangle voltage levels just to get a feel for what is happening and how it works.

Another idea, and this isn't really cheating, is to try all the values for the two diodes and see what combinations make sense. This isn't as nice though as actually calculating everything outright and probably doesn't teach you as much. The idea here is simple, replace each diode with either it's 'on' or it's 'off' characteristic, then analyze the circuit for that part. Because there are two diodes involved for each possibility, that means there are a total of 4 possibilities for those diodes. Either the first one is on and second one off, or the first one is off and second one on, or the two are both on, or the two are both off. This gets a little tricky though as you also calculate the input triangle voltage level also.

Amazing as it may sound right now, it is possible to determine the optimum values for the resistors such that you get a sine wave output with the lowest THD, or one of the lowest possible THD values. That's of course a bit more difficult.

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#### MrAl

Joined Jun 17, 2014
11,466
Hello again,

I re-read your post and see that you already calculated the first breakpoint (they call the places where the topology changes breakpoints).
That is when Vs=3v, and that is correct when we consider Vd of any diode to be zero. Note that if Vd was 0.7 we would get a different solution because of that diode drop.

So now you replace the diode with a short, period. You don't place a constant 3v anywhere because that 3v will also change as Vs rises. All you do is change the topology of the circuit, which means now the diode is shorted and that's all you need to know for the next step.
Once the diode is shorted, the resistor associated with that diode is now in the circuit and stays in the circuit throughout, until the voltage of Vs starts to fall and then you have to do some more calculations.

I think you got the basic idea right though. You solve for when the voltage at the top of the diode is equal to the bottom of the resistor with that diode or the bottom of that diode, and that gives you the solution. That will give you a particular value for Vs.

On the way down, you just do the same thing only this time the topology will change by open circuiting a diode.

It's nice to see you got this far already.

The way i would do this is calculate the voltages at all the nodes (in that one half section) with both diodes open, then one diode shorted, then the other diode shorted, then go from there. You can also place a resistor in series with the other diode and that means you can just solve the circuit once and when you want to open a diode just let it's associated resistor value go toward infinity. When both diodes are open you let both associated resistors go to infinity, and that is the result when no diodes are turned on yet.

It would also be nice to plot the output, but we would need the peak value of Vs(t) to do that and i guess we assume it is centered at zero volts.

#### RitterTree

Joined Dec 18, 2022
23
So now you replace the diode with a short, period. You don't place a constant 3v anywhere because that 3v will also change as Vs rises. All you do is change the topology of the circuit, which means now the diode is shorted and that's all you need to know for the next step.
Once the diode is shorted, the resistor associated with that diode is now in the circuit and stays in the circuit throughout, until the voltage of Vs starts to fall and then you have to do some more calculations.
Yes, sorry for not being clear in the question. Every time I draw a new topology of the circuit I implicitly consider myself working within the boundaries of the values of $$V_s(t)$$ for which the topology is valid, meaning, for which the ideal diodes hold their bias.

The way i would do this is calculate the voltages at all the nodes (in that one half section) with both diodes open, then one diode shorted, then the other diode shorted, then go from there.
I tried to do that but, still, I can't figure out how to calculate the new value of $$V_{K1}$$ when $$D_1$$ is off (open circuited) and $$D_{2}$$ is on (short circuited).

It would also be nice to plot the output, but we would need the peak value of Vs(t) to do that and i guess we assume it is centered at zero volts.
Yes, I omitted the specifics about $$V_s(t)$$ but the exercises states that it is a $$(+12V; -12V)$$ triangular wave centered at $$0V$$.

Even if I finished the analysis for this portion of the circuit I still wouldn't be able to explain why the symmetrical opposite holds true for its other half.

I've never worked with symmetrical circuits and if my professor wouldn't have told me that this was one of them I would have never guessed it. Not only I couldn't see symmetry, I couldn't even have imagined that a circuit can be symmetrical not relative to an axis of the transcharacteristic but to one of its bisectors..

#### MrAl

Joined Jun 17, 2014
11,466
Yes, sorry for not being clear in the question. Every time I draw a new topology of the circuit I implicitly consider myself working within the boundaries of the values of $$V_s(t)$$ for which the topology is valid, meaning, for which the ideal diodes hold their bias.

I tried to do that but, still, I can't figure out how to calculate the new value of $$V_{K1}$$ when $$D_1$$ is off (open circuited) and $$D_{2}$$ is on (short circuited).

Yes, I omitted the specifics about $$V_s(t)$$ but the exercises states that it is a $$(+12V; -12V)$$ triangular wave centered at $$0V$$.

Even if I finished the analysis for this portion of the circuit I still wouldn't be able to explain why the symmetrical opposite holds true for its other half.

I've never worked with symmetrical circuits and if my professor wouldn't have told me that this was one of them I would have never guessed it. Not only I couldn't see symmetry, I couldn't even have imagined that a circuit can be symmetrical not relative to an axis of the transcharacteristic but to one of its bisectors..
Hello again,

Once you short out D2 you have a different circuit with Vs already at some level above zero. It becomes a two source circuit, maybe that is what is the problem. Do you know how to solve two source and multiple resistor circuits? This is a crucial point in solving this next step.
Once you draw the circuit with D2 shorted you should be able to see that it is just a matter of finding the next Vs where D1 turns on.
If you still have trouble i'll give you a hint. But it is important that you know how to solve a two source resistive circuit.

If you think about the symmetry for a minute, what is happening to the other half of the circuit while you are solving this half? It's out of the picture and that is because neither diode is conducting on that side. Same thing will happen when you allow Vs to go negative, except on the side you are now working on, isn't that right? Does this make sense to you now?

Some things take a little longer to learn than others and sometimes that is just because it "looks" too hard, when really it isn't. A few key points and it gets easier to understand.

#### RitterTree

Joined Dec 18, 2022
23
Once you short out D2 you have a different circuit with Vs already at some level above zero. It becomes a two source circuit, maybe that is what is the problem. Do you know how to solve two source and multiple resistor circuits?
I think I can see it now, and I also think I messed up badly by simplifying that branch (when $$D_1$$ is off and $$D_2$$ is on) using Millman the way I did. I did it again but using the Thevenin theorem:

$$R_{th} = \left(\frac{R}{2} // \frac{R}{4}\right) + \frac{R}{3} = \frac{R}{2}$$
$$E_{th} = \frac{\frac{R}{4}}{3 \frac{R}{4}} 9V = 3V$$ (Since no current passes through $$\frac{R}{3}$$ when I clip the terminals)
$$therefore \; V_o ( = V_{A2} = V_{A1}) = \frac{2}{3}V_s + 1V$$

Now, about $$V_{K1}$$: since it depends on two voltage sources I need to use the superposition principle to derive its value (right?). I thought about using Millman again, and this time taking into account all the branches (as I should have done before). So, considering this circuit:

I can write: $$V_A = \frac{\frac{12}{7R}V_s+\frac{2}{R}9V}{\frac{12}{7R}+\frac{2}{R}+\frac{4}{R}} = \frac{2}{9}V_s + \frac{7}{3}V$$

$$KVL) \; V_A = 9V - \frac{R}{2}I$$
$$therefore \; I = \frac{2}{R} \left( 9V - V_A\right) = \frac{2}{R} \left( \frac{20}{3}V - \frac{2}{9}V_s\right)$$
$$therefore \; V_{K1} = 9V - \frac{R}{4}I = \frac{4}{27}V_s + \frac{41}{9}V$$
(I feel like something is wrong..)

$$therefore \; D_1 : ON \; iff \; V_{A1} \gt V_{K1} \; iff \; \frac{2}{3}V_s + 1V \gt \frac{4}{27}V_s + \frac{41}{9}V \; iff \; V_s \gt \frac{48}{7}V \simeq 6.86V$$
(Yes, something is most definitely wrong..)

Is this, at least, the right approach?

If you think about the symmetry for a minute, what is happening to the other half of the circuit while you are solving this half? It's out of the picture and that is because neither diode is conducting on that side. Same thing will happen when you allow Vs to go negative, except on the side you are now working on, isn't that right? Does this make sense to you now?
How can you say that no diode is conducting in the other half without analyzing it at least for $$V_s = 0V$$?

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#### MrAl

Joined Jun 17, 2014
11,466
I think I can see it now, and I also think I messed up badly by simplifying that branch (when $$D_1$$ is off and $$D_2$$ is on) using Millman the way I did. I did it again but using the Thevenin theorem:

View attachment 296269

$$R_{th} = \left(\frac{R}{2} // \frac{R}{4}\right) + \frac{R}{3} = \frac{R}{2}$$
$$E_{th} = \frac{\frac{R}{4}}{3 \frac{R}{4}} 9V = 3V$$ (Since no current passes through $$\frac{R}{3}$$ when I clip the terminals)
$$therefore \; V_o ( = V_{A2} = V_{A1}) = \frac{2}{3}V_s + 1V$$

Now, about $$V_{K1}$$: since it depends on two voltage sources I need to use the superposition principle to derive its value (right?). I thought about using Millman again, and this time taking into account all the branches (as I should have done before). So, considering this circuit:

View attachment 296271

I can write: $$V_A = \frac{\frac{12}{7R}V_s+\frac{2}{R}9V}{\frac{12}{7R}+\frac{2}{R}+\frac{4}{R}} = \frac{2}{9}V_s + \frac{7}{3}V$$

$$KVL) \; V_A = 9V - \frac{R}{2}I$$
$$therefore \; I = \frac{2}{R} \left( 9V - V_A\right) = \frac{2}{R} \left( \frac{20}{3}V - \frac{2}{9}V_s\right)$$
$$therefore \; V_{K1} = 9V - \frac{R}{4}I = \frac{4}{27}V_s + \frac{41}{9}V$$
(I feel like something is wrong..)

$$therefore \; D_1 : ON \; iff \; V_{A1} \gt V_{K1} \; iff \; \frac{2}{3}V_s + 1V \gt \frac{4}{27}V_s + \frac{41}{9}V \; iff \; V_s \gt \frac{48}{7}V \simeq 6.86V$$
(Yes, something is most definitely wrong..)

Is this, at least, the right approach?

How can you say that no diode is conducting in the other half without analyzing it at least for $$V_s = 0V$$?
Hi again,

First, i can say that no diode is conducting in the other half because the voltages at the bottom of those two diodes is negative, and since we are dealing with the positive half cycle of Vs, that means the cathodes are positive. When the cathode is positive and the anode is negative, no diode conducts.

Now about the calculation for the next voltage breakpoint.
Whatever method you use should work as long as you make sure you do it right.
For superposition with two sources, you would short one source out and calculate the node voltage with the one remaining source and that would give you part 1 of the total node voltage. Then short out the other source and calculate the node voltage with that second source now, and that would give you part 2 of the total node voltage. You add parts 1 and 2 together and that gives you the total node voltage.
You then subtract that from Vs and equate that to zero, and then solve for Vs.
If you did that and you get a result of say 5v, then check to see that everything works as far as the current in each branch and the voltage at each node. If something doesn't work you'll get a node where the sum of currents is not zero, and that would defy the most basic rule about nodes and currents. That would mean you made a mistake but probably only in the calculations not in the technique.
I do not think anything around 6v works for the second breakpoint i think it is in the neighborhood of 8v, but I'll double check that and get back here.

[A LITTLE BIT LATER]
Ok the second breakpoint is above 8v. See if you can get the exact value.
The interesting thing here is that as Vs rises both the anode voltage AND the cathode voltage rise so it's just a matter of finding the point where they both reach the same value as Vs rises, with Vs being the unknown variable.

Again, i cant stress enough that once you short out D2, you again reduce the circuit to a two-source circuit with several resistors. That's really the same thing you did before just that now the topology is slightly different. You will solve for a new Vs.
There may be cases where there is no solution. For example, what if the source Vs only went up to 7v and the solution was exactly 8v (but it's not 8v here). That would mean that the second breakpoint would never be reached, and although that could be a solution in itself it wouldn't help the sine shaping scheme very much.

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#### RitterTree

Joined Dec 18, 2022
23
Again, i cant stress enough that once you short out D2, you again reduce the circuit to a two-source circuit with several resistors. That's really the same thing you did before just that now the topology is slightly different. You will solve for a new Vs.
Didn't I do that? Millman implicitly takes into account the superposition principle, doesn't it?
With the topology given when $$D_1$$ is open and $$D_2$$ is shorted, I calculated the expressions for:
• $$V_{A1}$$ by first simplifying the the branch with the Thevenin theorem so that I could apply Millman and directly derive it.
• $$V_{K1}$$ by first using Millman to calculate the voltage $$V_A$$ ($$A$$ is just the label I gave to the node, it has nothing to do with the diode anodes) so that I could calculate the current passing through the resistor near the $$9V$$ generator and therefore derive $$V_{K1}$$ from the $$KVL) \; V_{K1} = 9V - \frac{R}{4}I$$.

The interesting thing here is that as Vs rises both the anode voltage AND the cathode voltage rise so it's just a matter of finding the point where they both reach the same value as Vs rises, with Vs being the unknown variable.
I calculated the value of $$V_s$$ for which $$V_{A1} \gt V_{K1}$$ (same information as $$V_{A1} = V_{K1}$$ because I know that from that value of $$V_s$$ onward $$D_1$$ will be shorted).

Isn't that right? My problem now is that $$V_s \simeq 6.86V$$ and not around $$8V$$ like you said.. I wonder what's wrong between my calculations, the expressions I derived or both..

For superposition with two sources, you would short one source out and calculate the node voltage with the one remaining source and that would give you part 1 of the total node voltage. Then short out the other source and calculate the node voltage with that second source now, and that would give you part 2 of the total node voltage. You add parts 1 and 2 together and that gives you the total node voltage.
I tried to calculate the expression of $$V_{K1}$$ with the superposition principle instead of using directly Millman and I got a different one: $$V_{K1} = \frac{10}{3} V_s + \frac{57}{8}V$$, which is totally wrong but I don't know where I'm making my mistake.

You then subtract that from Vs and equate that to zero, and then solve for Vs.
Isn't this wrong? $$V_{A1}$$ doesn't equal $$V_s$$ anymore. Since $$D_2$$ got shorted, it depends on the contribution of the $$9V$$ generator too, doesn't it?.

#### MrAl

Joined Jun 17, 2014
11,466
Didn't I do that? Millman implicitly takes into account the superposition principle, doesn't it?
With the topology given when $$D_1$$ is open and $$D_2$$ is shorted, I calculated the expressions for:
• $$V_{A1}$$ by first simplifying the the branch with the Thevenin theorem so that I could apply Millman and directly derive it.
• $$V_{K1}$$ by first using Millman to calculate the voltage $$V_A$$ ($$A$$ is just the label I gave to the node, it has nothing to do with the diode anodes) so that I could calculate the current passing through the resistor near the $$9V$$ generator and therefore derive $$V_{K1}$$ from the $$KVL) \; V_{K1} = 9V - \frac{R}{4}I$$.

I calculated the value of $$V_s$$ for which $$V_{A1} \gt V_{K1}$$ (same information as $$V_{A1} = V_{K1}$$ because I know that from that value of $$V_s$$ onward $$D_1$$ will be shorted).

Isn't that right? My problem now is that $$V_s \simeq 6.86V$$ and not around $$8V$$ like you said.. I wonder what's wrong between my calculations, the expressions I derived or both..

I tried to calculate the expression of $$V_{K1}$$ with the superposition principle instead of using directly Millman and I got a different one: $$V_{K1} = \frac{10}{3} V_s + \frac{57}{8}V$$, which is totally wrong but I don't know where I'm making my mistake.

Isn't this wrong? $$V_{A1}$$ doesn't equal $$V_s$$ anymore. Since $$D_2$$ got shorted, it depends on the contribution of the $$9V$$ generator too, doesn't it?.
Hi,

Refer to the two drawings in the attachment. The nodes are numbered as well as the resistors.

Yes, I misspoke when I said that last statement, obviously I meant that you subtract the node at the cathode from the node at the anode. When they become equal we enter the state where the topology changes again.
Also be aware it is not when the voltage at node 1 becomes greater than the voltage at node 3, it is when these two are EQUAL, meaning Vn1=Vn3. That gives you an easy way to write the equation because then it is just an equality. Vn1 will never be greater than Vn3 (Vn1>Vn3) because as soon as they become equal D1 is shorted, meaning there can be no voltage drop across it.

What I think we should do is first number all the resistors for that section, then redraw with only the 9v source 'shorted out', then draw again with only Vs shorted out. That will give you two schematics to work with directly and you will be working with individual resistors like R1, R2, etc., rather than R/4 and R/3 and the like. You may be able to see the right way to do the analysis using superposition this way.
I suggest we do comparisons to Millman's after you get the right results using superposition.
Also, just so you know, the next value for Vs for breakpoint #2 is somewhat above 8v, but you may be referring to node 1 which will be somewhat less. I don't want to specify any exact values yet though until you had a chance to find these yourself.

In the attachment I numbered all the resistors in the first diagram, then in the second I shorted out D2.
We can now refer directly to the node numbers. Node 1 will be equal to node 3 with diode D1 open, and that is the criterion for breakpoint #2. You can ignore R21 as that is not part of the original circuit, so just short that out.

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#### MrAl

Joined Jun 17, 2014
11,466
Hello again,

You may want to read post #9 first.

Here is a trick you can use when you have two voltage sources that are connected in series via a ground connection and nothing else connects to ground.

In the diagram in the first circuit, you will notice that if we remove the ground then V1 and V2 are connected directly in series, as in the second circuit, and in reality we don't even have to remove the ground to see that they are actually really connected in series. Since the two are connected with their negative terminals, that means the voltages subtract (as shown in gray with the "2v" text).
Since for this circuit we now have 2v across two resistors we simply have a resistive voltage divider although that may be different for other circuits, as long as nothing else connects to ground.
The ground is now at the bottom of R2, so that means we subtracted 8v from that node. If R2=R1 the output is now 1v, but because we subtracted that 8v previously we now have to add that back. This means the output voltage of the first circuit is 1v+8v=9v when R2=R1.

With any of these, including Millman, we have to be very careful when we rearrange the circuit because if we make one little mistake in the application of that idea we can get a very wrong result. That is why it is probably better to learn a very general way to solve these because then we will always be doing the same thing using the same procedure.

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#### RitterTree

Joined Dec 18, 2022
23
Also be aware it is not when the voltage at node 1 becomes greater than the voltage at node 3, it is when these two are EQUAL, meaning Vn1=Vn3. That gives you an easy way to write the equation because then it is just an equality. Vn1 will never be greater than Vn3 (Vn1>Vn3) because as soon as they become equal D1 is shorted, meaning there can be no voltage drop across it.
I missed that completely, thanks!

What I think we should do is first number all the resistors for that section, then redraw with only the 9v source 'shorted out', then draw again with only Vs shorted out. That will give you two schematics to work with directly and you will be working with individual resistors like R1, R2, etc., rather than R/4 and R/3 and the like.
Considering the circuit:

I proceed to calculate $$V_{K1}$$ when the $$9V$$ voltage generator is shorted:

Using Thevenin's theorem i simplify the circuit and derive $$V_{K1}$$ with a voltage divider.

$$with\;R_1 = R_2 = R_3 = R_4 = R$$
$$with\;V_{th} = \frac{R}{2R+R_{22}}V_s$$
$$with\;R_{th} = \left( R + R_{22}\right)//R = \frac{R\left( R+R_{22}\right)}{2R+R_{22}}$$
$$therefore\;V_{K1} = \frac{R}{2R+R_{th}}V_{th} = \frac{R\left(2R+R_{22}\right)}{5R^2+3RR_{22}} \frac{R}{2R+R_{22}}V_s =\frac{R^2}{5R^2+3RR_{22}}V_s$$

$$with\;R = \frac{R}{4}\; , \; R_{22}= \frac{R}{3}$$
$$then\;V_{K1} = \frac{1}{9}V_s$$

Now I proceed to calculate $$V_{K1}$$ when the $$V_s$$ is shorted:

I simplify the resistors, calculate the current that runs through the mesh and then derive $$V_{K1}$$.

$$with\;R_1 = R_2 = R_3 = R_4 = R$$
$$with\; R_{eq} = \left( R + R_{22}\right)//R = \frac{R\left( R+R_{22}\right)}{2R+R_{22}}$$
$$with\; KVL)\; 9V = I\left(2R + R_{eq}\right)$$
$$therefore\; I = \frac{9V}{\left(2R + R_{eq}\right)} = \frac{2R+R_{22}}{5R^2+3RR_{22}}9V$$

$$with\;R = \frac{R}{4}\; , \; R_{22}= \frac{R}{3}$$
$$then\; I = \frac{40}{3R}V$$

$$with\; KVL)\; V_{K1} = 9V - RI$$
$$then\;V_{K1} = 9V - \frac{R}{4} \frac{40}{3R}V = \frac{17}{3}V$$

Putting the two results together, for the superposition principle:
$$V_{K1} = \frac{1}{9}V_s + \frac{17}{3}V$$

$$therefore\; V_{A1} = V_{K1}\; iff\; \frac{2}{3}V_s + 1V = \frac{1}{9}V_s + \frac{17}{3}V\; iff\; V_s = \frac{42}{5}V = 8.4V$$

Is that correct?

You may be able to see the right way to do the analysis using superposition this way.
I suggest we do comparisons to Millman's after you get the right results using superposition.
I feel like it's safer but also a lot more time consuming, I don't know if I'd be able to be efficient or not make any errors doing so many calculations in a short period of time like during an exam. In my last reply I followed the same exact procedure but skimmed through it and, as you saw, got the wrong result..

I don't want to specify any exact values yet though until you had a chance to find these yourself.
I want to thank you for the immense patience you are having with me!

#### RitterTree

Joined Dec 18, 2022
23
Here is a trick you can use when you have two voltage sources that are connected in series via a ground connection and nothing else connects to ground.

In the diagram in the first circuit, you will notice that if we remove the ground then V1 and V2 are connected directly in series, as in the second circuit, and in reality we don't even have to remove the ground to see that they are actually really connected in series. Since the two are connected with their negative terminals, that means the voltages subtract (as shown in gray with the "2v" text).
Since for this circuit we now have 2v across two resistors we simply have a resistive voltage divider although that may be different for other circuits, as long as nothing else connects to ground.
The ground is now at the bottom of R2, so that means we subtracted 8v from that node. If R2=R1 the output is now 1v, but because we subtracted that 8v previously we now have to add that back. This means the output voltage of the first circuit is 1v+8v=9v when R2=R1.
This could come in handy when Millman is not directly applicable, thanks!
So, for the calculation of $$V_{K1}$$ I could have directly used that and limited the analysis to one configuration, right?

With any of these, including Millman, we have to be very careful when we rearrange the circuit because if we make one little mistake in the application of that idea we can get a very wrong result. That is why it is probably better to learn a very general way to solve these because then we will always be doing the same thing using the same procedure.
But still, I can't figure out where I got my Millman application wrong, if I look back at it I can't see anything..

#### MrAl

Joined Jun 17, 2014
11,466
I missed that completely, thanks!

Considering the circuit:

View attachment 296307

I proceed to calculate $$V_{K1}$$ when the $$9V$$ voltage generator is shorted:

View attachment 296308

Using Thevenin's theorem i simplify the circuit and derive $$V_{K1}$$ with a voltage divider.

$$with\;R_1 = R_2 = R_3 = R_4 = R$$
$$with\;V_{th} = \frac{R}{2R+R_{22}}V_s$$
$$with\;R_{th} = \left( R + R_{22}\right)//R = \frac{R\left( R+R_{22}\right)}{2R+R_{22}}$$
$$therefore\;V_{K1} = \frac{R}{2R+R_{th}}V_{th} = \frac{R\left(2R+R_{22}\right)}{5R^2+3RR_{22}} \frac{R}{2R+R_{22}}V_s =\frac{R^2}{5R^2+3RR_{22}}V_s$$

$$with\;R = \frac{R}{4}\; , \; R_{22}= \frac{R}{3}$$
$$then\;V_{K1} = \frac{1}{9}V_s$$

Now I proceed to calculate $$V_{K1}$$ when the $$V_s$$ is shorted:

View attachment 296309

I simplify the resistors, calculate the current that runs through the mesh and then derive $$V_{K1}$$.

$$with\;R_1 = R_2 = R_3 = R_4 = R$$
$$with\; R_{eq} = \left( R + R_{22}\right)//R = \frac{R\left( R+R_{22}\right)}{2R+R_{22}}$$
$$with\; KVL)\; 9V = I\left(2R + R_{eq}\right)$$
$$therefore\; I = \frac{9V}{\left(2R + R_{eq}\right)} = \frac{2R+R_{22}}{5R^2+3RR_{22}}9V$$

$$with\;R = \frac{R}{4}\; , \; R_{22}= \frac{R}{3}$$
$$then\; I = \frac{40}{3R}V$$

$$with\; KVL)\; V_{K1} = 9V - RI$$
$$then\;V_{K1} = 9V - \frac{R}{4} \frac{40}{3R}V = \frac{17}{3}V$$

Putting the two results together, for the superposition principle:
$$V_{K1} = \frac{1}{9}V_s + \frac{17}{3}V$$

$$therefore\; V_{A1} = V_{K1}\; iff\; \frac{2}{3}V_s + 1V = \frac{1}{9}V_s + \frac{17}{3}V\; iff\; V_s = \frac{42}{5}V = 8.4V$$

Is that correct?

I feel like it's safer but also a lot more time consuming, I don't know if I'd be able to be efficient or not make any errors doing so many calculations in a short period of time like during an exam. In my last reply I followed the same exact procedure but skimmed through it and, as you saw, got the wrong result..

I want to thank you for the immense patience you are having with me!

Hello again,

YES! Congratulations! Very good indeed.

Yes, it is safer and unfortunately more time consuming, but when you have a problem you cant figure out you have to go back to the most clear way of looking at the circuit and that usually helps if not solves the problem. It's not easy but sometimes it's the only way because then you start to see the solution more clearly and things pop out at you that didn't before. As you can see, it helps.

You're welcome, and i guess i always found these kinds of diode circuits interesting. I remember long time ago i tried to find different ways to analyze them and one way was to use a curve fitting program, strange as that sounds. It gets modified to do non linear analysis (in the case of the actual exponential model of the diode). Nodal analysis is pretty straightforward too.

#### MrAl

Joined Jun 17, 2014
11,466
This could come in handy when Millman is not directly applicable, thanks!
So, for the calculation of $$V_{K1}$$ I could have directly used that and limited the analysis to one configuration, right?

But still, I can't figure out where I got my Millman application wrong, if I look back at it I can't see anything..
Hello again,

Can you describe in words the way you went about doing it that way. That is, what did you change to get everything in parallel, which i assume you did, but if not what did you do then.

These more unusual theories have to be handled perfectly or something always goes wrong, mostly because they are not as direct.

#### RitterTree

Joined Dec 18, 2022
23
Can you describe in words the way you went about doing it that way. That is, what did you change to get everything in parallel, which i assume you did, but if not what did you do then.
I've already shown the passage in a previous reply, I'm going to copy and paste it again here:

Now, about $$V_{K1}$$: since it depends on two voltage sources I need to use the superposition principle to derive its value (right?). I thought about using Millman again, and this time taking into account all the branches (as I should have done before). So, considering this circuit:

I can write: $$V_A = \frac{\frac{12}{7R}V_s+\frac{2}{R}9V}{\frac{12}{7R}+\frac{2}{R}+\frac{4}{R}} = \frac{2}{9}V_s + \frac{7}{3}V$$

$$KVL) \; V_A = 9V - \frac{R}{2}I$$
$$therefore \; I = \frac{2}{R} \left( 9V - V_A\right) = \frac{2}{R} \left( \frac{20}{3}V - \frac{2}{9}V_s\right)$$
$$therefore \; V_{K1} = 9V - \frac{R}{4}I = \frac{4}{27}V_s + \frac{41}{9}V$$
(I feel like something is wrong..)

$$therefore \; D_1 : ON \; iff \; V_{A1} \gt V_{K1} \; iff \; \frac{2}{3}V_s + 1V \gt \frac{4}{27}V_s + \frac{41}{9}V \; iff \; V_s \gt \frac{48}{7}V \simeq 6.86V$$
(Yes, something is most definitely wrong..)

#### MrAl

Joined Jun 17, 2014
11,466
I've already shown the passage in a previous reply, I'm going to copy and paste it again here:

Now, about $$V_{K1}$$: since it depends on two voltage sources I need to use the superposition principle to derive its value (right?). I thought about using Millman again, and this time taking into account all the branches (as I should have done before). So, considering this circuit:

View attachment 296360

I can write: $$V_A = \frac{\frac{12}{7R}V_s+\frac{2}{R}9V}{\frac{12}{7R}+\frac{2}{R}+\frac{4}{R}} = \frac{2}{9}V_s + \frac{7}{3}V$$

$$KVL) \; V_A = 9V - \frac{R}{2}I$$
$$therefore \; I = \frac{2}{R} \left( 9V - V_A\right) = \frac{2}{R} \left( \frac{20}{3}V - \frac{2}{9}V_s\right)$$
$$therefore \; V_{K1} = 9V - \frac{R}{4}I = \frac{4}{27}V_s + \frac{41}{9}V$$
(I feel like something is wrong..)

$$therefore \; D_1 : ON \; iff \; V_{A1} \gt V_{K1} \; iff \; \frac{2}{3}V_s + 1V \gt \frac{4}{27}V_s + \frac{41}{9}V \; iff \; V_s \gt \frac{48}{7}V \simeq 6.86V$$
(Yes, something is most definitely wrong..)
Hello again,

What i meant was explain what you did in your own words so i can follow your logic. Unfortunately, math alone does not always tell you this information because it is after the fact. This network is particularly troublesome because several of the resistors are labeled with the same label.
What i need is a description of what you actually did to arrive at your conclusion. For example, "I placed R4 in parallel with R3 in order to ... ", etc.

I have included the schematic with resistors numbered to help with this. In this way it will become clear what you did with each resistor and voltage source. This way i can follow your thinking.

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#### MrAl

Joined Jun 17, 2014
11,466
Hello again,

You may want to look at post #16 first.

I went ahead and proved that you will get the same result with Millman as you would with Norton, as you would with superposition. Thus, the error must be in the way you applied the theorem. Sorry to have to say that, but if you go over the theorem and go over your work again you may find the error. Either that or explain how you went about solving this.
For example, "I added R1 and R2 and called that R12, then added R4 and R5 and called that R45, etc., etc.
This will help greatly to pinpoint the error.

If you want to prove using Norton, convert the left side into a current source in parallel with a resistor Rx1, then the other side to a current soruce in parallel with another resistor Rx2, then put Rx1, Rx2, and R3 in parallel and the two current sources in parallel. This way you end up with one current source and one resistor, and that gives you the voltage at A. To calculate the output, you would just use the voltage divider formula, the formula with two resistors and two voltage sources.

The schematic i provided helps, noting that if I say I added R/4 and R/4, you don't know which resistors I actually added together: was it R1 and R4, or R4 and R5, or ..., etc. There's no way to follow if I applied the theorem correctly, or at least if I started to follow the theorem correctly and then somewhere later in the process I messed up.