Diode Operating Points

Looking at a diode datasheet, we see that there is a huge spread in forward voltage. Recall that:

• Small $\Delta V_D$ causes a big $\Delta I_D$
• Regular $\Delta I_D$ causes very small $\Delta V_D$

Circuit diagram:

Datasheet:

Problem - change in current

What if $I_D$ goes from $20\text{ mA}$ to $100\text{ mA}$?

Normally, we would just check the datasheet for $I_D$ vs $V_D$ curves.

In this case, we will it mathematically solve it:

$$\begin{array}{r}{I}_{D_1}=I_S{e}^{\frac{V_{D_1}}{V_T}}\\ I_{D_2}=I_S{e}^{\frac{V_{D_2}}{V_T}}\\ \frac{I_{D_1}}{I_{D_2}}=e^{(V_{D_1}-V_{D_2})}/V_T\end{array}$$

or:

$$V_{D_1}-V_{D_2}=V_T\cdot \ln \left(\frac{I_{D_1}}{I_{D_2}}\right)$$

Solving, we get $V_{D_1}=3.3\text{ V}$ and $V_{D_2}=3.168\text{ V}$. Basically, a 200x change in $I_D$ only resulted in a $4\%$ change in $V_D$. This is why mostly assume that $V_D$ is constant when we’re doing problems with a diode in a circuit.

Problem - picking R

Choose E24 R when $V_{DD}=5\text{ V}$ for nominal operating conditions. What is the max current given 5% tolerance on this resistor? We determine the worst-case power considering the min and max $V_F$ using the $R$ chosen above.**

We have:

$$R=\frac{V}{I}=\frac{1.7}{20\text{m}}=85\Omega$$

We choose 91 ohms for R (based on E24) as this limits current better.

Solving for maximum current based on 5% tolerance for a 91 ohm resistor:

$$\begin{array}{r}{R}_{min}=(0.95\times 91)=86.45\\ I_{max}=\frac{V}{R_{min}}=\frac{1.7}{86.45}=19.67\text{ mA}\end{array}$$

Keeping $R=91\Omega$, consider $V_D$:

$$\begin{array}{r}{P}_{min}=V_{DD}\times I_F=5\left(\frac{V_R}{R}\right)=5\left(\frac{5-3.80}{91}\right)=66\text{ mW}\\ P_{max}=V_{DD}\times I_F=5\left(\frac{5-2.80}{91}\right)=121\text{ mW}\end{array}$$

The 3.80 and 2.80 values are from the operating forward voltage of the diode from the datasheet above.