What does the resistor do when the second switch is closed?

When the first switch is closed, the capacitor charges to 5V. When the second switch closes, capacitor discharges and Vo will rise upto 0.25V, I calculated. However, my question is, when switch 2 is then opened and switch one is closed, resistor is open. What happens to the 0.25V across it? how does it drop off? suddenly or gradually? why? Thanks in advance.

• Why would Vo rise to (just) 0.25V? Commented Feb 16, 2016 at 16:01
• @VladimirCravero Switch closed for 0.5 micro seconds only. Commented Feb 16, 2016 at 16:02
• If switch 2 is open, then the current through the resistor must be zero, so the voltage across the resistor must also be zero. Commented Feb 16, 2016 at 16:05
• @uint128_t It must be zero, but I am worried about the voltage transient. Will it drop off suddenly, or will it follow some exponential? Commented Feb 16, 2016 at 16:05
• When switch(es) 2 are closed, the instantaneous voltage across the resistor will be 5V, this will decay a little before switch(es) 2 open, at which point the voltage across the resistor drops immediately to 0V Commented Feb 16, 2016 at 17:21

0.25 V doesn't appear at output anytime in this circuit ,not for the timescale provided .(I am getting a time constant of 10 microseconds).I suggest you draw the circuits corresponding to the switches closed and open separately.First draw the circuit for the first 0.5 microseconds.What does it look like?And when you go to the second circuit which is in effect from 0.5-1 microseconds what changes?The approximation provided is a helpful hint ,not just for calculation but for the nature of the problem(the second circuit mostly of all). As far as the value of the current across the resistor is concerned it can change suddenly(provided the resistor is ideal,no reactive elements).If you had a simple Resistor connected to a source and the voltage were in the form of periodic pulses(A square wave for a example ,or any signal that is zero for a finite time and this is suppose WLOG periodic.),what waveform of current do you expect through the circuit?There is no scope for a transient in the resistor here as has been pointed out by Vicente Cunha.

• I already did what you asked me to do. In the first circuit, I get 0 time constant as R=0 for the first circuit. Then from 0.5micro to 1micro the time constant is 10 micro seconds. So, I calculated, as an approximate, V=0.25 =(5*(1/20)) Commented Feb 21, 2016 at 11:01
• The approximation goes like $$exp(-t/(\tau))=1-t/\tau$$ t=0.5ms and $$\tau =10ms$$.You have calculated the change in voltage instead of the value. Commented Feb 21, 2016 at 11:05
• So, Yes you are right. 1-0.5/10 = 19/20. $1-exp(-t/(\tau))=t/\tau = 1/20$ So, It means, voltage across resistor = 5(1/20)=1/4. Commented Feb 21, 2016 at 11:14
• Now you know the value of the voltage at the end of the second section,use it to sketch the waveform. Commented Feb 21, 2016 at 11:16
• I sketched the waveform as a ramp when the second switch is closed, rising from 0 to 0.25V , and 0V when the first switch is closed. Commented Feb 21, 2016 at 11:18

Since this is clearly a homework question, I won't provide a complete answer, but rather some hints. You are currently way off track.

• Pay attention to the polarity of the voltage across the capacitor. It cannot change when one set of switches opens and the other closes.

• Keep in mind that when the second set of switches is closed, the voltage across the resistor is the same as the voltage across the capacitor. How much current is flowing, and how does it change with time?

A more interesting question: What happens if you connect a second capacitor, say, 10 nF, in parallel with the resistor?

Here's a simulatable copy of the circuit:

simulate this circuit – Schematic created using CircuitLab

Run the simulation to get a feel for what's happening qualitatively, and this should give you some insight into what you're doing wrong with the equations. Try it with SW5 both open and closed.

• I am still unable to get anywhere close to the answer... can you give some insight? By the way, this is not a homework question. I am trying to make a negative charge pump and am stuck . My prof asked me to figure this out in order to know my mistake in the circuit. Commented Feb 21, 2016 at 9:54
• I also would like to know, why 0.25V is wrong.. The capacitor should be having 5V keeping aside the polarity, and the time constant is 1e-5 .. in half a micro second, it will be 5*(1/20)=1/4V is how I got it. Commented Feb 21, 2016 at 9:58

A resistor does not "store" energy under any form (a resistor has no reactance), only dissipates what it is currently connected to. Once switch 2 opens, the resistor is no longer connected to the capacitor and you can safely expect an immediate voltage step from 0.25V to 0V across the resistor.

• If you specify that your answer applies perfectly to an 'ideal' resistor (free from any capacitance/reactance), it would be quite accurate. As all 'real world' components do have some degree of 'parasitic' capacitance & inductance, there will be some degree of 'discharge time' across the resistor. However, whether or not OP's real, or theoretical, equipment would be capable of measuring this transient is most definitely an arguable point. ;) Commented Feb 16, 2016 at 17:35
• This is a poor (or at least misleading) answer, because it fails to point out that the voltage across the resistor is at no time equal to 0.25 V. Commented Feb 16, 2016 at 20:09
• @DaveTweed You are definitely correct, simply looking at the circuit one can observe that. I just gave blind trust in OP's calculation: "When the second switch closes, capacitor discharges and Vo will rise upto 0.25V, I calculated."; then answered his question directly: "What happens to the 0.25V across it? how does it drop off? suddenly or gradually? why?". I also gave the proper justification I believe is due. I answered his question, not the "homework". Commented Feb 17, 2016 at 10:16

When the switches controlled by phi1 close the capacitor gets charged to 5V. Since there is no series resistor this happens instantaneously. The left terminal of the capacitor is now at +5V, the right terminal at 0V. The output voltage is 0V since the resistor is disconnected and there is no current flowing through it. Hence the voltage drop across the resistor is zero.

When the switches controlled by phi2 close the left terminal of the capacitor gets connected to 0V and right terminal to the resistor. At the beginning of phi2 the voltage across the capacitor (from left to right) is 5V. For this reason the right terminal is now at -5V. This voltage also appears across the resistor, so at the start of phi2 we have -5V at the output. During phi2 the resistor discharges the capacitor. Phi2 is only 500us, therefore the capacitor is only discharged to 1-t/(tau) ~ 95% or (-4.75V) at the end of phi2.

This happens over and over again since with phi1 the capacitor gets charged again to 5V and the voltage across the resistor drops immediately to zero as the switch interrupts the current flow.