ideal miller integrator

ntetlow

Joined Jul 12, 2019
63
Hello,
attached are two screenshots. The first is downloaded from the internet and shows a schematic of an ideal miller integrator in its bare bones state.
The formula given is V2/II = -K/(1+KSC). The second screenshot is of my simulation in LTSPICE which I have created to try and prove the aforementioned
formula correct. The text underneath shows nodular workings I believe are correct and from which I was hoping to find the proof.
I can't solve the formula and was hoping that someone else might do so. Of course, the formula may be incorrect, it's not unheard of.

Notes on my workings:- I've used the value of K to be 10. I've used the value of C as 0.1592 and the frequency as 1 (so that SC equals 1). My II value is 2 from source transformation.
Please can someone either prove or disprove the formula.

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crutschow

Joined Mar 14, 2008
34,673
Why did you add R1 to the circuit (which does nothing in the ideal integrator circuit)?
There is no resistor in the ideal schematic.

ntetlow

Joined Jul 12, 2019
63

crutschow

Joined Mar 14, 2008
34,673
which implies that there is a resistor there, maybe it should no;t be there.
There is a resistor but its value is 1 tera-ohms so it has no essential effect on the simulation.
Why did you use a value of 2 ohms?

Also the ideal op amp integrator has the input signal going to the (-) input on the op amp.
Why is it going to the (+) input of your simulation?

And the netlist file shows an EGAIN of 1000.
Why did you use only 10?

You do not seem to know how to read a Spice netlist.

Last edited:

ntetlow

Joined Jul 12, 2019
63
I chose the values I used to keep the arithmetic simple. I have not simulated the one using the op amp as this is irrelevant to what I am looking for. The simulation should be sufficient to prove or disprove the formula.

WBahn

Joined Mar 31, 2012
30,230
If you look at http://www.ecircuitcenter.com/Circuits_Audio_Amp/Miller_Integrator/Miller_Integrator.htm which is where the schematic came from you will see at the bottom
* IDEAL INTEGRATOR
I__1 0 31 AC 1 PWL(0US 0A 0.01US 1MA 1US 1MA 1.01US -1MA 2US -1MA 2.01US 0MA)
RIN 31 0 1e12
C__ 31 32 500PF
EGAIN 0 32 31 0 1000
which implies that there is a resistor there, maybe it should no;t be there.
Notice that the resistor has a value of one million megohms!

It is there to satisfy the requirements that all nodes have a DC path to ground in order for the simulator to be able to converge on an initial operating point solution.

WBahn

Joined Mar 31, 2012
30,230
Your workings have a number of issues, some of which make it impossible to even tell what you are attempting to do.

You use "Vin", yet your schematic has three nodes: 0, V, and Vout. What is Vin?

You way that

(Vin-Vout) = I(R + 1/S.C)

I assume you are using the period to mean multiplication?

You have messed up order of operations. What you have written is

(Vin-Vout) = I(R + (1/S).C)

You are adding (or attempting to add) the two reactances as if they are in series. They aren't. Furthermore, you can't just add reactances, you need to use impedances. In addition, the reactance of a capacitor is negative.

In the next line, you have

Vin = II.R

Again, what is Vin?

What is II? You only have I defined. Are you multiplying I by I? If so, you have voltage on the left side and power on the right. Or did you mean I1 (the current source)? In which case you are claiming that ALL of the current from the current source is going through the resistor, which means that NONE of it is flowing through the capacitor.

In the last line, you have

(Vin-V)/R

What is this? Again, what is Vin? Whatever it is, it is NOT the voltage across the resistor, which is just V (since the other end of the resistor is tied to the reference node).

You final equation is trying to say that the current in the resistor AND the current in the capacitor are BOTH equal to I, but these are in parallel, which means that the current out of the current source must be 2*I, which is not possible.

In short, you need to step back and go learn fundamental current concepts before attempting to tackle op-amp circuits.

ntetlow

Joined Jul 12, 2019
63
Hello,
I'm sorry to have misled you, it's difficult to explain these things without face to face contact.
I've attached another screenshot which makes more sense. Basically it's the same schematic except for the source transformation.
The II mentioned in the formula equates to 2 amps (the result of the source transformation).
Once again, sorry for not sending the second screenshot.

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WBahn

Joined Mar 31, 2012
30,230
You are still trying to sum resistances and reactances, which can't be done. You need to use impedances.

The impedance of a capacitor is -j/(wC). That '-j' is critically important.

Your VCVS is using positive feedback, which will make it rail.

ntetlow

Joined Jul 12, 2019
63
Albut hand it l I want to know is, is the formula used in this piece on the internet, correct or not correct, I know imaginary numbers and reactance come into it but you're just splitting hairs. I did n't write the document and you can't contact them for an explanation so what am I supposed to do but hand the problem over to someone who does.

WBahn

Joined Mar 31, 2012
30,230
Interesting that you see the difference between positive and negative feedback as hair splitting. Or trying to add things that can't be added and expecting to get a correct result.

Perhaps if you had paid more attention to hair splitting, you'd be able to analyze their circuit yourself.

The first question to ask is whether the units work out. If the units don't work out, you know the result is wrong.

What do the units of V2/I1 need to be?

What are the units of the numerator (-Kv)?

What are the units of the denominator (1 + Kv·s·C)?

ntetlow

Joined Jul 12, 2019
63
What are you saying? I've ben a computer programmer for 40 years so I think I know what binary is. In the schematics I have sent, where is the negative feedback, I don't know? Once again, it's not my document.
Put yourself in my position, I'm keen on electronics, I wish I'd studied it at Imperial college instead of stupid Chemical engineering. I'm looking at understanding amplifiers, I don't have any professors to cry to, all I've got is the internet and any books I might buy. The website looks good, but the formula doesn't look right. As for paying attention, I don't need to be told by some pedant, what to do. I think you've insulted my enough, for the moment, are you going to help or not?

crutschow

Joined Mar 14, 2008
34,673
I have not simulated the one using the op amp as this is irrelevant to what I am looking for.
It is totally relevant to your simulation since it shows what the polarity of the voltage-controlled voltage-source should be, which is incorrect in your simulation.

crutschow

Joined Mar 14, 2008
34,673
I don't need to be told by some pedant, what to do. I think you've insulted my enough, for the moment, are you going to help or not?
With that crappy attitude, certainly not me.

ntetlow

Joined Jul 12, 2019
63
crutschow you're the one with the crappy attitude, I don't know who the hell you think you are.

WBahn

Joined Mar 31, 2012
30,230
What are you saying? I've ben a computer programmer for 40 years so I think I know what binary is. In the schematics I have sent, where is the negative feedback, I don't know? Once again, it's not my document.
Put yourself in my position, I'm keen on electronics, I wish I'd studied it at Imperial college instead of stupid Chemical engineering. I'm looking at understanding amplifiers, I don't have any professors to cry to, all I've got is the internet and any books I might buy. The website looks good, but the formula doesn't look right. As for paying attention, I don't need to be told by some pedant, what to do. I think you've insulted my enough, for the moment, are you going to help or not?
I am trying to help you, you are simply refusing it. I have asked you three very direct questions to help you answer your own question, you choose to ignore them as pedantic. Fine. You don't want to learn, I can't force you to learn.

Since all you want to know is whether the formula on the internet is correct or not, the answer is that the first one has one of the same errors that you made. But since you think that you making that error doesn't matter and that people pointing it out is pedantic and hair-splitting, then I can only assume that you are consistent in your position and are opposed to anyone pointing out the error on that internet site, since doing so would be pedantic and hair-splitting. So just ignore the error, like you expect everyone else to ignore yours.

ntetlow

Joined Jul 12, 2019
63
Oh dear. I have touched a nerve. I think that having treated me like a child, you expected me to just put up with it.So you think the internet is wrong, Do you know what the correct answer is? I've worked thru the previous example
V2 (gm - s·C)·R1·R2
---- = ---------------------------------------
I1 s·C·(gm·R1·R2 + R1 + R2) + 1
and rhat is correct.
As regards the three questions, V2/II should be ohms which the answer would be with, as they say, a large Kv value (1/(SC)).

ericgibbs

Joined Jan 29, 2010
18,982
hi netlow,
Sorry, the thread has gone a little sour.

Can the participants take a deep breath and try to get back on topic.

E.

ntetlow

Joined Jul 12, 2019
63
Ok. Will do.

ericgibbs

Joined Jan 29, 2010
18,982
hi ntetlow
That's good to hear, drinks all round later at the Hop Pole Hotel, OK.?
E