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Solved Examples · Example 1

Q.A silicon power diode has VjV_j (the drop across the p+ n- junction) of 0.4 V, RONR_{ON} (ohmic drop) in the drift region of 0.002 Ω\Omega and IFI_F = 75 A. Determine VAKV_{AK}.

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[!TLDR]

Adding the junction drop to the ohmic drift-region drop gives VAK=0.55V_{AK} = 0.55 V.

In a power diode the forward voltage drop has two parts. The first is the drop VjV_j across the p+n−p^+ n^- junction itself. The second is the ohmic drop VRDV_{RD} across the wide, lightly-doped n−n^- drift region that a power diode needs in order to block large reverse voltages; this ohmic term is VRD=RONIFV_{RD} = R_{ON} I_F, and it is what makes the forward V-I characteristic of a power diode more linear than that of an ordinary signal diode.

Given Vj=0.4V_j = 0.4 V, RON=0.002 ΩR_{ON} = 0.002\ \Omega and IF=75I_F = 75 A:

VAK=Vj+VRD=Vj+RONIFV_{AK} = V_j + V_{RD} = V_j + R_{ON} I_F

VAK=0.4+(0.002×75)=0.4+0.15=0.55 VV_{AK} = 0.4 + (0.002 \times 75) = 0.4 + 0.15 = 0.55\ \text{V}

[!ANSWER]

VAK=0.55V_{AK} = 0.55 V

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