An 8-V source is to be used in conjunction with two standard resistors to design a voltage divider that will output 5 V when connected to a 100-μA load. While keeping the consumed power as low as possible, we wish to minimize the error between the actual output and the required 5 volts.

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Accepted Answer

The divider can be modeled as shown in Fig. 3.32. Applying KCL at the output node yields the equation

[latex]\frac{V_{S} - V_{o}}{R_{1}} = \frac{V_{o}}{R_{2}} + I_{o}[/latex]

Using the specified parameters for the input voltage, desired output voltage, and the current source, we obtain

[latex]R_{1} = \frac{3R_{2}}{5 + (100μ)R_{2}}[/latex]

By trial and error, we find that excellent values for the two standard resistors are [latex]R_{1}[/latex] = 10 kΩ and [latex]R_{2}[/latex] = 27 kΩ. Large resistor values are used to minimize power consumption. With this selection of resistors the output voltage is 5.11 V, which is a percent error of only 2.15%.

Question 3.23: An 8-V source is to be used in conjunction with two standard...

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