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Current Flow in the OFF Time of a Switching Power Supply

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I am having a difficult time picturing the flow of current and the role of the capacitor during the OFF time of a buck converter. Here is my understanding so far. During the ON time the magnetic field of the inductor grows and positive charge is being stored on the top plate of the capacitor as well as current being delivered to the load. The load sees a somewhat smooth but ramping voltage because the capacitor is resisting a change in voltage by sucking some current to charge up.

When the switch is OFF, the polarity reverses, and the collapsing magnetic field continues to supply current to the load. Here is where I can't visualize how current is flowing through the flywheel loop - I picture the charged capacitor with the positive charge on the top plate and the negative charge on the bottom plate like the picture below:

Capacitor Charge

Now how does the positive charge supplied by the inductor flow through the capacitor, through the diode, and back to the negative side of the inductor when there is a dielectric blocking it? Additionally, how is the current supplied to the load during this OFF period? Does the capacitor supply current to the load in tandem with the collapsing magnetic field, or is it either the inductor or the capacitor that is supplying current to the load? If the voltage is decreasing during this OFF time based on that classic zig-zag output shape, then it seems to me that at least the capacitor is definitely supplying current to the load.

Why should this post be closed?

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It's not clear whether you are asking about what the capacitor does in a buck converter, or how a capacitor works internally. I'll therefore go with:

I am having a difficult time picturing the flow of current and the role of the capacitor during the OFF time of a buck converter.

At this level, the internal details of the capacitor are irrelevant. It is simply a component which stores charge proportional to the voltage on it. The capacitance is the proportionality constant.

Your confusion seems to be about the basic operation of a buck converter. To talk about that, we need to define it. Here is a basic buck converter:

Image

Vout is always a lower voltage than Vin. Let's assume things are up and running in steady operation with a fixed load on Vout. There will be some ripple on Vout at the switching frequency, but that ripple is a small fraction of Vout by design. Even with a resistive load (current proportional to voltage), the Vout load current is essentially constant.

Let's also assume ideal parts for now. The inductor has no series resistance, the diode no forward voltage drop, etc.

The switch is being closed and opened regularly. Let's start the analysis just as the switch closes. At this point, the voltage Vin-Vout is applied across the inductor. By definition of what an inductor does, the current thru it ramps up linearly.

When the switch opens, the same current continues to flow thru the inductor immediately after than was flowing immediately before. This current now flows thru D1. Since D1 is an ideal diode, it has no voltage across it. The top end of the inductor is therefore now held at ground. The net effect is that now -Vout voltage is applied to the inductor. This causes the current thru the inductor to ramp down.

There are two possibilities what happens next, depending on whether the inductor current drops all the way to 0 before the switch closes again or not. If it does, then the converter is running in discontinuous mode, else it is in continuous mode.

Continuous mode

For now, let's assume continuous mode, meaning the inductor current is still flowing when the switch closes again. "Continuous" refers to the inductor current being continuous (never stops).

Now we are back to where we started with the switch closing. The current thru the inductor is a triangle wave. It increases linearly when the switch is closed, and decreases linearly when the switch is open.

Now let's look at the capacitor current. Since this analysis has been of steady operation, the average inductor current, which is half way between the peaks, must be equal to the load current. The capacitor current is the inductor current minus the load current. The load current is essentially constant, so the capacitor current is the same triangle as the inductor current, shifted so that the average is 0.

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When the capacitor current is above 0 (the dashed line), the capacitor is being charged up. The current coming out of the inductor exceeds the load current. The capacitor makes up the difference by absorbing the excess. During this time, the voltage on the capacitor increases.

When the capacitor current is below 0, the capacitor is being discharged. The current coming out of the inductor is less than the load current. The capacitor makes up the difference by sourcing the remainder. During this time, the voltage on the capacitor decreases.

The voltage on the capacitor is the integral of the current, by definition of what a capacitor does. Since the current is in linear segments, the voltage is in quadratic segments. The peaks of the voltage waveform are at the current zero crossings.

Discontinuous mode

In discontinuous mode, the inductor current has three phases: Ramp up, ramp down, and zero. However, the basic principle of what the capacitor does is still the same. The average inductor current still matches the load current. The capacitor makes up the instantaneous difference, sourcing current when the inductor current is below the load, and absorbing the excess when the inductor current is above the load.

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Thank you, this was very clear. ‭jm567‭ about 2 months ago

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