Resistor Grounding in a 555 Timer Circuit Explained

[RED119]

Homework Statement

So I am solving a 555 timer questions, and there is a resistor that is grounded on both terminals at one point. My question is what does this do to the resistor?

Homework Equations

Vc(t) = A + Be^(-t/(R*C))

The Attempt at a Solution

I know that if you ground one terminal of a capacitor nothing much happens, or if you try to discharge a capacitor directly to ground you get infinite voltage which is a bad thing. solving the above equation for discharge and a 0 for resistance... it doesn't turn out pretty... but I am pretty sure that's wrong?

 

[trurle]

You get surge current limited by capacitor`s ESR and ESL, not the infinite voltage.

 

[CWatters]

In the homework section you should post the whole question word for word.

Hint: if the voltage on both sides of a resistor as the same what is the voltage across the resistor going to be? Therefore what current is flowing through it? Will that damage the resistor?

 

[RED119]

In the homework section you should post the whole question word for word.

Hint: if the voltage on both sides of a resistor as the same what is the voltage across the resistor going to be? Therefore what current is flowing through it? Will that damage the resistor?

So maybe I am very off in even my approach... to answer your question, if you have the same voltage across a resistor then there is no electric potential, and the resistor is effectively shorted? you can change voltage instantaneously across a resistor without damaging it, but not a capacitor or battery?

With regards to the question and my approach, if the Q bar is set high from the start then the transistor is closed, so the capacitor has 0V, so then when the trigger input goes up as says in the timing diagram that is given the capacitor begins charging when Q bar goes to zero and the transistor closes. So then capacitor is charging up to a pint until it is greater than 4V, at which point it will be grounded again and drop to zero almost instantly? I don't know if I am close or just totally on the wrong track, sorry for the paragraph, but that is my reasoning here.

Summary is: ln(2) * Rb * C = discharge time,

so, since Rb is zero, the discharge time is also zero... and you can't change the voltage across a capacitor instantaneously... so what's up?...

 

[CWatters]

The capacitor voltage doesn't fall instantaneously to zero, just fast. The transistor can't carry infinite current and there will be some small but finite resistance in the path. The capacitor isn't an ideal capacitor either, it has some resistance - look up Equivalent Series Resistance - for some types of capacitor it can be significant.

If the 555 is configured as a monostable the discharge time limits how soon you can retrigger it but that might not be a problem. If it's set up as an oscillator it limits how fast it can oscillate.

 

[CWatters]

The time scale on the timing diagram looks to be around 1cm=10ms. So on that scale the capacitor discharge can be shown as a vertical line. eg <1ms.

 

[CWatters]

So maybe I am very off in even my approach... to answer your question, if you have the same voltage across a resistor then there is no electric potential, and the resistor is effectively shorted? you can change voltage instantaneously across a resistor without damaging it, but not a capacitor or battery?

Correct. If the voltage across a resistor is zero the power dissipation in it is zero so its all fine.

You can't change the voltage across an ideal capacitor instantly as that implies/requires infinite current.

Edit: There isn't a resistor in the circuit with both ends connected to 0V.