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By Dietz P.H.

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We could also have calculated this directly from the voltage divider equation: R2 6kΩ V out = V -------------------- = 5V ---------------------------- = 3V R1 + R2 6kΩ + 4kΩ It is also good to verify that KVL works. VR1 + VR2 = 2V + 3V = 5V. How Do I Solve More Complex Resistive Circuits? By using a voltage divider, we can generate virtually any voltage between 0V and 5V using two carefully chosen resistors. However, the problem for this chapter asks us to build a system that will produce a number of different voltages under program control.

This is a pretty serious limitation. The output impedance of a linear circuit is just its Thevenin resistance. Basically, this tells us how much current we can draw before we start to change the output voltage significantly. For our DAC, this was not very much current. Before we leave this topic, we should note that there is another circuit that can create all possible linear I-V curves. It is the dual of the Thevenin equivalent circuit, and is called a Norton equivalent circuit. This circuit consists of a current source in parallel with a resistor, as shown in the figure.

A simple circuit will be used to convert the resistance measurement problem into a time measurement problem - something the Stamp does easily with precision. 0 - ©1998 Paul Henry Dietz - All rights reserved. 48 What You Need to Know What You Need to Know In order to solve this problem, here are a few things you need to know: • • • • What is a serial interface? What is a capacitor? How do I use a capacitor in a circuit? What is an oscilloscope? What is a Serial Interface? When driving the LED displays in our previous problems, we used one Stamp I/O line for each LED segment.

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