# Trying to solve an integral involving an LRC circuit

1. Sep 21, 2011

### danham1979

Well here goes for my first post...

Currently I am trying to build an active integrator.

The circuit is similar to this one...

http://www.art-sci.udel.edu/ghw/phys245/05S/classes/images/opamp-integrator.gif

However, the voltage source is an inductor, and the voltage out goes to an oscilloscope (50 Ohms). Using a Marx Generator, a pulse current induces a voltage across my inductor. Using Lenz's Law, the voltage to abhor the flux is of the equation

V_in = L*di/dt + R*i + V_op_amp

and the Voltage Out

V_out = -V_capacitor + V_op_amp

After applying the ideal OpAmp we get

V_in = L*di/dt + R*i & V_out = -V_capacitor

The capacitor is set up the same across the inverting input. My challenge is trying to solve an integral across the capacitor in order to find the integrator time constant.

For now, solving the difference equation for i(t) is...

i(t)= (V_in(t))/r*(1-exp(-(r*t)/l))

When I integrate this equation through the capacitor, I am having trouble with the integral

int(-infinity to t) (1/(r*c))*V_in(t')*exp(-(r*t')/l) w.r.t. t'

If I could somehow preserve the integral of V_in and pull out that exponential function then I could find the integrating factor! I know that with the current being generated (1MA) and the pulse duration (risetime of 90ns) the circuit will have a lot of problems, but the practice of the circuit is not a concern; the theory part is my main challenge. If you want to know the V_in(t) it is...

V_in(t) = (l/n)*A*sin(w*t)*sin(w*t)

where

l=inductance (60 nh)
n= number of turns on inductor (4)
A= Amplitude (1MA)
w= angular frequency (Pi/180ns)
t= time (t = [0, 90ns])

If I may put my problem aptly, it is what technique should I go through to solve that integral with the product of the voltage source and exponential function? Laplace? Fourier? Nevermind the back emf, or saturation across the OpAmp; I just need to know how to solve that integral.

Thank you for your time and any help and/or advice would be appreciated!

2. Sep 24, 2011

### jsgruszynski

Two ways:

1. Convert the derivatives and integrals to their Laplace transform equivalents and then solve using complex algebra (d/dt = s, int() = 1/s on operated functions)

2. Use a complex exponential as the solution for and solves directly in time domain - this automagically incorporates the sinusoidal time domain solutions.

Both are strictly equivalent.

3. Sep 25, 2011

### danham1979

Thanks! I will try that out. I've gotten nowhere on it, so I appreciate the help. I may return with more questions soon lol...

4. Oct 2, 2011

### jim hardy

maybe you're making it harder than necessary ...

for inverting opamp, basic rule is that input and feedback currents are equal (and opposite)
I in flows through resistor left to right
I feedback flows through capacitor left to right

so, since I in = I feedback

and I in = V in / R ;
I feedback, which is current through capacitor, is the same,
so voltage across capacitor should be integral of current,
(integral of ) Vin / R , from time = zero to whenever you stop the integration. Plus the initial condition of course and multiplied by 1/C also of course.

does that evaluate to anything simpler than your expression?

remember the polarity swap - there's a minus sign missing from my arithmetic.

i am not sure i understand your input -
is it really the voltage across the inductor produced by a current pulse?
That formula near end of your post looks more like a half cycle of AC voltage applied instead of a current pulse...
is it really sin^2(wt) ? even that should integrate readily.....

perhaps your inductor is in SERIES with R ? it wasn't shown in your drawing.
same logic applies, solve for input current and integrate.

regards,

old jim

5. Oct 5, 2011

### psparky

Active integrator?

Off the top of my head.....put a resistor in parallel with an capacitor in the feedback......and a resistor in the input of the omp amp. RF over RA will be your gain.

The break frequency will occur at 1/(2*pi*RF*C)

All frequencies over the break frequencies will be attenuated at a rate of 20 db per decade.

Done.

Also, the picture you show above is simply an integrator....not an active integrator. Active implies gain.....passive implies no gain like your picture.

But if you really just want an integrator....the picture you show above is perfect. It will integrate all day long.
Switch the resistor and capacitor and you have a differentiator.

Last edited: Oct 5, 2011
6. Oct 5, 2011

### jim hardy

should you actually build that circuit you'll soon find the trouble with an ideal integrator is - it truly integrates.
It will integrate to its physical limit
even if it's just integrating its own input error
so it needs a pushbutton switch in parallel with cap so you can periodically reset it to zero.

if you build it use good film capacitor
and FET op-amp , electrometer grade if you can.
error of one millivolt per hour is achievable with 1 uf polypropylene cap and OPA128.

7. Oct 7, 2011

### danham1979

Hey guys, I am back.

Jim,
The inductor is used as a B-dot probe. Derived here...

emf = turns*-d(flux)/dt = turns*area*d(B)/dt= L*dI/dt

Or better yet... this is better
http://www.umich.edu/~peplweb/diagnostics/b-dot.html

With that, the A*sin^2(wt) is just a test signal for my pulse with t=[0,90ns] (at all other times it is zero). I have numerically integrated the derivative of this, and now, when I compare my voltage across my capacitor, I need to alter my resistance and capacitance values (all others are fixed, R and C are variable) to match the voltage. With this, I hope to find my integrating constant of

V_out=K*∫(V_in) dt

If I make K equal one, which I believe will hold L,R,C and the oscilloscope values (50 Ω) then I will be doing okay... I think.

The goal of this is to find the magnetic field from the pulse, and the best way to it is to measure the voltage. But since the inductor only picks up the change in current, I must integrate it to get to the voltage for the magnetic field.

Questions.

What is RF and RA and how does gain play into this?
What should I look into about frequencies?

I am new to this, and being a physics student, I figured EE's could point me in a better direction than Physicists at the moment... Thank you very much and thanks for your time.

8. Oct 7, 2011

### psparky

RF and RA are often used when referring to op-amps.

RF is the feed back resistor.....RA is the input resistor.

If you just have a simple inverting op amp with these to resistors, the gain will be RF/RA.

Op amps are much simpler to figure out than transistors. But op amps take up a lot more real estate...so your smart phone is filled with transistors rather than op amps. However, your stereo system or guitar amplifier is certainly using op amps.

U can solve most if not all op amp problems by using V=IR. Since V+ = V- and no current enters the op amp (ideal amplifiers).....using simple node equations and using your head will do the trick.

As far as frequencies, you can attenuate unwanted frequencies by using differentiators and integrators. They both have the RF and RA...just the active integrator (low pass filter) puts a capacitor in parrallel with RF. The break frequency will occur at 1/(W*RF*C). That RC should look familiar....it's the same RC in this eqaution of e^-(t/RC) The transients of a RC circuit and the op amp described above are identical.

The differntiator attenuates low frequencies and has the capactor in series with RA.

Regardless, you can use the RF/RA in inverting amps to figure the transfer function keeping in mind that capacitors have a reactance of (1/JWC)

Plug in an frequency (radians in this case), and you will be able to find the gain anywhere from zero to infinity. Keep in mind that your limits, zero and infinity can often be used to help you get the general "feel" for you filter.

9. Oct 7, 2011

### jim hardy

Hi Dan

i had a nice response typed out and previewed, as was reaching for the 'submit' key suffered a power failure. Electric company just got us back on, so if this post seems scrambled it's because i'm trying to re-create it. And i'm of a sorta scrambled nature anyway.

a flux detector? What fun. in the power plant i used ten turns of small wire wound into a 1/10 square meter coil taped to cardboard for a detector, and a battery powered oscilloscope. I was looking for stray 60 hz fields. Worked very well.

I have one slight worry about your scheme but i am sure with your obvious math skills you can work through it. Later on that.

"What is RF and RA and how does gain play into this?"
i think Sparky meant:
RF as feedback resistor, which you dont have in your sketch but he suggested adding.
and RA as R in your sketch

for simplicity let us just start with your ideal integrator sketch with only R and C and the amplifier. Then we'll look at RF.

Voltage on a capacitor = 1/C * (time integral of current)

You want K to be 1
and
(how do you guys make an integral sign appear? let me use instead intf()...)

since voltage on a capacitor = 1/C * intf(current )

Vout will be 1/C * intf(feedback current), let me call feedback current Ifb

and Ifb will equal I input, which = Vin/R

so Vout = 1/C * intf(Vin/R)
since R is constant take it outside the integral
Vout = 1/(R*C) * intf(Vin)

so to make K=1 you want RC product to equal 1 , it's really that easy.
1 megohm and one microfarad would do that.

Use a good film capacitor, as i mentioned i had good results with polypropylene capacitors and an electrometer grade opamp.
Read the application notes for TI's OPA128
to get good hold time you should make the connection of R and C to pin 2 not on the board but hang them in the air for extreme insulation
and connect the outside foil of capaitor (the end with stripe) to pin 6 not 2

Now here's my worry - your input includes a sine function...
so when you dot-ed flux you multiplied it by omega , and your omega of pi/180ns is a substantial number.
Your Flux moves fast but it may be a pretty small amplitude.
So when you un-dot it with this integrator, your integral may turn up pretty small.
(That's what Sparky is getting at, an integrator has low frequency reponse cutoff so may not give much output for such a short duration input signal ).
your math will tell you that

i think you may well need to increase your K to make enough signal for you to see .
That's as simple as making R and C both smaller numbers.
Trouble is, smaller C makes your drift worse.
Smaller R gives more input current, and since input current flows through your sense coil it'll try to cancel out the flux you are trying to measure. Lenz warns us of that.

Those two things are the worry i mentioned. So experiment and find a K that gives you reasonable hold time and reasonble amplitude.

One other worry - precision opamps like electrometers are not particularly fast and you're in the microsecond range. If you have the luxury, try the scheme with slower signal for proof of principle. If you need a faster opamp, well,
Perhaps there's a good RF person in attendance here who could suggest a faster opamp still with ultra low input bias current and offset? Tall order, i know.
but i suspect you'll be okay.
might even get away with plain old LM324 opamp if you can capture result quickly before it drifts.

Now - that integrator will need to be reset to zero every so often. To that end place a pushbutton switch across the feedback cap C.. (place a few kohms in series with switch so it's not slamming the cap right across input...)
pressing the switch will zero your integrator.

Now, to RF:

IF instead of a switch you connected a high ohm resistor across the feedback capacitor,
you would change the transfer function and give it DC gain of Rfeedback/Rinput.
(Rfeedback is what i think Sparky meant by RF
and Rinput is your R at pin 2)

It would no longer be a true integrator. But so long as RF >>> R it will approximate a true integrator for first few% of first RF * C time constant
keep Rfeedback 100X Rinput and you should be in pretty good shape.

wow, what a ramble - i'm a real mess, huh?
sorry for the colloquilaisms and informality.

old jim

10. Oct 12, 2011

### danham1979

Jim,

Thanks for the well informed response. Unfortunately, I won't be able to test out the integrator on a slower pulse due to the machine. However, you're response has been very helpful. I am pretty sure in the near future I will be responding back with more questions. I suppose my main challenge at the moment is solving that integral (obtw, I just used the integral on the quick symbols bar!) with an exponential decay and sine function, but it looks like a Laplace transform should solve the problem. You're RF advice is very helpful (and makes me worried a bit to continue on) and I will make sure to take it into account! Thanks!

Dan