(pics fixed)very basic questions on capacitors/Clamping Circuit - Page 1
Forum section: electronics / beginners

author: questron
author_rank: Regular Contributor
timestamp: March 25, 2014, 07:12:29 pm
content: thank you linux-works and maurish for your great help about inserting pics!ok, this is a kindergarten level question. please excuse the ignorance, i need to get this clarified in order to understand how a Clamper Circuit works, from its physical side, as opposed to its mathematical side.if we charge up a capacitor in a way it can get charged,the two sides of it will acquire an equal amount of charges but of opposite types, positive and negative. yeah?this situation,will never happen provided there is current path to both sides, because that is not the way a capacitor behaves, correct?now, if a circuit is setup this way,one side of C1, netA, will have 5V, but this has nothing to do with capacitor charging, because the other side of C1, netC, is open, therefore there is no current path to it for current flow, and there won't be any charging happening. C1 is superfluous to the circuit, its charging/discharging function as a capacitor is not being used.voltage of netC in this case is uncertain.is this correct?i'm reading a text book now, which gives a Clamper Circuit like this,assuming the initial condition is such that the sinusoidal signal starts from 0V to Vpk, Vin=Vout=0V, forward biasing voltage and forward biasing resistance of D1 are both 0V as well, then D1 is not conducting, open circuit there, and C1 is in the same situation as the previous pic, i.e. it is not functioning as a capacitor and no charging/discharging can be talked about, because the Vout side of it has no current path for current to flow.this text book also gives a set of curves,which shows C1 is charged to a constant voltage of Vpk. it merely says "capacitor voltage", i'm not clear on whether that means Vin or Vout with both referencing ground, or referencing each other, or whatever.in all cases, how can this be? the other side of C1, Vout, is an open circuit at all times, its voltage has nothing to do with how Vin varies, and because of this D1 will never be turned on.nonetheless, LTSpice simulation shows this circuit is really working!where did i go wrong in my above reasoning?thank you all for helping!

author: cyr
author_rank: Frequent Contributor
timestamp: March 25, 2014, 08:15:53 pm
content: The potential at netC in your first case will depend on the charge on the capacitor. If the capacitor is discharged it will be equal to netA.In your second example it's the same. If the capacitor starts out discharged it has zero volts across it. As soon as vin reaches the threshold voltage of the diode, so does vout and the diodes starts conducting, thereby charging the capacitor and increasing the voltage across it.

author: questron
author_rank: Regular Contributor
timestamp: March 25, 2014, 10:06:09 pm
content: thank you cyr! this is exactly what got me. i know what you said must be right, i just don't understand it.LTSpice does show what you said, from t=0, voltage of netC does follow that of netA to about 0.7V, then D1 turns on. then while voltage of netA rises from 0.7V to Vpk, that of netC is pulled down to 0.7V and kept there, because D1 is on, i suppose this is the so called clampping action itself at work. this is the flat part of the red curve, for netC, in the first T/4 of the input signal. i understand that only if i accept what you said by faith. but i want to undertstand why you said that.at t=0, voltage of netA and netC are both at 0V, D1 is off. so when the voltage of netA rises from 0V to Vpk, netC is just a piece of broken wire, not connected to anything and just happen to be lying next to netA, there is no current path to it, how could it follow the rising voltage of netA from 0V to 0.7V? in order to follow, current need to flow in netC, yeah? which it is not the case, D1 is off, open circuit there, netC is like a piece of broken wire lying there by itself. so without current flowing, there is no capacitor action to be talked about.even if C1 is functioning as a capacitor during the first T/4 of the input signal, as the voltage of netA go up, that of netC should go down, that's how a capacitor works, a capacitor does not have both sides going up and down in the same direction at the same time, as shown in the first pic, yeah?this Clamper Circuit is clearly working, i just couldn't figure out where i went wrong in my reasoning.

author: Everton
author_rank: Contributor
timestamp: March 26, 2014, 03:01:39 am
content: Look at it this way.When the voltage is rising and the diode is conducting, the capacitor will charge to the value of the source minus the diode drop. Now the capacitor has approx Vpk -0.7V accross it (with the positive being closest to the voltage source).Now when the source signal starts droping back towards 0, the voltage on the other side of the capacitor (node C) is going to be Vsource-Vcap which causes the diode to turn off and no current flows. As such the output is always going to be Vsource -Vcap which is the same as Vsource - (Vpk-0.7).Hope that helps.

author: cyr
author_rank: Frequent Contributor
timestamp: March 26, 2014, 05:38:38 am
content: NetC isn't floating, it is connected to one side of the capacitor and the capacitor has a very well defined voltage across it (proportional to the charge on the plates). There is no path for current to flow through the capacitor, so there is no way for the amount of charge on the cap to change, and therefore no way for the voltage across the cap to change. NetC *must* "follow" netA, and there is no need for any current to flow for this to happen.Note that this is a theoretical circuit, in real life there is no such thing as a perfectly floating net, there will always be tiny current paths (especially if you try to measure anything).

author: questron
author_rank: Regular Contributor
timestamp: March 26, 2014, 10:35:27 pm
content: Quote from: cyr on March 26, 2014, 05:38:38 amoh, i actually meant to say that without a current path to the other side of D1, that side of D1, that plate, that electrode of D1, and netC taken as one single net of the circuit are no different than a floating piece of wire, because the open circuit will render C1 a non-capacitor when there is no current path to the other side. but that's not important, let's return to the question.i need to back up a little here.very well defined voltage, sure, but that's if the capacitor is functioning as a capacitor, with current paths to both sides. can a capacitor get charged with one side open, how does that work? where is it going to suck electrons in from, or pump electrons out to, in order to get charged?Quotewhy "NetC *must* "follow" netA, and there is no need for any current to flow for this to happen"?for the sake of simplicity, let's just say 'voltage of netC can change, and there is no need for any current to flow for this to happen'.how does that work? voltage change is not related to current change, regarding the same capacitor? so no current is necessary in charging up a capacitor to some voltage?Quoteunderstand that. if that has no bearing on the functioning of this circuit in theory, it's good enough to ignore that point for now. let me understand how the Clamper Circuit works first.

author: questron
author_rank: Regular Contributor
timestamp: March 26, 2014, 11:30:07 pm
content: thank you very much for the help Everton! i know what you said must be right, otherwise LTSpice won't show this Clamper Circuit as working. i just don't understand why it is working, or how it works.the text book i have gives the same equation as yours, Vout=Vsource-Vcap, but this rests upon a premise, Vcap exists. this is what the conversation with cyr boils down to as well.with reference to the theoretical "Clamper Circuit", at t=0, Vsource=Vout=0V, so D1 is off. it follows that one side of C1 is open. if the way a capacitor works is that no charging will occur if one side of it is an open circuit, then C1 will never be charged no matter how Vsource varies, and D1 will never be on. then nothing works, which is obviously wrong.now let's say net Vout is an open circuit, so it stays in its initial condition, 0V, and net Vin can vary in every which way without any effect on net Vout since the open circuit will make C1 not behaving as a capacitor anymore, then at some point in time between t=0 and t=4/t, D1 will turn on. as there is no open circuit anymore, C1 can charge/discharge now, and when Vin goes up, Vout should go down, as what a capacitor should do. the problem of this scenario is that this is not what LTSpice shows. LTSpice shows that Vout follows Vin from t=0 up to Vin=~0.7V, so Vout follows Vin from the very beginning. what state does C1 have to be in for this to happen, D1 on, D1 off, both sides of a capacitor go in the same direction during charging, the two sides go in opposite directions during charging, or charging/discharging is irrelevant here? and how is it that this circuit provides the working state, whatever it needs to be, to C1? i couldn't figure this one out.i need to get over this hurdle first, which in my mind is the premise of the aforementioned equation.