Why should in a buck converter the average voltage across the inductor be equal to zero to have a DC current?
Could someone help me understand this in a simple way by mathematical equation?
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\$\begingroup\$ KCL at the output node: iL(t) = iC(t) + ILood, now simply take the average of both side and remember that average current flowing via the capacitor in a period is zero so the average inductor current equals the load current. \$\endgroup\$– emnhaCommented Apr 24, 2023 at 4:37
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3 Answers
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The integral form of defining relation for inductance is: $$i_{L}(t)=\frac{1}{L}\int_{0}^{t}v_L(t)dt+i_{L}(0^{-})$$
\$i_{L}(0^{-})\$ is the dc term. In a buck converter the integral term must be zero to reveal the dc current.
The integral can be expanded over the period T as two piecewise linear terms.
$$i_{L}(t)=\frac{1}{L}\int_{0}^{t_{1}}v_L(t_{a})dt_{a}+\frac{1}{L}\int_{t_{1}}^{T}v_L(t_{b})dt_{b}+i_{L}(0^{-})$$
The two integral terms must sum to zero.
For this to happen, the product of the voltage and the associated time interval must be equal the same in each time interval ta, and tb. This is called the volt-second balance.
$$v_a(t_1-0)=v_b(T-t_1)$$
For a buck converter \$v_a=V_{in}-V_{OUT}\$ and \$v_b=-V_{OUT}\$ for an ideal diode. So:
$$(V_{in}-V_{OUT})(t_1-0)=V_{OUT}(T-t_1)$$
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An inductor integrates the voltage across it according to: $$V=L*di/dt$$ or $$I=1/L*\int V(t)dt$$ So you can see that if there's any average voltage across the inductor the current will increase or decrease depending on the direction of the voltage.
Since the output current equals the average inductor current in a buck converter, (from here)
if there's a DC steady state load on the output the inductor current can't increase or decrease (on average) so the average voltage must be zero.
Of course during switching the inductor current ramps up and down in accordance with the duty cycle, and the voltage changes from approximately Vin-Vout to -Vout, causing a ripple current in the inductor. It's only the average voltage that works out to zero in steady-state.
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Mathematically, why in a Buck converter the average voltage across the inductor must be equal to Zero to have a dc current
It has nothing specifically to do with buck converters; any inductor in a circuit MUST have an average voltage of zero across it's terminals else this happens: -
- For an ideal inductor the current will become infinite
- For a non-ideal inductor, the core might saturate
- Fuses/breakers may blow
This has a natural equivalence to what is not permitted for a real capacitor circuit; for a capacitor, the average current through it must be zero. If this didn't happen, infinite voltage would eventually be attained and, we know that this can't happen.
