Let's set $K$ (the number of diodes connected in parallel) to 8 . We do this since it is easy to remember $\ln 8=2$. Solving for $R$ using Eq. gives

$$
R=\frac{1 \cdot 0.026 \cdot 2}{10^{-6}}=52 k
$$


The voltage drop across the resistor is $1 \mu A \cdot 52 k=52 m V$. Using Eq. , we can solve for $L$ as 48 (so $L \cdot R=2.5 M \Omega$ ). The simulation results are seen. At room temperature, the reference voltage is roughly 2.5 V . Note how, as temperature increases, the reference voltage (PTAT) increases. Using Eq. , we can estimate the reference voltage's change with temperature as

$$
\frac{\partial V_{R E F}}{\partial T}=\frac{n k \cdot L \cdot \ln K}{q}
$$

Using the numbers from this example with $k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$ and $q=1.6 \times 10^{-19} \mathrm{C}$ gives

$$
\frac{\partial V_{R E F}}{\partial T}=\frac{1 \cdot 1.38 \cdot 10^{-23} \cdot 48 \cdot 2}{1.6 \times 10^{-19}}=8.26 \mathrm{mV} / \mathrm{C}
$$


For every $25^{\circ} \mathrm{C}$ increase in temperature, we expect $V_{R E F}$ to go up by 206 mV .