Do the capacitors always charge exponentially?

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    Capacitors Charge
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SUMMARY

Capacitors charge differently based on the source used. When charged from a constant voltage source, capacitors exhibit exponential charging behavior. Conversely, when charged from a constant current source, the charging is linear, as demonstrated by the relationship Q = C × V. The discussion also highlights that the presence of components like transistors and zener diodes can influence the charging characteristics, with transistors potentially acting as constant current sources under specific conditions. Understanding these principles is essential for accurate circuit analysis and design.

PREREQUISITES
  • Understanding of capacitor charging principles
  • Familiarity with constant voltage and constant current sources
  • Basic knowledge of circuit components like transistors and zener diodes
  • Proficiency in using equations related to capacitors, such as Q = C × V
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  • Study the behavior of capacitors in RC circuits
  • Learn about the role of transistors in current sourcing
  • Explore the effects of zener diodes on voltage regulation
  • Investigate the mathematical modeling of capacitor charging using differential equations
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Electronics students, circuit designers, and engineers looking to deepen their understanding of capacitor behavior in various charging scenarios.

  • #31
Low pass filters, wiki article on logic gates...general official references to have with me, since I can't use google during the test :(
 
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  • #32
Yes. It's best to be prepared!Tbh the circuit doesn't look as if it makes much sense.

Since Vb=0, I think Ve will become 0.7V always.
At first the capacitor would charge, up to 0.7V (exponentially), and then it would stop.

So no, not a constant current source.
 
  • #33
In that case I'll just print two of your replies

It depends on how the capacitor is charged.

If the capacitor is charged from a constant voltage source, it will be charged exponentially as you say.

If the capacitor is charged from a constant current source, it will be charged linearly (as in your recent thread).
The current source will have to work harder and harder to keep its current constant though. ;)

If the capacitor is charged in another way, it will yet be different.

If you have a symbol in your circuit with the text "current source" next to it, it is also a current source. ;)

Note that your mediating transistor also needs constant voltage between the base and the emitter, before it will behave like a current source.
In your previous threads you have seen a couple of examples of that (for instance a zener diode).

What you need is that the current is the same whether the capacitor is full or empty.
If you can calculate the current in both those cases and it is the same, you have a current source.

Should do the job :) thanks
 
  • #34
Don't forget to trust your visual cues! ;)
 
  • #35
Good point!
 
  • #36
Not points - cues!

If you see points, you have to connect the dots.
 
  • #37
How to recognise a constant voltage source? The volts are the same for a big range of load resistance values. i.e. you have a low (enough) source resistance. What does "same" mean? It means small enough not to matter in the context that the source is used. Most electronic measurements are 'good enough' if they are within 1% (more than enough for many purposes). If the source resistance is 1/100 of the load resistance, you are in the right ball park. (And electronics is very often a 'ball park' topic) It would be adequate for charging many types of battery at constant voltage, for instance.

You can apply precisely the same sort of criterion to determine whether you have a constant current source. Does the current vary significantly over the range of load resistances you plan to use? If you are after 1% accuracy, a high enough voltage source and a series resistor 100 times that of the load will also give you a 'useful' current source.
Those are examples of "rules of thumb". On some occasions, they would be entirely inadequate, of course.

Once you include some gain and feedback, your accuracy can easily be much better and you wouldn't actually need a vast voltage and high series resistance.
 

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