Electrical – Why does parasitic capacitance influence high frequency Dc-Dc switching

dc/dc converterfrequencyimpedance

From my understanding, Dc-Dc switching means turning something between 0 and some Dc voltage value. In doing that fast enough we obtain an equivalent Dc voltage value (pwm principle).

When talking about capacitor impedance, the relations is:

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In example in the image below, if we imagine that the voltage source is switching DC power source, instead of a sinus, that gives a signal as one on the image below:

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In that scenario, a current would never flow.

Doesn't that imply that the impedance of the capacitor is always infinite regardless of the frequency of switching?

If that is true, what does the suggestions not to do Buck-Boost circuits on breadboards?

Best Answer

In that scenario, a current would never flow.

Of course current flows through a capacitor each time the pulse rises and falls. For a capacitor Q=CV and differentiating on both sides to obtain current we find that: -

I = C dv/dt because dq/dt = current.

In other words when the voltage edge is sharp (high dV/dt) the current can be very large if not limited by some series resistance or an inductor.

It is for this reason that the current into a capacitor from a sine wave voltage source is shifted by 90 degrees into a cosine wave - it's all down to the maths behind differentiation of voltage (dv/dt).

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