Fanout Staging Scan - Electronic Circuits IV: Digital Logic Signals and Switch Networks Step by Step

A buffered logic handoff passes only when source and group currents all fit the same output limit. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Direct fanout multiplies one input current

Every input load draws the same model current, so direct fanout is checked by multiplying load count by input current.

I_{\text{total}}=N\cdot I_{\text{in}}

Three direct rows show below, equal, and over limit

The input current and output limit stay fixed. The load count alone moves the total current from pass, to equality, to fail.

\begin{array}{c|c|c|c|c}N&I_{\text{in}}&I_{\text{total}}&I_{\text{limit}}&\text{state}\\3&1\ \text{A}&3\ \text{A}&5\ \text{A}&\text{PASS}\\5&1\ \text{A}&5\ \text{A}&5\ \text{A}&\text{PASS}\\6&1\ \text{A}&6\ \text{A}&5\ \text{A}&\text{FAIL}\\\end{array}

Buffered rows split the same load count

A buffer tree has two checks: the source drives buffer inputs, and each buffer drives its own group. Both checks must pass.

\begin{array}{c|c|c|c|c|c}N&\text{groups}&I_{\text{first}}&I_{\text{group}}&I_{\text{limit}}&\text{state}\\6&3+3&2\ \text{A}&3\ \text{A}&5\ \text{A}&\text{PASS}\\8&4+4&2\ \text{A}&4\ \text{A}&5\ \text{A}&\text{PASS}\\8&6+2&2\ \text{A}&6\ \text{A}&5\ \text{A}&\text{FAIL}\\\end{array}

A buffer tree can still fail if one group is too large

The failed buffered row has the same total load count as the passing buffered row, but its first group exceeds the limit. Splitting is useful only when every stage passes.

6\ \text{A}>5\ \text{A}\Rightarrow\text{FAIL}