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

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SUMMARY

This discussion focuses on analyzing a diode circuit, specifically using a silicon diode with a forward voltage drop of approximately 0.6 volts. The analysis involves considering the diode in both its conducting (short circuit) and non-conducting (open circuit) states. With a 10-volt source and a 2.7 K resistor, the voltage across the resistor is calculated to be 9.4 volts, resulting in a current of approximately 3.48 mA. Variations in the diode's forward voltage drop have minimal impact on the current calculation.

PREREQUISITES
  • Understanding of diode characteristics, specifically silicon diodes
  • Basic knowledge of Ohm's Law and circuit analysis
  • Familiarity with voltage and current relationships in electrical circuits
  • Ability to perform calculations involving resistors and voltage sources
NEXT STEPS
  • Study the impact of different diode types on circuit behavior
  • Learn about the Shockley diode equation for more accurate diode modeling
  • Explore circuit simulation tools like LTspice for practical analysis
  • Investigate the effects of temperature on diode performance
USEFUL FOR

Electronics students, circuit designers, and hobbyists interested in understanding diode behavior and circuit analysis techniques.

tigertan
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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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