Determining the Voltage, Current, and Power at the MPPT Point

The $i_p$ versus $v_p$ characteristic of a PV cell can be described by two segments:
$i_{p1 }= -0.01v_{p1 }+ 1$

$i_{p2} = -3.5v_{p2 }+ 2.8$
Calculate (a) the voltage $V_{mp}$ , (b) the current $I_{mp}$ , and (c) the power $P_{max}$ .

Question

Determining the Voltage, Current, and Power at the MPPT Point

The [latex]i_p[/latex] versus [latex]v_p[/latex] characteristic of a PV cell can be described by two segments:

[latex] i_{p1 }= -0.01v_{p1 }+ 1[/latex]

[latex] i_{p2} = -3.5v_{p2 }+ 2.8[/latex]

Calculate (a) the voltage [latex]V_{mp}[/latex], (b) the current [latex]I_{mp}[/latex], and (c) the power [latex]P_{max}[/latex].

Accepted Answer

These segments represent the equation of a straight line of the form [latex]y = mx + C[/latex]. The constants are [latex]m_1 = -0.1, C_1 = 1, m_2 = -3.5[/latex], and [latex]C_2 = 2.8[/latex] at the intersections [latex]i_{p1} = i_{p2} = I_P[/latex] and [latex]v_{p1} = v_{p2} = V_P[/latex]. That is,

[latex] I_P = -0.1 V_p + 1[/latex]

[latex] I_P = -3.5V_p + 2.8[/latex]

Subtracting one equation from the other gives the voltage [latex]V_p[/latex]

[latex]V_P=\frac{C_2 - C_1}{m_1 - m_2}=\frac{2.8 - 1}{-0.1 - (-3.5)}=0.529 V[/latex]

Substituting [latex]V_P[/latex] in one of these equations gives the current [latex]I_P[/latex]

[latex]I_P = -0.1 V_P + 1 = -0.1 × 0.529 + 1 = 0.947 A[/latex]

The output power is

[latex]P_o = V_PI_P = 0.529 × 0.947 = 0.501 W[/latex]

Question 16.4: Determining the Voltage, Current, and Power at the MPPT Poin...

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