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In a scientific paper (Design Considerations for Current Transformer Based Energy Harvesting for Electronics Attached to Electric Motor J. Ahola et. al.) I came across the comparator circuit with hysteresis shown in the picture. The authors of the paper use this circuit to control the voltage at a capacitor. If the voltage at the capacitor exceeds a level Vdc,max the comparator should output high, if the voltage falls below the level Vdc,min the comparator should output low. The reference voltage of the comparator is set with Zener diode D3. R3 is used to limit the current through the Zener diode. I would now like to derive the equations for Vdc,min and Vdc,max given in the paper myself but fail at times.

As a first approach I tried to solve it using the voltage divider approach as soon as the voltage at the positive input is greater than Vz the comparator should switch, before that the output must be at Vcc-Low = GND. Therefore the voltage at R2 has to be determined. is this the right approach?

circuit and equation for switching values for Vdc

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    \$\begingroup\$ I have not yet decided if I love or hate the rounded corners and joints. \$\endgroup\$
    – pipe
    Commented Aug 24, 2023 at 11:41
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    \$\begingroup\$ @pipe I recommend hating them. ;) \$\endgroup\$ Commented Aug 24, 2023 at 12:42
  • \$\begingroup\$ Ziegi, just study the papers mentioned here. \$\endgroup\$ Commented Aug 24, 2023 at 12:43
  • \$\begingroup\$ Ziegi - Hi, Please remember the site rule which says that when a post includes content (e.g. text, image, photo etc.) copied or adapted from elsewhere, that content must be correctly referenced. For online content, the source webpage or PDF etc. should be linked as a minimum (references for printed books / articles should include title, author(s), publisher, edition, page numbers etc. - a typical citation). Therefore please can you edit your question to include the required reference for the paper, where that schematic originated? Thanks. \$\endgroup\$
    – SamGibson
    Commented Sep 8, 2023 at 3:35

4 Answers 4

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One way to immediately write down the answer is to do a Wye-Delta transformation on R1, R2, R4.

enter image description here

Then (picking the more complex one, because why not) we can immediately write:

\$ \frac{V_{MIN}}{V_Z} = \frac{R_5 ||R_{AC}}{R_{BC}} +1 \$

where \$R_{AC} = \frac{R_1 R_2 + R_1 R_4 + R_2 R_4}{R_2}\$ and \$R_{BC} = \frac{R_1 R_2 + R_1 R_4 + R_2 R_4}{R_1}\$

A bit of manipulation later and you'll see that the authors of the paper missed an added R1*R2 in both numerator and denominator, which might explain why you are having problems deriving their equation.

I'll leave the other derivation to you.

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    \$\begingroup\$ Thanks for your hint. My solution also contained the R1xR2 term you mentioned and therefore differs from the solution in the paper. This has already created quite a headache for me. \$\endgroup\$
    – Ziegi
    Commented Aug 24, 2023 at 20:30
  • \$\begingroup\$ Reading your solution a second time, I noticed that I don't understand why your solution is for Vmin. The other postings all assumed that Vmin is the voltage at which the output is connected to GND. Since you write R5 || Rac it means that the output is at Vdc. Shouldn't therefore Vmax / Vz be written? \$\endgroup\$
    – Ziegi
    Commented Sep 1, 2023 at 20:00
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    \$\begingroup\$ The minimum input threshold is when the output is high and contributing. \$\endgroup\$ Commented Sep 1, 2023 at 20:10
  • \$\begingroup\$ The output is high as long as Vmin does not fall below Vref, therefore R5 || RAC. If Vmin is undershot, the output changes to low. Then R5 || RBC. Sounds logical, all other contributions assume that for Vmin the output is negative and for Vmax the output is positive. This confuses me, there can be only one correct solution. \$\endgroup\$
    – Ziegi
    Commented Sep 4, 2023 at 10:43
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    \$\begingroup\$ My solution agrees with the original equations (except for those two terms). I am satisfied it is correct. I have not evaluated others' answers in detail, you might want to raise it with them. Perhaps they have a different definition of min and max. In any case, one will be higher than the other. \$\endgroup\$ Commented Sep 4, 2023 at 10:55
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This is how I would approach the problem. Below is the circuit drawn with nodes Z and X, where Z has fixed reference potential \$V_Z\$, and X has potential \$V_X\$ that is a function of both output \$V_{OUT}\$ and the supply \$V_{DC}\$:

schematic

simulate this circuit – Schematic created using CircuitLab

You have two conditions to consider, output high, \$V_{OUT} = V_{DC}\$, and low, \$V_{OUT} = 0V\$. In the former case, we have a network of resistances connected as shown below left, and the latter situation on the right:

schematic

simulate this circuit

The algebra is easier if we treat the supply DC and resistors R1 and R2 as their Thevenin equivalent:

schematic

simulate this circuit

where:

$$ V_{TH} = V_{DC}\frac{R_2}{R_1+R_2} $$

$$ R_{TH} = R_1 \parallel R_2 = \frac{R_1R_2}{R_1 + R_2} $$

The resistor networks in each situation (high and low output) are therefore equivalent to these:

schematic

simulate this circuit

These are simple resistor potential dividers, the only difference between the two being the presence/absence of the voltage source \$V_{DC}\$. The potentials at X are trivial to write:

$$ V_{X(H)} = V_{DC} + (V_{TH} - V_{DC})\frac{R_5}{R_4 + R_5 + R_{TH}} $$

$$ V_{X(L)} = V_{TH}\frac{R_5}{R_4 + R_5 + R_{TH}} $$

You may have noticed that R4 is redundant. It's effectively in series with the other two resistances R5 and Rth. You can obtain exactly the same values for \$V_X\$, in both states, by removing it altogether, and adjusting Rth and/or R5 to compensate. I won't do that here, it's just a heads-up.

Now you can substitute back in the expressions for \$V_{TH}\$ and \$R_{TH}\$ from before. The last step is to equate \$V_{X(H)} = V_Z\$ and \$V_{X(L)} = V_Z\$, which I think you can handle from here. You should end up with expressions for \$V_{X(H)}\$ and \$V_{X(L)}\$ in terms of only the resistances and \$V_{DC}\$.


Update 1 - choosing resistances

Here's an answer I wrote about this, which will help to choose an order of magnitude for the resistances in the circuit.

If you start by specifying explicit potentials for the upper and lower switching thresholds, say \$V_{DC}=8V\$ and \$V_{DC}=6V\$, and the zener voltage, say \$V_Z=4.7V\$, you obtain two simultaneous equations that can be rearranged to yield any two of the resistances. The problem is that there are four resistances to find, and only two simultaneous equations. That's because there are infinitely many solutions for those resistances that would satisfy the conditions described by the equations.

Note: As I mentioned before, I consider \$R_4\$ and \$R_{TH}\$ to be a single resistance, and it will greatly simplify things if you set \$R_4 = 0\Omega\$. You could always re-introduce \$R_4\$ again later, and adjust \$R_{TH}\$ to compensate, but I see no reason to do that. \$R_4\$ is completely redundant here.

It seems impossible to solve, but all you need to do is randomly (well, not randomly, follow the guidelines from the linked answer above) select values for any two of them, say \$R_4 = 10k\Omega\$ and \$R_5=100k\Omega\$, and then solve for the other two.

At this stage it doesn't matter whether you guessed those resistances correctly or not, because the ratio of the resistances will be the same for all solutions. That is, if you multiply all the resistances by the same factor (say 2, or 0.5) they will still be valid solutions, and the thresholds remain the same. This is due to the linear nature of the system.

So if, for example, when you solve the system of equations using some randomly chosen value of R4 and R5 (or any of them), but you find the resulting resistances to be on the low side, then double them all (or multiply them all by whatever factor would yield more appropriate values), and they will produce the same voltages everywhere. This trick wouldn't work if the equations weren't linear, but in this case they are.


Update 2 - a walkthrough

$$ V_{X(H)} = V_{DC} + (V_{TH} - V_{DC})\frac{R_5}{R_4 + R_5 + R_{TH}} $$

$$ V_{X(L)} = V_{TH}\frac{R_5}{R_4 + R_5 + R_{TH}} $$

The Thevenin equivalencies:

$$ V_{TH} = V_{DC}\frac{R_2}{R_1+R_2} $$

$$ R_{TH} = \frac{R_1R_2}{R_1 + R_2} $$

Combining these:

$$ \begin{aligned} V_{X(H)} &= V_{DC} + \left( V_{DC}\frac{R_2}{R_1+R_2} - V_{DC} \right) \frac{R_5}{R_4 + R_5 + \frac{R_1R_2}{R_1 + R_2}} \\ \\ &= V_{DC} \left( 1 - \frac{R_1R_5}{(R_4+R_5)(R_1+R_2)+R_1R_2} \right) \\ \\ \\ \\ V_{X(L)} &= V_{DC} \left( \frac{R_2}{R_1+R_2} \right) \left( \frac{R_5}{R_4 + R_5 + \frac{R_1R_2}{R_1 + R_2}} \right) \\ \\ &= V_{DC} \frac{R_2R_5}{(R_4+R_5)(R_1+R_2)+R_1R_2} \\ \\ \end{aligned} $$

We are going to specify values of \$V_{DC}\$ where we intend switching to occur, corresponding to when \$V_X=V_Z\$ in each case, so substitute \$V_X\$ with \$V_Z\$:

Note: counter-intuitively, the upper threshold \$V_{DC(MAX)}\$ occurs when the output is low, and vice versa.

When the output is high:

$$ \begin{aligned} V_Z &= V_{DC(MIN)} \left( 1 - \frac{R_1R_5}{(R_4+R_5)(R_1+R_2)+R_1R_2} \right) \\ \\ V_{DC(MIN)} &= V_Z \frac{(R_4+R_5)(R_1+R_2)+R_1R_2}{(R_4+R_5)(R_1+R_2)+R_1R_2-R_1R_5} \\ \\ \end{aligned} $$

When the output is low:

$$ \begin{aligned} V_Z &= V_{DC(MAX)} \frac{R_2R_5}{(R_4+R_5)(R_1+R_2)+R_1R_2} \\ \\ V_{DC(MAX)} &= V_Z \frac{(R_4+R_5)(R_1+R_2)+R_1R_2}{R_2R_5} \\ \\ \end{aligned} $$

These don't look much like the equations quoted in your question. I don't want to spend ages reconciling your equations with mine. In any case, mine here are correct.

Since both of these share the same rather complex term, I'll write them like this:

$$ \begin{aligned} K &= (R_4+R_5)(R_1+R_2)+R_1R_2 \\ \\ V_{DC(MIN)} &= V_Z \frac{K}{K-R_1R_5} \\ \\ V_{DC(MAX)} &= V_Z \frac{K}{R_2R_5} \\ \\ \end{aligned} $$

Let's establish some targets. These are some example characteristics I am aiming for, using a 4.7V zener diode:

$$ \begin{aligned} V_Z &= 4.7V \\ \\ V_{DC(MIN)} &= 6V \\ \\ V_{DC(MAX)} &= 8V \\ \\ \end{aligned} $$

Unfortunately, here's where things get difficult, because to solve these two equations with four unknown resistances, requires us to "guess" two of them. If we choose unwisely the equations could yield a solution for the other two with negative resistances. Algebraically the solution would still be valid, but physically impossible to build.

Intuitively, I see that \$R_5\$ and \$R_4\$ together are responsible for hysteresis, the "gap" between the two thresholds. \$R_1\$ and \$R_2\$ on the other hand control the "midpoint" potential. We would have to take great care when setting the midpoint, which would be algebraically complicated, so I won't specify \$R_1\$ or \$R_2\$. I'd rather let the algebra we've already established deal with that.

Instead, I'll try choosing arbitrary values for \$R_4\$ and \$R_5\$. I am sure that \$R_4=0\$ will work just fine (which I recommend), but in the interest of keeping \$R_4\$ in this design, I'm going to set it to be much smaller than \$R_5\$, and non-zero. If \$R_4\$ is too large compared with \$R_5\$, we may inadvertently make it impossible for \$V_X\$ to ever equal \$V_Z\$, which would yield negative \$R_1\$ and/or \$R_2\$ in the solution.

$$ \begin{aligned} R_4 &= 10\Omega \\ \\ R_5 &= 100\Omega \\ \\ \end{aligned} $$

Clearly, those values are too small, but they keep the calculations simple, and permit me to show you later on how we can adjust and correct them.

Plug these values into the simultaneous equations from above:

$$ \begin{aligned} K &= (R_4+R_5)(R_1+R_2)+R_1R_2 \\ \\ &= 110(R_1+R_2)+R_1R_2 \\ \\ \end{aligned} $$

$$ \begin{aligned} V_{DC(MIN)} &= V_Z \frac{K}{K-R_1R_5} \\ \\ 6 &= 4.7 \frac{K}{K-100R_1} \\ \\ R_1 &= \frac{1.3}{600}K \\ \\ \end{aligned} $$

$$ \begin{aligned} V_{DC(MAX)} &= V_Z \frac{K}{R_2R_5} \\ \\ 8 &= 4.7 \frac{K}{100R_2} \\ \\ R_2 &= \frac{4.7}{800}K \\ \\ \end{aligned} $$

To solve these, substitute the expressions for \$R_1\$ and \$R_2\$ into the equation for \$K\$, and solve for \$K\$. Then use that value for \$K\$ to evaluate \$R_1\$ and \$R_2\$. We finally have all resistances:

$$ \begin{aligned} R_1 &= 19.57\Omega \\ \\ R_2 &= 53.07\Omega \\ \\ R_4 &= 10\Omega \\ \\ R_5 &= 100\Omega \\ \\ \end{aligned} $$

These are far too small, and will draw too much current from the supply and op-amp output. With these equations all being linear, we can simply scale all resistances by the same factor, without changing any voltages anywhere.

This means you can find a factor that will adjust any one of the resistances to become any value you choose, and when you scale all the resistances by that same factor, circuit behaviour doesn't change. I will multiply them all by 1000, to get resistances in the tens of kilohms:

$$ \begin{aligned} R_1 &= 19.57k\Omega \\ \\ R_2 &= 53.07k\Omega \\ \\ R_4 &= 10k\Omega \\ \\ R_5 &= 100k\Omega \\ \\ \end{aligned} $$

schematic

simulate this circuit

Here's a plot of \$V_{DC}\$ and \$V_{OUT}\$ as I sweep V1 from 0V to 12V and back to 0V. As you can see, the output rises when \$V_{DC}=8V\$, and falls again when \$V_{DC}=6V\$:

enter image description here

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  • \$\begingroup\$ Thank you for your explanation and the effort you have made, I think that this post is very informative for many beginners like myself. Now the procedure is clear to me, thanks for that. \$\endgroup\$
    – Ziegi
    Commented Aug 24, 2023 at 21:04
  • \$\begingroup\$ Following your approach, I derived the equations for Vx(H) and Vx(L). I now only have Vdc and the resistors. But I do not understand what is meant by equating Vx(H) = Vz and Vx(L) = Vz. \$\endgroup\$
    – Ziegi
    Commented Sep 1, 2023 at 20:23
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    \$\begingroup\$ @Ziegi You have two expressions for \$V_X\$, representing the conditions present for each output state. Switching occurs when \$V_X=V_Z\$ (the two comparator inputs have the same potential), so substitute \$V_X\$ with \$V_Z\$ in each equation, and rearrange to find the corresponding supply voltage \$V_{DC}\$. \$\endgroup\$ Commented Sep 2, 2023 at 5:09
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    \$\begingroup\$ @Ziegi I've appended some hints to my answer. As for R5 being over 100k, this is not necessarily true. Constrain and choose R5 according the op-amp's output current capability. All other resistances will change as a function of R5. and vice versa, so all resistor values need to be made suitable, as a group, and individually. My guidelines are all in my answer. \$\endgroup\$ Commented Sep 6, 2023 at 4:55
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    \$\begingroup\$ @Ziegi Actually, if you can wait a bit, I'll walk you through a complete solution. You've been good enough to follow my answer closely until now, I think you deserve further help. \$\endgroup\$ Commented Sep 6, 2023 at 5:05
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If the voltage at the capacitor exceeds a level Vdc,max the comparator should output high, if the voltage falls below the level Vdc,min the comparator should output low.

The comparator with hysteresis may not be a solution to your problem. Because with hysteresis, the max and min points change slightly around the original reference point (the reference without the hysteresis).

So window comparator might be what you need:


If the comparator with hysteresis what you need, here's your answer:

is this the right approach?

I couldn't understand your approach but here's how I calculate:

Without the extra resistor (from output to non-inverting input) the trigger point would be the reference voltage (voltage at the inverting input). Quite simple:

schematic

simulate this circuit – Schematic created using CircuitLab

With the extra resistor the division ratios will change. Assume the comparator gave an output of HI (VCC):

schematic

simulate this circuit

So if you calculate Vtrig_HI from the divider network above that'll be the max trigger point.

Likewise, assume the comparator gave an output of LO (GND):

schematic

simulate this circuit

Again, if you calculate Vtrig_LO from the divider network above that'll give you the min trigger point.


EXAMPLE:

VX = 10V, R1 = 10k, R2 = 1k, and comparator's supply is VCC = 5V.

Without any hysteresis the trigger point would be 2V.

But if we add an extra resistance of R3 = 20k for hysteresis, using the two diagrams above, the low trigger would be 1.92V and the high trigger would be 2.11V.

schematic

simulate this circuit

For HI, solve the following for Vtrig-HI:

$$ \mathrm{ \frac{10V-V_{trig-HI}}{4k}+\frac{5V-V_{trig-HI}}{20k}=\frac{V_{trig-HI}}{1k} } $$

And for LO, solve the following for Vtrig-LO:

$$ \mathrm{ V_{trig-LO} = 10V \cdot \frac{(1k \ || \ 20k)}{4k+(1k \ || \ 20k)} } $$


You can derive the equations if you want. During a design, I normally put the calculations into an Excel file and play with the values (cells) to see how trigger points change.

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  • \$\begingroup\$ Is not VX = VCC ... in the OP figure ? \$\endgroup\$
    – Antonio51
    Commented Aug 24, 2023 at 11:48
  • \$\begingroup\$ @Antonio51 It is, and it makes the things a bit more complicated. And VCC will no longer be taken as a constant parameter, instead it's going to be an unknown. \$\endgroup\$ Commented Aug 24, 2023 at 12:29
  • \$\begingroup\$ I made the calculations for your given values VX = 10V, R1 = 10k, R2 = 1k, and comparator's supply is VCC = 5V but for Vtrig-Lo I get 0,87V. I have also done a spice simulation with this values and set Vref to 2V. The Resistors R1, R2 and R3 I placed the same as you. Simulation only showed the right behavior when R1=1k and R2=10k. \$\endgroup\$
    – Ziegi
    Commented Sep 1, 2023 at 21:37
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Looking for the minimalistic solution...

This "scientific solution" caught my attention firstly with its many resistors, secondly with its single power supply, thirdly with the fact that the input signal serves as a power supply, and finally with the strange shape of its output signal which follows the input. I instinctively felt that it could be simplified and felt the desire to do so, albeit with less than "scientific methods". In the end, I was able to reduce the number of resistors by two. Here is how I did it.

5-resistor solution

In the original OP solution, the input voltage is scaled down twice with a voltage divider R2-R4 and applied to the input of a non-inverting comparator with hysteresis. Through the voltage divider R5-R3, a positive feedback is realized through which the hysteresis is obtained.

schematic

simulate this circuit – Schematic created using CircuitLab

STEP 1

4-resistor solution

If we make the voltage divider R2-R4 high enough resistive, it can also perform the functions of the input resistor R5; so we can remove the latter.

schematic

simulate this circuit

STEP 2

3-resistor solution

And finally, we can try to assign the functions of the R2-R4 voltage divider to the input resistor R5 (R2 in the schematic below).

schematic

simulate this circuit

In my opinion, the circuit should not work at its lower threshold because the reference voltage on the Zener diode ceases to be constant... but the graph shows hysteresis. This mystery should be cleared up...

STEP 3

Conclusion

In general, I think that supplying the op-amp from the input signal is not a good idea.

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