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The capacitor was allowed to discharge completely when the transistor was activated with the 3.4 V.

The capacitor is allowed to charge again when the conduction from the 5 V line through the resistor to the ground of the IC is open circuited because the transistor deactivated.

The transistor must deactivate at 0V to have the 5 second cycle. In the case that the transistor deactivated at 1.4 V across the capacitor, the 5 s cycle would be reduced to the time it takes to charge the capacitor from 1.4 V to 3.4 V.

So my question!:

If the transistor was activated at 3.4 Vin (3.4 V across the capacitor) and allowed the capacitor to discharge through it, then the voltage at the capacitor would be decreasing. V in of the transistor would not be 3.4 V anymore..

There is the mystery.

If The transistor activates at 3.4 V, then when does it close as it has current running through it from the capacitor to ground. The voltage across the capacitor would be 1 V at some time 't' , therefore the V in of the transistor would be 1V, which is less than 3.4 V, which is when it activates.

Does it deactivate at 0 V in - ~0 V in AFTER it has activated (opened/activated), and only activate at 3.4 V in?

A second question: If I replaced the cap with a resistor the IC would switch rapidly correct?

The schematic on the top left is the one I redrew on the wide ruled paper.

enter image description here

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  • \$\begingroup\$ Please edit your question for clarity: transistors "opening" and "closing" is very ambiguous, because some people use the valve analogy ("closed" = "not conducting") and some people use the switch analogy ("closed" = "conducting"). It is much clearer if you use "on" and "off", "closed circuit" and "open circuit", or "conducting" and "not conducting". \$\endgroup\$ – TimWescott Oct 17 '19 at 0:01
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You are getting wrapped up in matters that are irrelevant to the question. The question states that when the voltage reaches 3.4V, the transistor is turned on -- it doesn't say that when the voltage falls below that the transistor is turned off. So just assume that it happens by magic in another part of the IC, and answer the question at hand.

If The transistor activates at 3.4 V, then when does it turn off (ed)?

At some time that is not addressed in the problem, through some mechanism not described in the problem.

Does it deactivate at 0 V in - ~0 V in AFTER it has activated (opened/activated), and only activate at 3.4 V in?

That is both unknown, and immaterial to the problem at hand.

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  • \$\begingroup\$ Thanks. I would prefer an answer with possible possibilities. Maybe it works as I have described or in a way you are familiar with from experience with similar circuitry. This would help me explore further into interests I consist of. \$\endgroup\$ – Renzo M-Svartz Oct 17 '19 at 0:35
  • \$\begingroup\$ It's a made up problem, that very likely doesn't map back to reality. The author wants you to calculate a resistor value, and has made a story problem to fit what they want you to solve. Just imagine that the IC turns the transistor on for some finite time, long enough to bring the cap voltage to zero. Part of solving story problems (and even engineering problems) is to ignore the irrelevant and concentrate on what matters. \$\endgroup\$ – TimWescott Oct 17 '19 at 0:47
  • \$\begingroup\$ Thank you Tim. I appreciate your advice! \$\endgroup\$ – Renzo M-Svartz Oct 17 '19 at 2:01

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