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I have edited the question to make it easier to explain what I am asking for.To make it clear, I have deleted the part with the 555 timer. I have a question regarding the following:

schematic

simulate this circuit – Schematic created using CircuitLab

I know that once the battery is connected, the capacitor is going to have a varying voltage across it until it will be fully charged, reaching 7 volts. During this time, what will happen to the other two branches (the ones with the two resistors) ? According to the law of parallel connections, the voltage measured across any component will be the same, equal with the one of the battery. As the cap. charges, will the other branches have a varying voltage across them too, or will they have a potential difference of 7 volts from the beginning across them, as the capacitor?

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Have a look at the image on this page. It's one of the representations which are clearer on how the IC works. I'll reproduce the image here:

enter image description here

You can clearly see that the voltage divider is independent for any pin of the circuit (well, except Vcc and Gnd). So the 1/3 and 2/3 voltages are just derived from there, but the actual comparison is done by two comparators, which take the pin 2 and 6 voltages to set and reset the F/F (flip.flop)

EDIT: Just found this diagram from the Texas Instruments LM555 data sheet, which is also quite neat:

enter image description here

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  • \$\begingroup\$ I wanted to know more exactly if a capacitor , when placed in paralell with n branches , will determine the voltage across each of them to be equal with the voltage across it. \$\endgroup\$ – Daniel Tork Jul 6 '15 at 10:24
  • \$\begingroup\$ There is some confusion. In your original text, you say 'shuts down the FF' - The flip changes state to Off, but is not shut down (power is 'shut down', not a FF). I assumed that, with 'branches' you referred to the 3 R's in series. Now I'm lost. What do you mean by 'n branches'? \$\endgroup\$ – jcoppens Jul 7 '15 at 15:02
  • \$\begingroup\$ It really was a confusing question , so I have re-edited it.The part with the 555 IC was not too helpful , there is no more need to take it in consideration.Sorry for the confusion. \$\endgroup\$ – Daniel Tork Jul 7 '15 at 18:18
  • \$\begingroup\$ After coming back to the question, I realise I have left your answer out since I made the last edit, which is not fair. I want to give it a place here, but I'm mot sure how. What should I do? \$\endgroup\$ – Daniel Tork Jul 19 '18 at 6:52
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Initially, the discharged capacitor looks like a short circuit. All current will be flowing through it and none through the resistors, meaning that the voltage across all three devices must be 0V. Only once the capacitor has started charging will there be any current flowing through the resistors due to the voltage across the devices becoming non-zero.

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  • \$\begingroup\$ If a resistor was placed between the positive power supply and the capacitor , the process would take longer , wouldn't it?Would that be the only change? \$\endgroup\$ – Daniel Tork Jul 7 '15 at 19:53
  • \$\begingroup\$ If the new resistor was in series with the entire parallel subcircuit, yes. If in series with only the capacitor, the voltage across the other two resistors would be non-zero initially. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 7 '15 at 20:02
  • \$\begingroup\$ If the resistor was in series just with the capacitor an between it and the positive rail, how would the voltage of the two resistors be affected during the time when the cap. charges? \$\endgroup\$ – Daniel Tork Jul 7 '15 at 20:07
  • \$\begingroup\$ It would remain at 7V the whole time. Whatever the capacitor didn't drop, the resistor would. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 7 '15 at 20:11
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I'm guessing your question deals with the paradox where the voltage across the capacitor can't change instantaneously, but the resistors are shorted to the voltage source and the capacitor and the voltage across them can?

The voltage source and the capacitor have some resistance because they aren't ideal (as do the wires, but that isn't included in the circuit below), so the circuit in real life will look something more like this:

schematic

simulate this circuit – Schematic created using CircuitLab

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The voltage across the capacitor will be, indeed, the same as the voltage drop on the resistors.

The capacitor's voltage change can be analysed from two standpoints: theoretical and practical. Theoretically, there is some parasitic resistance (wire, etc.) which acts as a resistor placed in series with the capacitor.

The time it takes for a capacitor to charge up to 63% (63% of the source's voltage) is:

\$τ=R×C\$

The time it takes a capacitor to charge up to the supply voltage is:

\$5×τ\$

Resuming to the theoretical point of view, since both the capacity and the resistance are non-zero, the charging time will also be non-zero (as stated in llee94's answer). On the other hand, in practice, this parasitic resistance is so small that the charging time can be neglected. Since there is no resistor connected in series with the capacitor, R is zero and the charging time is 0 (the capacitor charges up to the supply voltage instantaneously). This means that all branches will always have 7V across them once the battery is connected, as there won't be, practically, any voltage transient present.

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