Simple LC circuit implementation

Question

I'm implementing a very simple LC circuit where the power supply is connected to an inductor (of 4 micro Henry specifically), and the inductor is connected through a diode to a capacitor going to ground (of 1 micro Farad). I was expecting that the capacitor will accumulate twice the input voltage, however I am consistently only reading that it accumulated exactly the input voltage.

I included below a schematic of the situation and the explanation of my thinking. I added a dummy load resistor \$R1\$ but in the derivation I assume to simplify matters that \$R1 = \infty\$.

I would be grateful for any explanation of why this circuit doesn't seem to work "in the real world".

^{simulate this circuit – Schematic created using CircuitLab}

We now have by Kirchoff's current law, and the definition of capacitor and resistance, (note that \$V3 = V2\$ so \$V3 / R1 = V2 / R1\$), $$ I1 = \frac{d V2}{dt} \cdot C1 + \frac{V2}{R1} $$ Differentiating this we get, $$ L1 \frac{d I_1}{dt} = \frac{d^2 V2}{dt} \cdot L1 \cdot C1 + \frac{d V2}{dt} \cdot \frac{L1}{R1} $$ Now using the definition of inductor the left-hand side is equal to \$V1 = Vin - V2\$. Therefore collecting everything together we get $$ Vin - V2 = \frac{d^2 V2}{dt} \cdot L1 \cdot C1 + \frac{d V2}{dt} \cdot \frac{L1}{R1} $$ It's easier to solve it if we assume that \$R1 = \infty\$. In that case we get $$ V2 = Vin + c_0 \sin \Big ( \frac{t}{\sqrt{L1 C1}} \Big ) + c_1 \cos \Big ( \frac{t}{\sqrt{L1 C1}} \Big ) $$ Now specializing we see that \$c_1 = - Vin\$ and differenating we see that \$C1 \cdot d V2 / dt = I2\$ which is also zero at \$t = 0\$ hence \$c_0 = 0\$. We conclude that $$ V2 = Vin - Vin \cdot \cos \Big ( \frac{t}{\sqrt{L1 C1}} \Big ). $$ Since we imagine that \$R1 = \infty\$ and there is a diode, we notice that the circuit will stop when the current \$I2\$ turns negative. When this happens \$V2 = 2 Vin\$.

Answer 1

If you simulate using a near ideal diode, you can see that it is a voltage doubler: -

• L = 10 mH
• C = 1 uF

The input DC voltage is 20 volts and the voltage across the capacitor (blue trace) becomes nearly 40 volts. The red trace is the inductor voltage and, at a little over 0.3 ms, it's energy is depleted having given most of it to the capacitor.

If you don't activate the supply quickly compared to the LC cycle time then you'll barely see any voltage magnification on the capacitor. Here's what happens when the power supply activates over a 5 ms period: -

In other words, there is very minimal magnification.

With L at 1 uH and an instant application of the supply we get this: -

If the supply ramps up over 4 us it becomes this: -

Circuit used: -