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What is the difference between a capacitor and a diode rectifier ?

I've learn to use a capacitor in debouncing circuit to absorb the current so that most momentary fluctuation that doesn't come from a solid closed contact would be absorbed by the capacitor charging.

Now, I also learned to use a diode rectifier with relay as the coil inside a relay will generate a voltage peak when the coil discharge. This discharge shoot a peak of high voltage current back in the opposite way of the circuit ( towards the + ), so the diode rectifier can absorb the surge and protect the circuit.

But how is this a different process then what a capacitor does ? Wouldn't a capacitor absorb the surge too ?

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    \$\begingroup\$ 'absorb any "extra electricity"' is a very very inaccurate way to describe how a capacitor can suppress the effect of switch bounce. Ditto for your description of the effect of a diode across a coil. If you want to draw conclusions, you will need a much more accurate understanding. \$\endgroup\$ – Wouter van Ooijen Jun 25 '13 at 18:21
  • \$\begingroup\$ A little bit better, but a button does not generate a current, bouncing or not. \$\endgroup\$ – Wouter van Ooijen Jun 25 '13 at 19:06
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    \$\begingroup\$ @WoutervanOoijen There are exceptions to that (en.wikipedia.org/wiki/Piezo_ignition) but I'm just being pedantic \$\endgroup\$ – W5VO Jun 25 '13 at 19:14
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    \$\begingroup\$ @Kaz ... But still, I was able to get 3 very good and helpful answers. The only mess I see here is being side tracked in justifications on the relevance of the question itself; feel free to start a meta, and stop cluttering this question, comments are not made for this. My question is not of a personal opinion, opinion-based, does not have multiple answers possible or require to be of an unreasonable length. \$\endgroup\$ – FMaz008 Jun 25 '13 at 19:43
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    \$\begingroup\$ @FMaz008 - exactly. Unfortunately, you have run up against people who dislike your question for reasons which are not official close menu choices, and respond by picking something random but inapplicable from that menu. \$\endgroup\$ – Chris Stratton Jun 26 '13 at 17:40
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When the relay coil is open circuited whilst current is flowing, without some "protection" device a high voltage spark can be generated that can destroy transistors. A capacitor across the coil can suppress this spike. The energy contained in the relay coil's magnetic field flows into the capacitor charging it up but, some of that energy is converted to heat in the resistance of the coil windings.

At some point the coil has pushed out all the available energy and the capacitor has a voltage across its terminals due to it "collecting" the energy and the cycle reverses with the capacitor (because it's in parallel with the relay coil) now forcing current to flow into the coil. But some of that energy transferred is again given off by heat in the coil resistance so a little less current is built-up within the coil.

At some point later the capacitor terminal voltage is zero meaning it has transferred all its energy into the current in the coil and that current starts the process of recharging the capacitor in the reverse direction.

The cycle repeats with a little less energy each time due to heat losses in the resistance of the coil and eventually (usually after a few milli seconds) it's done and there's no appreciable energy left in inductor or capacitor. The maximum voltage that the circuit reached would be (by design) no greater than twice the original supply voltage and probably a lot less - it all depends on how big the suppression cap is.

So, you apply power to the relay in the normal way and you are also supplying some energy to the cap but because it is dc, the cap soon charges up and doesn't draw current for most of the time the relay is energized.

Similarly, a diode doesn't draw any current when the power is applied to the relay because it is reverse biased but, when the relay opens, the current in the coil has to go somewhere (or make a big spark) and it suddenly finds it is connected across a forward biased diode. The energy circulates around coil and diode dissipating power from the diode and the relay coil and never attaining a voltage greater than supply voltage + 0.7V.

The two components protect the relay driver circuit but they are totally different parts and act very differently in this circuit.

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The impedance (AC resistance) of a capacitor varies inversely with its size (measured in Farads) and the frequency.

In other words for an alternating current a large capacitor acts as a very low resitance. The higher the frequency the lower the resistance becomes so it tends to become a SHORT CIRCUIT for the AC signal

BUT (and its a big BUT) at DC (direct current) the capacitor is an OPEN CIRCUIT.

So connecting a capacitor from a point in the circuit to ground it effectively short circuits the AC component but does not alter the DC value.

A diode only allows current to flow in one direction and changes AC into DC. When you use it across a coil (as in a relay) it effectively acts as a short circuit to the back e.m.f (high voltage spike) when you switch the current OFF through the coil. This spike is caused by the collapse of the magnetic field which induces a high voltage in the coil in the opposite direction to the normal current flow in the circuit.

A capacitor would not be able to charge fast enough to prevent the spike and may cause the circuit to oscillate several times exceeding the voltage ratings of the transistor.

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  • \$\begingroup\$ So the diode rectifier is like a fast capacitor ? What will come out of a diode in a DC circuit ? \$\endgroup\$ – FMaz008 Jun 25 '13 at 18:18
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A diode only allows current to pass through it in one direction. A capacitor, in the context you've described, acts as a low impedance path to GND for high frequency signals. So a diode protects current from going in an unintended direction whereas a capacitor resists voltage (across it's plates) changing quickly. I think that's the level at which you are seeking an answer without getting into a lot of circuit theory and math.

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  • \$\begingroup\$ "acts as a low impedance path to GND for high frequency signals" ... this is still Chinese for me. So if I get you right, a capacitor will provide current to the circuit when it will discharge, were a diode will dissipate all current so that it doesn't store anything and will not have a "discharge" phase ? \$\endgroup\$ – FMaz008 Jun 25 '13 at 17:49
  • \$\begingroup\$ I suggest drawing two circuits for comparison purposes. In a nutshell though, a capacitor wants to keep the voltage at the present level across its plates (and achieves that by discharging current to oppose changes in voltage). A diode just blocks current from flowing in one direction altogether. \$\endgroup\$ – vicatcu Jun 25 '13 at 18:04
  • \$\begingroup\$ But does the diode just block (like closing my toilet valve, so that the water don't go in my toilet, but still flow everywhere else in the house) or does it "consume" the current ? \$\endgroup\$ – FMaz008 Jun 25 '13 at 18:14
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    \$\begingroup\$ A diode just blocks the flow of current, so yes it's reasonable to think of it as a one-way-valve \$\endgroup\$ – vicatcu Jun 26 '13 at 17:22

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