RLC Circuit Question

[Zirad]

First let me say howdy, I am a newbie.
If you can refer to this website (a good one btw)
RLC Circuit
It is a 10 ohm resistor in series with a 1H inductor in series with a 15uF capacitor.
Once the switch is activated the circuit will charge and when the switch is opened the circuit will begin to oscillate.
Questions:
1. The voltage source is only 5V, but the peak voltage across the inductor can be as high as 12 V, why? Is this because of flyback?
2. If XL goes up or the resistance of the circuit goes up, will the voltage across the inductor also go up? Less current path and therefore voltage must go up across the inductor?
3. How can you calculate the VP across the inductor?
Thanks in advance

 

[ericgibbs]

First let me say howdy, I am a newbie.
If you can refer to this website (a good one btw)
RLC Circuit
It is a 10 ohm resistor in series with a 1H inductor in series with a 15uF capacitor.
Once the switch is activated the circuit will charge and when the switch is opened the circuit will begin to oscillate.
Questions:
1. The voltage source is only 5V, but the peak voltage across the inductor can be as high as 12 V, why? Is this because of flyback?
2. If XL goes up or the resistance of the circuit goes up, will the voltage across the inductor also go up? Less current path and therefore voltage must go up across the inductor?
3. How can you calculate the VP across the inductor?
Thanks in advance

hi,
Breaking it down into simple steps.

Assume the resistance of the inductor is 0R, as the applied voltage 5V is via a 100R to the inductor a current of 5/100 = 0.05A will flow thru the inductor.

The resonant freq of the loop = 1/[2pi sqr(LC)] = 1/ 6.28 sqr [ 1 * 0.000015] = 41Hz

The impedance of the RLC loop is Z= 10 + 2pi * f * L + 1/[2pi * f * C]

So the impedance of the RLC = 251R at 41Hz

If we now interrupt the current of 0.05A then VL= 251 * 0.05 = 12.55Vinst.

The amplitude of the VL voltage will decay exponentially due to losses in the loop.

 

[MrAl]

First let me say howdy, I am a newbie.
If you can refer to this website (a good one btw)
RLC Circuit
It is a 10 ohm resistor in series with a 1H inductor in series with a 15uF capacitor.
Once the switch is activated the circuit will charge and when the switch is opened the circuit will begin to oscillate.
Questions:
1. The voltage source is only 5V, but the peak voltage across the inductor can be as high as 12 V, why? Is this because of flyback?
2. If XL goes up or the resistance of the circuit goes up, will the voltage across the inductor also go up? Less current path and therefore voltage must go up across the inductor?
3. How can you calculate the VP across the inductor?
Thanks in advance

Hello there,

Yes, the inductor and capacitor are both energy storage elements, so they can act somewhat like little batteries for short periods of time. Being energy storage elements, they store energy and the way that energy is released depends not only on them themselves but also on the other components in the circuit. They are slightly different from each other though, with the cap acting more like a battery and the inductor acting more like a current generator. Im not sure if you covered current generators yet, but current generators generate a current of a set level that does not depend on the other resistances in the circuit. That means if you have a low resistance value you get a low voltage, but if you have a high resistance value you get a large voltage. That voltage could be several thousand volts even running from a 9v battery. The inductive "kick back" as it is sometimes called is better understood as the inductor simply acting like a current generator rather than a purely passive element.
If this seems hard to understand, consider a battery with a very small resistor connected to it. The low value resistor draws a ton of current, while a high value resistor only draws a small amount of current. Well the inductor is similar, except it puts out the same current level for short periods of time, and so acts like a current generator and in theory the current generator can put out an infinitely high voltage level.
To put it another way, the inductor absorbs energy, stores it, and then releases it the way it wants to release it according to other circuit components and their values. The current is controlled by the inductor, but not the voltage. With the capacitor, the voltage is controlled by the capacitor but not the current, so it's opposite from the inductor.
The main problem with understanding the inductor is when we first look at it we think of only a battery as being able to supply energy, and that energy voltage level is always capped by the voltage of the battery, but the inductor can also supply energy and the way that inductor supplies energy is totally different than the way a battery supplies energy once charged in that it's voltage is NOT capped, only it's current. To understand this better, study the "current generator" or "current source" in some detail. After that the inductor's behavior is much easier to understand.

 

[Zirad]

Thank you

Thank you Eric and Mr1, the formulas are exactly what I was looking for and also I understand what you mean about Inductors and Caps. Inductors where current is a constant and Caps where voltage is a constant.
Thanks again-
Steve from Oregon USA-

 

[Ratchit]

Zirad,

First let me say howdy, I am a newbie.

Hi newbie.

1. The voltage source is only 5V, but the peak voltage across the inductor can be as high as 12 V, why? Is this because of flyback?

The voltage across the coil is -L*(dI/dt). That means that you can make just about any voltage you want appear across the coil with even a current pulse of low peak value, as long as the rise and fall times are short enough. The current change in the exponential damped sine wave circuit you referenced is fast enough to exceed 5 volts.

2. If XL goes up or the resistance of the circuit goes up, will the voltage across the inductor also go up? Less current path and therefore voltage must go up across the inductor?

Resistance will have no effect on -L*(dI/dt), provided it is not so high that the circuit is overdamped or critical so as to prevent oscillation. Xl will increase because the coil inductance or the frequency increases. Both paramerters will increase the coil voltage value according to the given formula.

3. How can you calculate the VP across the inductor?

Is that peak voltage?
1)You can plot the voltage and see where it attains its highest value. In this case it is -12.47 volts at 0.00586 secs.

2)You can find the derivative of the current to determine where its current changes the fastest, then apply the coil voltage formula. A knowledge of differential equations and a study of RLC transients is helpful.

Ratch

 

[Zirad]

Thanks for the feedback. A follow up quesiton:
Using; L*(dI/dt)
What do we know?
L = 1H
And resonace freq is 41 Hz, or a period of 24.4ms, but is that the change in time that we can use? dt?
Also, for di, the change in current, all we know is max current in this case .05A.
finally how did you come up with -12.47 volts at 0.00586 secs?
thanks again-
Steve-

 

[Ratchit]

Zirad,

Thanks for the feedback. A follow up quesiton:
Using; L*(dI/dt)
What do we know?
L = 1H
And resonace freq is 41 Hz, or a period of 24.4ms, but is that the change in time that we can use? dt?
Also, for di, the change in current, all we know is max current in this case .05A.
finally how did you come up with -12.47 volts at 0.00586 secs?

I just love easy questions like you just asked.

I solved the differential equation for current using Laplace transforms. Then I plotted the current and saw that it was a underdamped oscillatory circuit. That means that it is a sine/cosine wave that decreases exponentially with every cycle. The first transistion through zero is the fastest rate of current change, because that is where the peak current is the highest. The peak of the sinusoidal is the slowest rate. I plotted the current, expanded the plot, and estimated the time within the first period where the current went to zero. That was 0.00586 secs. Next I found the derivative of the current at that point and multiplied by L to get the voltage of -12.47 volts.

Now let's do it without the fancy mathematics. We know the initial current through L was 0.050 amps. We know the circuit oscillates sinusoidally. That makes the current:

I = 0.050*exp(-t/TC)*cos(ω*t) . We can throw away exp(-t/TC) because we are only going to select the first point where the current transistions through zero. We are left with:
I Ξ 0.050*cos(ω*t) for times close to zero. Continuing, ω=2*pi*f , so I Ξ 0.050*cos(2*pi*41.09*t) . Now dI/dt = -0.050*2*pi*41.09*sin(2*pi*41.09*t) = -12.91*sin(2*pi*41.09*t) . The first maximum value of the sine occurs at a quarter of its period. One quarter of its period is (1/41.09)*0.25 = .00608 . The sine's value at that point is 1, so the voltage is L*(-12.91) = -12.91 volts at 6.08 msecs. These values compare favorably with what I read from the plot.

The important lesson from this exercise if that the maximum coil voltage is not determined by the energizing voltage, but instead by how fast the current through the coil changes.

This circuit is represent by a second order differential equation, because it has two storage elements. You can find lots of material on RLC transients, both with driving sources, or free oscillating like this circuit is. There are even analyses that use a minimum or no differential mathematics, and instead rely on stock formulas.

Ratch

 

[MrAl]

Hello there,


L*di/dt is the voltage across the inductor:
v=L*di/dt
and that basically states that:
The voltage across the inductor is equal to the time rate of change of current (di/dt) through the inductor times the value of the inductance (L) of the inductor. That's not really enough information to analyze the circuit though because there is also a capacitor in the circuit and we have to know what that is doing too.

The best approach is probably to use a general circuit analysis procedure to solve for the peak voltage across the inductor. Using a general method means we would be able to solve this circuit as well as any other one.

It also helps to know a little about how the parts work in general too. For example, if the switch is closed for a very long period of time the inductor becomes a short circuit and so that absorbs all of the energy from the cap through the 10 ohm resistor (that got there when the switch was closed for the first time) and also all the energy from the battery. That means after the switch is closed and a very long time after that the capacitor voltage is zero and the inductor current is dependent only on the battery 5v and series resistor of 100 ohms. Since 5/100=0.050 that means for this circuit condition the inductor has 50ma flowing through it and it is a constant current so di/dt=0 so L*di/dt=0 so the voltage across the inductor is clearly zero volts. The circuit at this point basically looks like an inductor in series with a resistor in series with a battery.
Next, the switch is opened up. This means the next circuit looks like an inductor with initial current equal to 50ma in series with a resistor in series with a capacitor. Since the initial voltage across the capacitor is 0v and the initial current through the inductor is 0.050, that puts 50ma through the 10 ohm resistor. Now 50ma through 10 ohms is 0.050*10=0.5v, so that is the voltage across the inductor at the instant 0+ the switch is first opened. The circuit at that very instant 0+ is just a 10 ohm resistor in series with a current source of 50ma (the capacitor is a short circuit at 0+ too).

So you see we solved part of the problem already just knowing a little bit about how the inductor and capacitor behave during the first instant of operation and after very long time periods. To proceed however we would have to use a little general circuit analysis im not sure if you want to learn or not, but with two energy storage elements it's not super difficult.
We can get around using calculus and just use algebra by knowing some circuit analysis and using a table of Inverse Laplace Transforms and possibly some techniques. In this way we first do some algebra on the circuit in the frequency domain and then transform the algebra into the time domain using Inverse Laplace Transforms.

Here is a solution to your circuit after the switch is opened after it had been closed for a very long time and we ground the connection between the capacitor and inductor so the inductor voltage looks negative:

vL(t)=-e^(-a*t)*(A*sin(t*w)+B*cos(t*w))
where
a=R/(2*L)
w=sqrt(4*C*L-C^2*R^2)/(2*C*L)
A=(2*L-C*R^2)/(20*sqrt(4*C*L-C^2*R^2))
B=R/20

Also note that f0=w/(2*pi)=41.085923826

and also:
vL=-12.537 approximately at t=0.00586 seconds.

If we just want the voltage across the inductor non polarized then just make the above negative voltage positive.


I almost forgot, if you would like to try using numerical approximation techniques take a look here for a simple example of a capacitor being charged through a resistor with a voltage source:
https://www.electro-tech-online.com...ff-when-learning-capacitor.113051/#post925171

 

[Zirad]

Thank you all very much, I really appreciate this information. I have a lot to digest and in school we are only taught the basics, but this information really helps the understanding of these crutial components. Every learning to be engineer should have these concepts down solidly.
I will return and report once I go through this information.
Thanks again
Steve-

 

[MrAl]

Hello again,

Ok great. BTW, your original question was about finding the peak value of the inductor voltage, and since that is a damped sinusoidal with actual sine and cosine components we cant just look at the first half cycle of a sine at the angular frequency of w. We have to actually look for minima and/or maxima in the function for vL(t). Doing this we get the true maximum value of the inductor voltage:
12.537v at time t=0.00586 seconds approximately.

Expressed as a forumla, the solution for t where the maximum peak occurs is:
t1=(1/w)*taninv(4*w*L*((C*R^2-L)/(2*C*R^3-6*L*R)))
(where w is defined in my previous post and again R is the 10 ohm resistor)

and so the max peak is:
VpMax=vL(t1)
(where vL is defined in my previous post)

Of course we could also find this numerically which is sometimes a lot easier then trying to determine the solution in the form of a formula. There are many many humps and dips in the response however so we have to be careful to get the biggest one.

Using the formula above or doing it numerically, you could make sure you are at a peak by calculating vL(t) at t right at the supposed peak and at t just slightly to the left and just slightly to the right, where in these last two the value of vL should decrease slightly as we choose a time t that is just a little different than the peak time t. To make sure we have the actual highest one, we could graph the function vL(t) and examine it.

I might also add that in order to determine the response as the switch is opened and closed more than one or two times we'd have to do a little more work in finding vL(t) to include the initial values of both the capacitor and inductor.