Testing voltage regulation of MC7805 using active load

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SUMMARY

The discussion focuses on testing the voltage regulation of the MC7805 voltage regulator using an active load circuit. The active load's resistance changes based on a controlling signal, with a square wave of 0-5V at 1KHz and 20% duty cycle. When the active load is "on," the resistance is approximately R18, resulting in a voltage of 5V at TP25 and a current calculated as 5/R18. Conversely, when the load is "off," the resistance is infinite, leading to a voltage of 0V at TP25 and zero current.

PREREQUISITES
  • Understanding of BJT operation and characteristics
  • Knowledge of voltage regulators, specifically the MC7805
  • Familiarity with active load circuits and their applications
  • Basic circuit analysis skills, including Ohm's Law
NEXT STEPS
  • Explore the characteristics of the MC7805 voltage regulator in detail
  • Learn about BJT transistor biasing and gain requirements
  • Investigate the design and implementation of active load circuits
  • Study the effects of varying duty cycles on active load performance
USEFUL FOR

Electronics students, circuit designers, and engineers interested in voltage regulation and active load testing methodologies.

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Homework Statement


The attachment depicts an active load (to the right is the equivalent circuit). It is called thus because this load (measured from TP27) changes its resistance according to the controlling signal connected to TP25. The load is switched “on” (minimum resistance) and “off” (maximum resistance) using a square wave, 0...5V/1KHz/20%D.C. I may assume that Q4’s resistance, when conducting, is roughly 0Ω.
I am asked to determine the active load’s resistance and the voltage in TP25 when the active load is “on”/"off". Also, if the MC7805 operates normally, what is the current flowing through the active load?

Homework Equations

The Attempt at a Solution


I think that when the active load is switched "on" the BJT is active and the resistance is R18. The voltage in TP25 would be 5V and the current, 5/R18. When the active load is switched "off" the BJT is in cutoff and the resistance would be infinite (hence current would be zero). The voltage in TP25 would be 0V.
Is that correct?
 

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Correct.

Aside: I know they told you that when the transistor is ON it's resistance can be assumed to be zero but as an exercise perhaps check what the minimum gain has to be to ensure the transistor is fully ON.
 

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