Testing voltage regulation of MC7805 using active load

AI Thread Summary
The discussion focuses on testing the voltage regulation of the MC7805 using an active load that adjusts its resistance based on a controlling signal. When the active load is "on," the BJT is active, resulting in a resistance equal to R18, a voltage of 5V at TP25, and a current calculated as 5/R18. Conversely, when the load is "off," the BJT is in cutoff, leading to infinite resistance, 0V at TP25, and zero current. The correctness of these assumptions is confirmed, with an additional suggestion to explore the minimum gain required for the transistor to be fully "on." This analysis aids in understanding the operational characteristics of the MC7805 voltage regulator.
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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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