Hi, I'm studing Diodes theory and now i'm in clippers and clampers topic, i understand clippers theory but i'm a little lost in clampers, i understand the basic but i'm the kind of people that wants to know how things really works. so what i need to know is a step by step explanation how the capacitor and the diode interact to shift the ac voltage, maybe a link to a site explaining this with diagrams, animations, etc? saddly this topic is empty in the pdf books in this site. any help? please excuse my poor english by the way i'm attaching a clamper circuit just to refresh your memory
assume that the capacitor and load resistor are large.. ie RC>>1/f,the frequency of I/P signal. that way the cap voltage essentially remains unchanged.. then apply kvl in the circuit... to find output voltage
In addition to haditya's approach you might consider the circuit with the diode disconnected and find the thevenin voltage and the thevenin impedance of the circuit. Once you have the thevenin equivalent circuit, connect the diode across the output and you will have a simple single loop circuit that is a little easier to evaluate. Then if you assume an ideal diode, you will see that C1 will charge up to the peak-to-peak thevenin voltage the first time the diode forward conducts and then discharge at a rate that is determined by the RC time constant. Once the capacitor is charged up, the output will follow the thevenin input voltage when the diode is reverse biased and then refresh any charge that was been lost across the capacitor each time the diode is forward biased. hgmjr P.S. Keep in mind that with a real diode the charge induced across the capacitor will be the magnitude of the thevenin voltage minus the diode's forward voltage drop. That forward voltage drop would be around 0.7 volts for a small signal silicon diode.
ok, thanks for your great help, that is what i understand after analyzing your replies pluss my book theory: first, the cap is uncharged, with the first positive ac source iteration the diode is forward biased, there's a short circuit and the cap charges almost intantaneous, then all the voltage(ac signal + cap v.) is droped by the capacitor , and at diode terminals we have 0 volts(short circuit), in the next voltage iteration (the negative one) the diode is reverse biased,or an open branch so the current flows trougth the load resistance, increasing the voltage seen by the load from 0(short circuit diode) to v max(cap voltage + ac voltage). is that correct? and again thanks for your great help
Sounds like you have got the behaviour down pretty good. In particular, your observation that the cap charges to its quiescent voltage (the Thevenin Voltage minus the diode's forward voltage drop) almost instantaneously due to the fact that the voltage source is being treated as an ideal voltage source and therefore has zero source resistance associated with it. hgmjr
Hi nanobyte, My experience with clamping circuits was with some equipment that processed baseband video signals. The clamping circuit removed the incoming DC component and clamped the negative sync tips to ground. It was then a very straightforward task to strip sync from the incoming video signal using a voltage comparator. Of course there are many application for clamping circuits. Maybe some of the other members can relate their own experiences. hgmjr
thanks for your help nanobyte, will tell you what my book states: "a clamper circuit adds a positive dc voltage to the sine wave. stated another way, the positive clamper shifts the ac reference level(normally zero) up to a dc level. the effect is to have an ac voltage centered on a dc level. both positive and negative clampers are widely used. for instance, television, receivers use a clamper to change the reference level of video signals. clampers are also used in radar and communication circuits". this is an extract from "Electronic Principles sixth edition" by Malvino. hope it helps and my translation is ok.