How to Analyse a Diode Circuit: Understanding Voltage and Current Relationships

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To analyze a diode circuit, start by considering the diode in both conducting (short circuit) and non-conducting (open circuit) states. For a silicon diode, assume a forward voltage drop of approximately 0.6 volts. In a scenario with a 10-volt source and a 2.7 K resistor, if the diode has 0.6 volts across it, the voltage across the resistor would be 9.4 volts, resulting in a current of about 3.48 mA. Variations in the assumed diode voltage (0.5 to 0.7 volts) have minimal impact on the calculated current, demonstrating the robustness of the analysis. Understanding these relationships is crucial for effective diode circuit analysis.
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I'm a quite confused as to how to analysie a diode circuit

How would I go about analysing the following? I know that I need to analyse the circuit with the diode on(short circuit) and off(open circuit).
 

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If you assume the diode is Silicon, you can assume about 0.6 volts will appear across it for smallish currents.

This changes, but it is near enough for this analysis.

OK say there is 0.6 volts across the diode, then there must be 9.4 volts across the resistor.

(10 volts minus 0.6 volts = 9.4 volts)

Now a 2.7 K resistor with 9.4 volts across it must be carrying a current of about 3.48 mA. 9 V / 2700 ohms = 3.48 mA.

Notice that even though we assumed 0.6 volts for the diode voltage, it would not have made much difference if we had assumed 0.5 volts or 0.7 volts. 9.5/2700 = 3.5 mA. 9.3/2700 = 3.44 mA
 

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