# Confusion over voltage & current waveforms of a capacitor in a buck converter

#### ANewYorker

Joined Feb 12, 2024
7
I am learning about the reasons behind the shapes of the capacitor's current and voltage waveforms.

What confuses me about the capacitor's current waveform is how capacitor knows when to start discharging to help maintain a constant output current. In comparison, inductor's current waveform in much easier to understand; it simply charges when the switching is on and discharges when the switch is off.

What confuses me about the capacitor's voltage waveform is that it seems to have no correlation with its current waveform. With inductor, I know v=L(di/dt), so voltage is the slope of the current waveforms. But with capacitor, when I zoom in, current does seems to be the slope of the voltage waveform like i=C(dv/dt) suggests.

#### ericgibbs

Joined Jan 29, 2010
18,997
Hi,
Please post the LTSpice asc circuit file.
E

#### ericgibbs

Joined Jan 29, 2010
18,997
Hi ANY,
Study this video of the Buck converter by Ben-Yaakov.

E

#### Ian0

Joined Aug 7, 2020
10,035
I am learning about the reasons behind the shapes of the capacitor's current and voltage waveforms.

What confuses me about the capacitor's current waveform is how capacitor knows when to start discharging to help maintain a constant output current.
There‘s nothing confusing about it, when you look at it from the right viewpoint.
The current in the load will be the voltage on the capacitor divided by the load resistance - simply Ohm’s law.
If that current is greater than the current being supplied by the inductor, the rest will come from the capacitor.

if (inductor current > load current) then (inductor current -load current) flows into the capacitor and the capacitor voltage increases.
if (load current > inductor current) then (load current-inductor current) flows out of the capacitor and the capacitor voltage decreases.

#### ANewYorker

Joined Feb 12, 2024
7
There‘s nothing confusing about it, when you look at it from the right viewpoint.
The current in the load will be the voltage on the capacitor divided by the load resistance - simply Ohm’s law.
If that current is greater than the current being supplied by the inductor, the rest will come from the capacitor.

if (inductor current > load current) then (inductor current -load current) flows into the capacitor and the capacitor voltage increases.
if (load current > inductor current) then (load current-inductor current) flows out of the capacitor and the capacitor voltage decreases.
OK. That makes sense. How do you explain the relationship between the output voltage and capacitor current? Is it governed by i=C(dv/dt)?

#### Ian0

Joined Aug 7, 2020
10,035
OK. That makes sense. How do you explain the relationship between the output voltage and capacitor current? Is it governed by i=C(dv/dt)?
Precisely.
But maybe more intuitive to think of V=(∫Idt)/C.

#### MrAl

Joined Jun 17, 2014
11,566
OK. That makes sense. How do you explain the relationship between the output voltage and capacitor current? Is it governed by i=C(dv/dt)?
Hi,

You should not limit yourself to looking at the individual currents like i=C*dv/dt or v=L*di/dt, what you need to do is look at them both together as a system because that it what it really is.

For the buck, we have the system of ODE's as:
C*dv/dt=i-v/R
L*di/dt=E-v

where
'i' is the inductor current (not the cap current),
'v' is the capacitor voltage (not the inductor voltage),
R is the load resistor value, and
E is the input voltage assumed to be constant.

In the above, you can see that the cap current is not 'i' it is i-v/R, and you can see that
the inductor voltage is E-v with v the cap voltage.
So this is a system, not individual components acting independently of each other.
If you solve that system you will get the actual inductor current and capacitor voltage, and the capacitor voltage is the same as the output voltage in the simple case.

It gets a little more involved as we get into a circuit that is closer to reality. For example, we have capacitor and inductor ESR which would be added to those two equations before solving.

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