The home page presentation provides insight into the operation of two ideal LC circuits: A series-resonant circuit (without a series resistor), and a parallel-resonant circuit (without a parallel resistor).  The author asserts that determining solution sets for ideal circuit configurations should precede attempts to analyze their parasitic-laden counterparts.  Gaining an intuitive understanding of ideal circuit operation ought to be a high priority.  Then, when parasitics are introduced, the practitioner will be better prepared to understand what is observed and, if necessary, synthesize what is needed for some particular application.

Circuit 1 (a series-resonant configuration)

Initial Conditions

1.  The charge on the capacitor is 0. Therefore its voltage, VC, is 0V
2.  No current is flowing in the loop, nor is the current changing.

The Circuit Elements

The battery is ideal (no resistance, no inductance, and the voltage is invariant with load)
The switch is ideal ( Ohms when open, 0 Ohms when closed, no inductance, no bounce)
The inductor is ideal (no resistance, no capacitance, and L is a fixed value)
The capacitor is ideal (no resistance, no inductance, and C is a fixed value)
The diode is ideal (no forward voltage drop, infinite reverse breakdown)
The wire is ideal (no resistance, no inductance)

The Event

At t=0 the switch closes, and remains closed

1.  Determine an expression for current in the loop.
2.  Determine an expression for voltage on the inductor.
3.  Determine an expression for voltage on the capacitor.

Observations

..No current will be flowing just after the switch is thrown because current cannot change instantaneously in an inductor.
•  The voltage will be 0V on the capacitor just after the switch is thrown because current has not begun flowing.
•  Therefore all the battery voltage will appear across L just after the switch is thrown.
•  The slope of VC just after the switch is thrown must be zero because i=C(dv/dt), and i=0.

Note

Readers familiar with switching regulators should recognize
Figure 1 as an idealized building block.  The diode is placed
to manage the current flowing in the inductor when the switch
opens (because current cannot change instantaneously in
an inductor).  In addition, there would exist some feedback
mechanism to control the switch in response to demands
placed on the load.  Highly efficient regulators would specify
particular diode types, or they may even replace the diode
with some synchronized switch circuitry.  The form this
synchronizing circuitry might take is not a subject of this work,
nor is the type of diode, nor the switch structure.

The Analytic Solution

If we write an equation for the voltage around the loop after the switch has closed, we get

…….VBatt – L(di/dt) – (1/C)idt = 0

Or, using an abbreviated notation, we have

…………VBatt – Li – (1/C)idt = 0

If we differentiate with respect to time, we see the following

………………..0 – Li – i/C = 0

Multiplying by -1, and dividing by L yields a classic 2nd-order differential equation

………………….i + i/(LC) = 0

The general solution is given by

……………………………i = Asinωt + Bcosωt

Where

……………………………ω = 1/SQRT(LC)

and where we note that

……………………………ω = 2πf

At time t=0 i is 0. Therefore, since sin0=0 and cos0=1, B must be 0, and we can write

……………………………i = Asinωt

We note that

…………………………..vL = Ldi/dt

Then

…………………………..vL = LAωcosωt

At t=0 VL equals VBatt.  The cosine of 0 is 1. So, we have

…………………………VBatt = LAω

Solving for A, we get

……………………………A = VBatt/(Lω)

Substituting 1/SQRT(LC) for ω permits A to be written even more precisely as

……………………………A = VBatt(SQRT(C/L))

which is the peak current. The complete expression for i is then given by

……………………………i = VBatt(SQRT(C/L))sinωt

The peak current occurs when sinωt is 1. This occurs many times (because the circuit is oscillating). The first time it occurs is when ωt equals
π/2. Solving for t, we have the following expression for the time of the first peak current

……………………………t = π
/2ω

Recalling that ω = 1/SQRT(LC), we can re-write the above equation, to give

……………………………t = π
(SQRT(LC))/2

which is the first occurrence of the peak current.

The voltage on the inductor is given by Ldi/dt.  Thus (with a modicum of algebraic manipulation), we arrive at

…………………………..vL = LVBatt(SQRT(C/L))ωcosωt

Then, substituting 1/SQRT(LC) for ω results in the following compact expression

…………………………..vL = VBattcosωt

Now, what about the peak voltage on the capacitor?  Well,

…………………………..vC = (1/C)idt

Doing the integration produces the following result:

…………………………..vC = VBatt(1/C)(SQRT(C/L))((-1/ω)cosωt + k)

Where k is a constant of integration. We have established that vC=0 at t=0.  Observing that cos0=1 leads us to the conclusion that k must be given by

……………………………k = 1/ω

Gathering and re-arranging terms allows us to write

…………………………..vC = VBatt(1/C)(SQRT(C/L))(1/ω)(1-cosωt)

But ω=1/SQRT(LC).  Substituting this in, and then grinding through the math, gives the following succinct relationship

…………………………..vC = VBatt(1-cosωt)

An easier way to get this solution would be to apply Kirkhoff’s voltage law on the loop, as follows:

…………………………..vC = VBatt – vL

Knowing that vL=VBattcosωt leads to an identical expression for vC (as it should).

cosωt ranges from +1 to -1.  Therefore vC ranges from 0 to 2VBatt (indicating that this circuit is a voltage doubler).  And when does it peak at 2VBatt?  That occurs whenever cosωt=-1, which occurs many times (because, again, this circuit is oscillating).  But the first time it occurs is when the following relationship is satisfied

…………………………..ωt = π

Solving for t, we get
……………………………t = π

which is twice the time it takes for the current to peak.  Thus

……………………………t = π
(SQRT(LC))

which is the first occurrence of peak voltage on the capacitor.

Solution Summary for the Case of Zero-volt and Zero-current Initial Conditions

…………………………iLoop = VBatt(SQRT(C/L))sinωt

…………………………..vL = VBattcosωt

…………………………..vC = VBatt(1-cosωt)

where

……………………………ω = 1/SQRT(LC)

Implications for Step-down Switching Regulators

Ordinarily the switch is opening & closing (regulating the output voltage to some level well below VBatt). But suppose (because of some circuit failure) it initially closes, and then does not re-open. The output will rise to twice the battery voltage. This could be catastrophic for the load, and must therefore be mitigated.

With the possible exception of circuit operation near Absolute Zero (0°K), there are no known zero-resistance circuits.  The switch (usually a MOSFET) and the inductor would be likely candidates for introducing enough series resistance to merit some investigation.  It should be apparent that adding a series resistance will introduce an additional term in the loop equation.  Interested readers are referred to Series RLC Circuitry for an analytical treatment of this condition.

The More General Case

Although closing the switch when vC=0, and when iLoop=0 is an important case (i.e., start-up for a switching regulator), two (likely simultaneous) initial conditions should also be considered.  Those initial conditions are:

1. A pre-existing loop current, I0 (flowing in L, C, and through the diode)
2. A pre-existing voltage on the capacitor, VC0

Recall that a regulated system has the switch opening & closing.  We now define t=0 to be the moment just after the nth closure of the switch.  We assume again that some anamolous condition removes the switch from its usual controlled state of operation.  That is, once closed it cannot re-open.  Then we look again at Figure 1.  We notice first that the diode is removed from consideration just as soon as it is forced to be in parallel with the battery.  Current in the inductor cannot change instantaneously; therefore I0 continues to flow in L & C, but is diverted entirely to the battery.  If we follow the original analysis it becomes apparent that we should now step in where, as before,

……………………………i = Asinωt + Bcosωt

If we set t equal to 0 it becomes apparent that B=I0 (because the sine of 0 is 0, and the cosine of 0 is 1)

We note (again) that

…………………………..vL = Ldi/dt
Then

…………………………..vL = LAωcosωt – Bωsinωt

Noting that vC(t=0) equals VC0 (the defined initial condition), we can write

………………………vL(t=0) = VBatt-VC0

And, because the cosine of 0 is 1, and the sine of 0 is 0, we can write

………………………VBatt-VC0 = LAω

A is now perfectly defined.
……………………………A = (VBatt-VC0)/Lω

Substituting 1/SQRT(LC) for ω permits A to be written even more precisely as

……………………………A = (VBatt-VC0)(SQRT(C/L))

So, we now have a complete solution for i using only the applied voltage, circuit values, and the given initial conditions.

……………………………i = (VBatt-VC0)(SQRT(C/L))sinωt + I0cosωt

As before, the voltage on the inductor is given by Ldi/dt. So, we can write

…………………………..vL = L((VBatt-VC0)(SQRT(C/L))ωcosωt – I0ωsinωt)

Expanding ω to 1/SQRT(LC), we can simplify vL to the following form

…………………………..vL = (VBatt-VC0)cosωt – I0SQRT(L/C)sinωt

At this point we express vC as the battery voltage less the inductor voltage, as follows

…………………………..vC = VBatt – (VBatt-VC0)cosωt + I0SQRT(L/C)sinωt

Rearranging gives
…………………………..vC = VBatt(1-cosωt) + VC0cosωt + I0SQRT(L/C)sinωt

If (as a sanity check) we set VC0 to 0 and I0 to 0 we get precisely the same expression for vC that derives from the start-up case (as we should). So, we now have a general solution summary for all cases.

Solution Summary for the General Case of Non-zero Initial Conditions

…………………………iLoop = (VBatt-VC0)(SQRT(C/L))sinωt + I0cosωt

…………………………..vL = (VBatt-VC0)cosωt – I0SQRT(L/C)sinωt

…………………………..vC = VBatt(1-cosωt) + VC0cosωt + I0SQRT(L/C)sinωt

where
……………………………ω = 1/SQRT(LC)

Circuit 2 (a parallel-resonant configuration)

Three ideal circuit elements in parallel:

• A current source (iSource = 0 for all t < 0, iSource = I for all t > 0)
• An inductor, L
• A capacitor, C

Initial Conditions

The charge on the capacitor is 0. Therefore its voltage, VC, is 0V
No current is flowing anywhere in the network, nor is any current changing

The Circuit Elements

The current source is ideal (infinite impedance, no capacitance, and the current is invariant with voltage)
The inductor is ideal (no resistance, no capacitance, and L is a fixed value)
The capacitor is ideal (no resistance, no inductance, and C is a fixed value)
The wire is ideal (no resistance)

The Event

At t=0 the current source switches from 0A to I.

Determine an expression for the voltage across the network.
Determine an expression for current in the capacitor.
Determine an expression for current in the inductor.

Observations

The voltage will be 0V just after the current steps to I because voltage cannot change instantaneously on a capacitor.
iL = 0 just after the current steps to I because current cannot change instantaneously in an inductor.
iC = I just after the current steps to I because no current is flowing in the inductor.
Even though the magnitude of VC at t=0 is 0V, the slope of Vat t=0 must be >0 because iC=C(dv/dt), and i is I.

The Analytic Solution

Kirkhoff’s current law must be satisfied (the sum of the currents into any node must be zero). Thus,

…………………..I – iL – iC = 0

Recognizing that iC=Cdv/dt, and then re-writing, we get

……………….I – iL – Cdv/dt = 0

Differentiating with respect to time, we have

……………..0 – diL/dt – Cv = 0

But LdiL/dt is the voltage on the inductor, which is the same as the voltage on the capacitor, which is v. So, we can write

………………..0 – v/L – Cv = 0

Multiplying through by -1, dividing by C, and then rearranging, gives

………………….v + v/(LC) = 0

In form, this is identical to the switched-voltage case.  Therefore, the solutions for v, iC, and iL will be sinusoidal, and the analysis proceeds in a manner similar to the series-resonant case.  Without going through it step-by-step, the solution summary is given below.

Solution Summary for the Case of Zero-volt and Zero-current Initial Conditions

……………vCurrentSource = vL = vC = I(SQRT(L/C))sinωt

…………………………..iC = Icosωt

…………………………..iL = I(1-cosωt)

where

……………………………ω = 1/SQRT(LC)

Implications

The current peak will be 2I in the inductor.  This circuit may therefore be referred to as a current doubler.  For that to predictably occur the current source must not saturate (if structured from a bipolar device) or go into the linear region of operation (if structured from a MOSFET).  Therefore the peak voltage must be controlled by managing I, L, and C.

With the possible exception of circuit operation near Absolute Zero (0°K), there are no known zero-resistance circuits. It should be apparent that adding a parallel resistance will introduce an additional term in the Kirkhoff equation.  Interested readers are referred to Parallel RLC Circuitryfor an analytical treatment of this condition.

What if the Input Source Does Not Differentiate to 0?

It may be that the input source is not a simple voltage or current step (each of which nicely differentiates to 0 for t>0).  If that is the case, an appeal to one of the References will lead to the concept of the “particular” integral (solution).  Or, applying an online search engine to one of the following strings may lead an investigator to a circuit with the input source of interest.

• LC Circuits
• RLC Circuits