Question 23.5: Design a nominally 2.500 V voltage reference (at room temper......
Design a nominally 2.500 V voltage reference (at room temperature of 27 °C or 300 °K) PTAT voltage reference. Use the topology seen in Fig. 23.25 with a bias current of 1 μA. Simulate the operation of the design.
Let’s set K (the number of diodes connected in parallel) to 8. We do this since it is easy to remember In 8 = 2. Solving for R using Eq. (23.31) gives
R={\frac{n V_{T}\cdot\ln K}{I}}\;\mathrm{{or}}\;I={\frac{n k\cdot\ln K}{q R}}\cdot T (23.31)
R={\frac{1\cdot0.026\cdot2}{10^{-6}}}=52kThe voltage drop across the resistor is 1 μA · 52k = 52 mV. Using Eq. (23.32), we can solve for L as 48 (so L · R = 2.5 MΩ). The simulation results are seen in Fig. 23.26. At room temperature, the reference voltage is roughly 2.5 V. Note how, as temperature increases, the reference voltage (PTAT) increases. Using Eq. (23.32), we can estimate the reference voltage’s change with temperature as
V_{R E F}=I\cdot L\cdot R=\frac{n k\cdot L\cdot\ln K}{q}\cdot T (23.32)
\frac{\partial V_{R E F}}{\partial T}=\frac{n k\cdot L\cdot\ln K}{q} (23.33)
Using the numbers from this example with k=1.38\times10^{-23}{J}/K~~\mathrm{and}~~q~=~1.6\times 10^{-19}C gives
{\frac{\partial V_{R E F}}{\partial T}}={\frac{1\cdot1.38\cdot10^{-23}\cdot48\cdot2}{1.6\times10^{-19}}}=8.26~m{{V}}/CFor every 25 °C increase in temperature, we expect V_{R E F} to go up by 206 mV.
