
#1
Feb711, 03:34 AM

P: 63

A series LC circuit at resonance has a higher total voltage than the applied voltage due to voltage magnification.
If a diode were inserted between L and C it would prevent the back and forth oscillation of energy, But would it prevent voltage magnification if the circuit were driven at resonance? Does resonance still exist in an LC circuit containing a diode? 



#2
Feb711, 04:50 AM

P: 247

A capacitor effectively blocks DC current flow except during the period of time it is initially charging, so, placing a diode between an inductor and a capacitor would effectively stop AC current flow, and there would be no resonance.
In the case of a power supply, there is frequently an inductor a diode and a capacitor in series; however, the "load" is in parallel with the capacitor (typically referred to as a "filter capacitor") and this fundamentally changes the circuit behavior. Such a circuit can be designed as "step up", "step down" or "isolation". You might take a look @ http://en.wikipedia.org/wiki/Boost_converter http://en.wikipedia.org/wiki/Buck_converter To see how "resonance" is achieved using an inductor, capacitor, diode, switch and load. Notice the placement of the switch is different for step up (boost) and step down (buck). Fish 



#3
Feb711, 01:49 PM

P: 63

Ok, I think I understand.
The boost converter steps up the voltage by switching the circuit at resonance (creating the resonat voltage rise) The buck converter does not operate at resonance so the inductor and capacitor have an impedance which causes a voltage drop. Ok if I understand that right then you answered my question! So, Thanks 



#4
May1912, 12:53 PM

P: 9

Diode in a series LC circuit and voltage at resonance?
A diode may block AC type movement (swinging) and turn the circuit more into a DC like system...(one direction).. so it's almost a paradox to call it resonant if it's not really resonant? Unfortunately the word resonant is overloaded and can be used to mean different things. It's also a buzzword.




#5
May2012, 07:02 AM

P: 661

Let me grab these two comments once again: When the resonant series RLC circuit reaches steadystate, the total circuit impedance is resistive (current is a max). VLVc= jXLIjXcI=0. The voltage drop across either component could be gigantic but the voltage drop across the two components (L&C) in series is zero. What happens?? The capacitor's magnetic field energy oscillates with the electric field energy of the inductor. Not from the source! He is not talking about a parallel circuit, but let's still mention it..... When the parallel RLC circuit reaches steadystate, the total circuit impedance is resistive. The same thing happens except the capacitor and inductor currents are 180 degreess out of phase. Equal inductive and capacitive reactances in a parallel circuit at resonance act as an open circuit. That's the only way it will work since all components in a parallel circuit have the same voltage drop. Above it is said that the diode turns the circuit into DC in regards to the capacitor for example. Well, I don't quite agree with this. It's definitely not AC.....but it's not exactly DC either. The diode acts as a half way rectifer......which it's waveform is not a flat line like a battery. Due to the C*dv/dt=it nature of the capacitor.....the cap is still operating due the change in voltage every half cycle. So due to this and the diode, there will be no oscillation, but I believe there will be pulsing in one direction due to the change in voltage every half cycle. I'm not totally sure on this one.....I would like to hear what the bigger guns have to say. 



#6
May2012, 07:21 AM

P: 349

Would the capacitor just charge up.......this is my first guess.
If not....why not? 



#7
May2012, 08:07 AM

P: 661

Maybe it does.....but keep in mind the change in voltage over time with the half wave rectifier effect. Wouldn't it just take the half wave rectifed voltage wafeform....shift it by 90 degrees......and then that will be your current output waveform? The magnitude will obviously change.... 



#8
May2012, 12:14 PM

Sci Advisor
P: 4,006

The capacitor would charge to the peak value of the incoming AC signal (minus the diode drop) and then it would be unable to discharge because of the diode.
After that, the inductor and capacitor would be effectively isolated from each other by the reverse biased diode, so there would be no resonance effects. 



#9
May2012, 03:57 PM

P: 661

Or give some insight on how it gets to steady state? When you are talking C*dv/dt=it... As the sin wave builds from zero...there is a current heading clockwise in a circuit for example. Does this current change direction at the top of the sin wave? In other words is it the negative slope of the sin wave that changes the current direction or is it when the sin wave goes on the negative side of the axis? If the current changes direction on negative slope then I would have to agree the cap will just charge. But if the current changes direction in a cap upon passing the 0 axis....then I would agree that it pulses. 



#10
May2012, 04:48 PM

PF Gold
P: 369

in terms of the voltage current relationship, the current starts at a maximum since there is the greatest change of voltage over time at that point. the rate of voltage change decreases as the cap draws current, until it crests and the rate of change is zero, in which case the cap draw 0 current. Now, as the voltage decreases, the current goes negative, since now the capacitor acts as the source, and feeds current (negative) to the source. But it will not be able to do so, since the diode is reverse biased in the circuit. So, I think the first 90 degrees of the voltage, is all that will charge the capacitor, then it will not be able to discharge, for the diode?




#11
May2012, 04:55 PM

PF Gold
P: 369

Yeah sparky, it changes current direction as the voltage waveform peaks, because if you think about two black boxes, whichever has the highest voltage, will send current to the other. So, when the waveform for voltage is increasing, we know the capacitor voltage is lower than the source and so it draws current. As the capacitor and source have equal voltage, this is the top of the waveform, where the slope (dv/dt) is 0. It draws no current because the voltage are equal, then the voltage begins to fall, but the cap is still charged, so normally without a diode this is where the current changes direction. The equation for capacitor current yields the graph for the ninty degree phase shift I didn't really notice it before just took it at face value.
I think you were right when you said it will just charge, because it is at that point the capacitor tries to send current back to the source, in which case it can't I try to remember the terminology for caps and inductors based on these principles. For a capacitor, it is a two terminal passive device that stores energy in an electric field, and periodically returns energy to the circuit (based on frequency) For an inductor, same sentence except with magnetic field. The changing of the voltage sine wave means the capacitor is at a different potential than the source, and so can consume and deliver power from / to it. The returning of energy is the nature of capacitive reactance, in the same way a magnetic field collapsing and sending voltage back to the source is the nature of inductive reactance. A little away from the point... but I agree it will just charge.. 



#12
May2012, 08:25 PM

Sci Advisor
P: 4,006

Yes, FOIWATER has covered it.
The actual result you get depends a bit on the setup, especially on how the signal generator is connected. Assuming the circuit is driven at its resonant frequency, you can get something like this: This gives the classic half wave rectifier output. The output voltage is the green trace and the voltage at the anode of the diode is the orange trace. However, with different values of the resistors at left, there can be magnification of the first input pulse and the diode charges the capacitor to this value, which may be higher than the peak input voltage. The effect is the same, though, and the diode is effectively an open circuit after a few cycles of input. 



#13
May2112, 05:51 AM

P: 661

Beautiful gentleman.
It does get confusing because a basic sin wave changes current direction thru a resistor when it passes the zero axis. On our cap here it changes directions upon change in slope. Learned something once again....way to ask a good question, HMS776! Keep them coming! 



#14
May2112, 06:05 AM

P: 661

But let's go one step further here to be perfectly clear. Let's forget we have the diode.
With the cap, The current changes direction at the top of the sin wave at 90 degrees. Does it keep this same direction even as it passes 0 axis? In, other words, does it keep this same direction as the slope is negative until it reaches the very bottom at 270 degrees? In other, other words.......positive slope of input voltage makes for positive (say clockwise for example) current thru capacitor.....and negative slope of input voltage makes for negative (say counterclockwise for example) current thru capacitor? I want to make sure this is 100% correct. If not, please correct me. 



#15
May2112, 06:44 AM

PF Gold
P: 369

well I would draw it out and shift it ninety degrees if I were you, 90 degrees ahead that is for a capacitor which I know you know.
Assume we are discussing the voltage right now > the voltage waveform starts at 0. While the voltage waveform is at zero, the maximum current will be drawn from the source by the capacitor, since the greatest difference in voltage exists at this point (dv/dt) times C = instantaneous current value. As the voltage value rises, the current value is decreasing, which makes sense, since the capacitor is assuming the voltage of the source, the difference is becoming smaller (again relate back to the equation) The slope of the voltage is constantly decreasing, so the current drawn is constantly decreasing in value. Eventually, the slope becomes 0, that is to say, the capacitor is fully charged, and is accepting no more current, at this value of voltage, the current value is zero. It is about to cross the Xaxis, meaning the current is going negative. The only way for this to be true, is for the capacitor voltage to be higher than the source voltage, which is the case as the voltage sine wave begins to drop off (considering the cap was fully charged at Vslope = 0, IE peak) Now the capacitor begins to return energy to the circuit! or, OPPOSE the source voltage, in the very same way a resistor OPPOSES voltage. This is why reactance is at a ninety degree phase shift and resistance is not. This is why a capacitors voltage and current are not in phase. And  it is also why we must add them vectorally (that a word?). So, the patter continues, of source voltage being higher than cap voltage, charging, cap voltage being higher than source voltage, discharging, so energy is returned to the circuit every period, which is why capacative (and inductive!) reactance values are frequency dependant! and also, if you can imagine the periodicity in time of a sine wave, it explains the omega (angular velocity portion as well) multipling frequency by two pi gives us total radians covered by the circuit in a second, multiplied by the capacitance, we have a TOTAL value for reactance. I think it's pretty amazing. I had a hard time conceptualizing how a cap stored energy in an electric field (and an easy time with the concept for an inductor and a magnetic field) but really it simply accepts and returns energy based on voltage levels, it's great. A bit off topic but.... Yeah psparky, well the current doesn't keep negative slop until 270, rather 180, because it is at this point the voltage crosses the xaxis (IE, is zero) with zero voltage, we have no current flow once again, and as the voltage crosses the xaxis (IE, the voltage applied is negative, the current delivered by the capacitor becomes less and less, until 270 degrees, where the vltage peaks negative, and begins to increase towards zero, at the 270 degree mark the current is once again being delivered to the capacitor, and the value is increasing as dv/dt does. until eventually at 360 the voltage crosses the xaxis again (IE, is a maximum) and max current is drawn by the capacitor again and the process repeats. Remember this is for a purely capacitive circuit. I'm sure you know that I may have messed something in that last paragraph but you get the idea, just pull up the graph or draw it. 



#16
May2112, 08:12 AM

P: 661

Whew......reading that can definitely smoke your brain....but by drawing it to scale I think I see it.
Let me correct what I said above and restate.....in regards to a AC source and just a capacitor. If I take voltage as a sin wave......and the current thru the capacitor as the cos wave......and put them on top of eachother (90 degrees cap leading) If I take the first 90 degrees of the voltage.....the voltage is climbing and the current thru the cap is heading in one direction (say positive direction) and decreasing as the voltage climbs. At 90 degrees or the peak, the current passes the zero axis and CHANGES direction (say negative direction) and is actuallly at zero amps at this point. From 90 to 180... current starts to build in the negative direction....as voltage falls from it's peak to zero volts at 180.....current is at it's negative maximum. From 180 to 270.......Voltage starts at zero and heads to it's maximum negative value. Current starts falling from it's max negative value and heads toward zero amps all in the negative direction. 270 to 360.......voltage is at max neg at 270 current at zero.......current changes direction at 270 +. As the max negative voltages goes back up to zero......the current goes back up to it's maximum value in the positive direction. At 360 complete process repeats........ So current will only change direction upon the change in slope in voltage waveform from positve to negative slope....or negative to positive slope. Or you could say, when the voltage slope is positive, current flows in positive direction. When voltage slope is negative, current flows in negative direction. This obviously relates to rise over run....or the dv/dt in the equation. I think we got it! To go further, I need to absorb the way you put it in the post above.....that will take a bit more time, but I needed to get the first part right. Ironically, this whole conversation can be summed up with C*dv/dt=it.....but apparently that equation is clear as mud! lol. 


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