The figure shows a straight-line approximation for the forward characteristics of a power diode. A continuous on-state current of 15 A is flowing through the diode.
What is the power loss in the diode?
Show Hint
Fit the straight line as V0 plus I times Ron using the two marked points on the graph, then multiply the voltage drop at 15 A by the current.
Step 1: Turn the graph into an equation.
The line starts conducting at $V_0=1.0$ V and reaches $50$ A at $1.8$ V. The rise over run of the line gives the conductance:
\[ \frac{1}{R_{on}}=\frac{50}{1.8-1.0}=62.5\text{ A/V} \implies R_{on}=0.016\ \Omega \]
Step 2: Model the diode as a battery plus a resistor.
A piecewise linear diode behaves like a fixed voltage source $V_0$ in series with $R_{on}$, so at any forward current $I$:
\[ V_D=V_0+IR_{on} \]
Step 3: Plug in the actual operating current.
\[ V_D=1.0+15(0.016)=1.24\text{ V} \]
Step 4: Multiply to get dissipated power.
\[ P=V_DI=1.24\times15=18.6\text{ W} \]
\[ \boxed{18.6\text{ W}} \]
