Analysis of a diode L-C circuit

[samiran_dam]

Hi All,

I am trying to understand the behavior of the following circuit. The simulated waveform is also attached alongside.



I am not clear about why the voltage at node vx (red line) suddenly jumps to vout (green line) from vin (black line) when the inductor current (blue line) is crossing the zero.

Is it because the current through the inductor is just about to change the direction, the inductor opposes that change and enforces the voltage at vx to go up to vout?

 

[pancho_hideboo]

I am not clear about why the voltage at node vx (red line) suddenly jumps to vout (green line) from vin (black line)
when the inductor current (blue line) is crossing the zero.

It is due to self induction effect of L.

Is it because the current through the inductor is just about to change the direction

Wrong.
Current keeps same direction when current is zero.
L1 release current.

vx jumps at I(L1)=0.

Assume on/off switch instead of diode.
Switch is off, then vx will jump up.

 

[albbg]

When V(in) rise, the current starts flow through L, charging C. So the output voltage is increased since the current is integrated by C.
Remember that Vc=1/C*integral(I*dt).

When the capacitor is fully loaded, the current goes to zero, but we know that the voltage across an inductor is given by VL=L*dI/dt
thus when the first derivative of the current is zero (that is the current i constant and, in our case, this constant is zero), Vl will be zero that means
V(Out)=V(x)

With a switch you cannot have the same effect.

 

[KlausST]

Hi,

(albdg was faster)

at the beginning:
* voltage across capacitor is 0V
* current through L is 0mA
But there is voltage across the L. Left side positive.
--> this causes current
--> and this current charges C

with time...
The voltage across L decreases, because the voltage across the capacitor increases.
The current through L is the integral of the voltage across L divided by L: C_L = integral ((V_in - V_C)dt / L )
And the energy stored in L: W = 0.5 x I^2 x L
V_in = source, L = sink, C = sink

The energy increases until the voltage across the capacitor is the same as the input voltage....Then the voltage across L becomes inverted, but the current direction remains the same.
This means the energy previously stored in L decreases and gets shifted into the capacitor. The current decreases.
(for energy calcualtions: mind that during this phase the power supply still delivers energy, becuase the voltage remains the same and the current still goes the same direction)
V_in = source, L = source, C = sink
This goes on until the energy in L becomes zero and thus the current becomes zero.

Now with ideal devices you have a stable condition.
This is independent of values of C and L. In the end the voltage across C is always 2 x V_in.
The values don´t change waveform, but current magnitude, energy magnitude and timing.

Klaus

added:
(albdg was faster)

 

[pancho_hideboo]

With a switch you cannot have the same effect.

You are right.
During 1.4nsec~2.0nsec, dI(L1)/dt < 0.0 and vx-vout =L1*dI(L1)/dt < 0.0.

 

[samiran_dam]

Hi,

(albdg was faster)

at the beginning:
* voltage across capacitor is 0V
* current through L is 0mA
But there is voltage across the L. Left side positive.
--> this causes current
--> and this current charges C

with time...
The voltage across L decreases, because the voltage across the capacitor increases.
The current through L is the integral of the voltage across L divided by L: C_L = integral ((V_in - V_C)dt / L )
And the energy stored in L: W = 0.5 x I^2 x L
V_in = source, L = sink, C = sink

The energy increases until the voltage across the capacitor is the same as the input voltage....Then the voltage across L becomes inverted, but the current direction remains the same.
This means the energy previously stored in L decreases and gets shifted into the capacitor. The current decreases.
(for energy calcualtions: mind that during this phase the power supply still delivers energy, becuase the voltage remains the same and the current still goes the same direction)
V_in = source, L = source, C = sink
This goes on until the energy in L becomes zero and thus the current becomes zero.

Now with ideal devices you have a stable condition.
This is independent of values of C and L. In the end the voltage across C is always 2 x V_in.
The values don´t change waveform, but current magnitude, energy magnitude and timing.

Klaus

added:
(albdg was faster)

Perfect, thanks for the detailed explanation. However, I have still one doubt to be clarified.

Lets say there is no diode, then the Vout waveform keeps on oscillating (sustained oscillation when L and C are ideal) b/w 2Vin and 0 at frequency = 1/(2*pi*sqrt(LC)).

Now, bring the diode back into picture. When the capacitor is fully charged and the inductor is entirely devoid of any energy, then the cap should start charging the inductor back and there should be a reverse flow of current through the inductor. So, my thought was that similar to the situation when there is no diode, the Vout (with diode case) starts decreasing according to the cosine curve until it reaches vin and then it stays there for eternity.

My question was why the voltage at Vx jumps to Vout when the energy of the inductor is just exhausted and the current just becomes zero?

 

[KlausST]

Hi,

Now, bring the diode back into picture. When the capacitor is fully charged and the inductor is entirely devoid of any energy, then the cap should start charging the inductor back and there should be a reverse flow of current through the inductor

No, the diode prevents the current flow back.

My question was why the voltage at Vx jumps to Vout when the energy of the inductor is just exhausted and the current just becomes zero?

The energy in the L is the driving source now.
As long as there is energy in the L, then the voltage across the L is in a way that the capacitor becomes charged.
L_voltage: left side: V_IN because of current flow through the diode. right side: V_Capacitor = up to 2 x V_IN. The difference is up to V_IN

As soon as there is no energy in the L, the L becomes passive, but low impedance, though.
On the right side of L is the capacitor voltage. No current flow driven by the L. Therefore on the left side of L is also the capacitor voltage. The voltage across the diode makes the diode high impedance. Again: no current flow through diode. --> steady state

Klaus

 

[samiran_dam]

Hi,


No, the diode prevents the current flow back.


The energy in the L is the driving source now.
As long as there is energy in the L, then the voltage across the L is in a way that the capacitor becomes charged.
L_voltage: left side: V_IN because of current flow through the diode. right side: V_Capacitor = up to 2 x V_IN. The difference is up to V_IN

As soon as there is no energy in the L, the L becomes passive, but low impedance, though.
On the right side of L is the capacitor voltage. No current flow driven by the L. Therefore on the left side of L is also the capacitor voltage. The voltage across the diode makes the diode high impedance. Again: no current flow through diode. --> steady state

Klaus

Thanks....everything is clear now. It was my mistake. How could I miss the point that "the diode prevents the current flow back" ??