Why inductor don't discharge completely even when the duty cycle is 50%?

Question

1] My simple buck convertor circuit description

I made this buck convertor on my breadboard- Falstad simulation

• The battery voltage is 12 V
• And the MOSFET is a N-channel enhancement mode MOSFET
• The PWM generator's frequency is 50 kHz and the duty cycle is 50%
• [EDIT] And the PWM generator output voltage is 30 V.
• The inductor is a toroidal iron core inductor of inductance 990 µH
• The resistor is of value 100 Ω
• (note: Actually I am using ATmega328p as a PWM source and hence the actual frequency is 62.5 kHz, But here for convenience I added 50 kHz frequency so the graph (scope) looks good. And not only that I removed the capacitor intentionally to understand inductor behavior properly.)
• Additional calculated values-
  • ON-time = 10 µs
  • OFF-time = 10 µs
  • Time constant τ = 9.9 µs

2] My question

My question is not about the above buck convertor circuit but it is about the behavior of the inductor in it.

As we can see here in this scope that inductor is dropping voltage in this particular way-

The frequency is 50 kHz so the one cycle period is 20 µs each horizontal block in the above image is of 10 µs because the duty cycle is 50%. So the one 10 µs block shows the ON-time and the other one 10 µs block shows the OFF-time of the PWM wave.

(ON-time) The first 10 µs block is understandable Inductor is in charging state so after 10 µs the inductor would be charged up to 63.2% because the value of tau(τ) is approximately 10 µs in our case (9.9 µs in reality). And the current will also be 63.2% of the final current value. And by the value of current after 10 µs we can calculate the voltage drop of the inductor also which will be -

\begin{align} Current \space after \space 10\mu \space seconds &= 0.12(1-e^{\frac{-t}{\tau}}) \\\\ &= 0.12(0.632) \\\\ &= 0.07584 \\\\ Voltage \space drop \space across \space the \space resistor &= (0.07584)(100) \\\\ &= 7.5 V \\\\ Voltage \space drop \space across \space the \space inductor &= 12 - 7.5 \\\\ &= 4.416 V \end{align}

So we get the value 4.416 V for the voltage drop across the inductor after 10 µs, it is not the case in reality! If we look at the scope then it says- voltage drop across the inductor after 10 µs is 3.5 V. I guess it is 3.5 volts instead of 4.416 V because the inductor is not discharging completely in the OFF-time of the PWM input wave.

(OFF-time) The next 10 µs block is confusing- This is my question- Why the inductor is not discharging completely? In simple words here is what should happen - inductor gets charged up 63.2% for 10 µs then Inductor should get discharged completely in next 10 µs because in the first 10 µs it was charged only for 10 µs / 63.2%.

What intuitive understanding I am lacking about inductors? I want to understand inductor behavior intuitively so little math and simple explanations would be great.

Answer 1

What you see is a typical steady-state transient response of the LR circuit.

As you have noticed at the beginning we first apply 12V across the circuit. And after \$10\mu s\$ the voltage across the inductor will be 36.8% of input voltage. This means that we have 63.2% across the resistor.

Thus, you will have \$V_L(1) = 4.41V\$ and \$V_R = 7.58V\$ and the current \$75.8mA\$.

But now we have a OFF state and the discharge phase begins. But the current through the inductor wants to flow in the same direction as before. This means that inductor voltage must now be equal to \$-7.58V\$ (Vin = 0V)
And again after \$10\mu s\$ the induced will drop to 36.8% of the previous value ( -7.58V) \$V_L = -2.8V\$ and the voltage across the resistor will be \$2.8V\$.

But now once again we have an ON state and we apply 12V. So the VL will jump to \$-2.8V + 12V = 9.2V\$ to keep the current value unchanged.

And again we have the same situation as before, with the new initial values (VL = 9.2V and VR = 2.8V).
So, after the end of a charging phase (another 10us), the induced voltage will drop by 36.8% to \$V_L = 3.38V\$ and resistor voltage to \$V_R = 12V - 3.38V = 8.62V\$ and the current to \$86.2mA\$. And we have another off time. And the induced voltage must now be equal to \$ 3.38V - 12V = -8.62V\$ to keep the current value unchanged.
And at the end of a off state, we will have \$V_L = -8.62V * 0.368 = -3.1V\$ and \$V_R = 3.1V\$, and the new ON state begins. And the \$V_L\$ jumps to new value \$V_L = -3.1V + 12V = 8.9V\$ and charging phase begines. At the end of a charging phase (after another 10u) we will have: \$V_L = 8.9V * 0.368 = 3.27V\$ - \$V_R = 12V - 3.27V = 8.7V\$

And that concludes my chaotic attempt at describing what’s happening in this circuit.

simulation

How can there be a current without a voltage?

Answer 2

I only see a voltage curve, and that one gets messed up by the potential change when the transistor opens and closes.

I'm sure you can set up the simulation tools to plot the current curve flowing through the inductor - that one should make a lot more sense.

Answer 3

I'll answer your question with another question:

How could the current be discontinuous?

Consider an inductor charged to some current, then discharging into an ideal diode and resistor. The current decreases exponentially over time. The current never goes to zero, it merely gets arbitrarily close. So CCM is the default case here.

Real diodes have an excess voltage drop, which can be read in two ways: either current can go discontinuous because that voltage drop can discharge the inductor through zero current (of course it stops at ~zero being a diode), or alternately: the diode is a nonlinear two-terminal element, a non-ohmic resistor, which starts as a low impedance while current is high, but as current runs out, the impedance rises, and the exponential slope steepens -- current discharges much more quickly towards zero, though still never precisely to zero. (Actually, in a real circuit, there may be ringing due to node capacitance and other effects. We might well say that current becomes discontinuous at the point where ringing becomes relevant, or other thresholds.)

You may find this analysis helpful: What happens when you insert an inductor between Vcc and a MOSFET's drain? it's not about this circuit exactly, but very similar dynamics apply, and the same method can be adapted to derive the behavior of the circuit.