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Are Shunt Capacitors in the Signal Path?


mboxler

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Yes, but with reservations. The amplifier is the driving force but the parts in it's path determines the amount of power the speaker will see. The shunt enables more of the power of the amplifier to go to the speakers. Without the shunt the power is limited by the part. Hope this makes sense. 

 

Informative link speaker builders would enjoy. The one thing that is not mentioned in the video is the phase shift caused by the extra parts in higher order crossovers. Any time you add a part it achieves what the designer is aiming for but it also brings some negative into the equation as well. There are always interactions when another part is added. Thanks for posting the link. I enjoyed watching it. 

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A better way to describe what is happening is to use water. Often used describing electronics. You have water coming to your house with a lot of pressure and volume but the faucet (valve) limits the amount of each you will get when you turn on the faucet. The electronic parts control the amount of power you will get at your speakers. You put one component in you may get more of this available power out or possible just the opposite depending on the circuit. This may make what is happening better understandable.

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A circuit consists of a 4mH inductor in series with a 6 ohm resistor.  A 100uf capacitor is across the resistor.  Frequency is 170Hz.

 

At that frequency, the impedance of the parallel resistor capacitor is 5.05 ohms @ -32.65 degrees, and the impedance of the inductor is 4.273 ohms @ +90 degrees.  Add the two together and you get 4.525618 ohms @ +19.99 degrees for the entire circuit.

 

2.83 volts (amplifier) into 4.525618 ohms (circuit impedance) equals 0.62532896 amps. 

 

0.62532896 amps into 5.05 ohms (capacitor/resistor impedance) equals 3.1579 volts.

 

The voltage across the resistor is higher than the amplifier voltage.

 

Danny would say the extra voltage comes from the capacitor(?).  I think it's because the impedance of the capacitor/resistor (5.05) is greater than the impedance of the entire circuit (4.525618).  Or are we both saying the same thing in a different way?

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3 minutes ago, mboxler said:

The voltage across the resistor is higher than the amplifier voltage.

 

Danny would say the extra voltage comes from the capacitor(?). 

 

The extra voltage comes from the interaction between the capacitor and the inductor. In a resonant circuit, the voltage across the individual reactive elements can be much higher than the driving voltage. The specifics depend upon the configuration of the circuit and the resonant Q.

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Danny says "that's not how it works". Yes it is. Text books usually assume a perfect voltage source i.e. zero output impedance and infinite current capability. Also the load is resistive only. So those 1st, 2nd,3rd and 4th curves he showed are correct given those parameters. Unfortunately real world amplifiers are not perfect voltage sources and speakers are not purely resistive loads. My take is anything in the filter circuit that effects the output to the speaker is in the circuit path. The shunt parts are less critical maybe but they effect the signal surely otherwise why have them there.

And yes Mike I think you both are saying the same thing in different ways.

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30 minutes ago, babadono said:

The shunt parts are less critical maybe but they effect the signal surely otherwise why have them there.

 

 

Is the choice between an electrolytic and a film capacitor based more on the "life" expectancy of the caps rather than the "quality" of the attenuated voltage across the driver?

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View it in terms of current and not voltage.

 

The capacitor and the woofer coil are in parallel. The current flowing through the capacitor depends on the frequency, the higher the frequency the more current passes through the capacitor. The current from the amp passes through the inductor and through the capacitor right back to the amp, the current through the capacitor does not pass through the woofer voice coil.

 

If the capacitor was in series, like with the tweeter network, all the current that passes through the tweeter must also pass through the capacitor. The lower the reactance of the capacitor the less voltage is across it's two terminals and therefore has very little effect on the current because impedance is so low. As frequency decreases and reactance increases along with impedance, the current will create a voltage drop across the capacitor and it's in this region where the effects of the capacitor will effect the current passing through the tweeter, this is where the distortion of the capacitor will show through the tweeter.

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For the smarties in the group, you will realize that the inductor will eventually reach an impedance and limit the amount of current through the capacitor as frequency increases. At 100Hz you will have around 500mA of current though the capacitor where at 1kHz it increases to 800mA. If the inductor wasn't there the current would be higher than 800mA.

 

Regardless you can see current though capacitor passes back to the source and is completely separate of the current passing through the woofer and back to the source. The source needs to supply enough current for both "loads" since they are in parallel.

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23 minutes ago, captainbeefheart said:

For the smarties in the group, you will realize that the inductor will eventually reach an impedance and limit the amount of current through the capacitor as frequency increases. At 100Hz you will have around 500mA of current though the capacitor where at 1kHz it increases to 800mA. If the inductor wasn't there the current would be higher than 800mA.

 

Regardless you can see current though capacitor passes back to the source and is completely separate of the current passing through the woofer and back to the source. The source needs to supply enough current for both "loads" since they are in parallel.

 

Are you using the component values I stated above to get to these current levels?  I seem to be coming up with something different.

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A layman's contribution from me, but one that might continue to stimulate discussion. Besides my active xover/DSP for my Underground Jubilees, I have a passive xover from Bob Crites which of course can only work for a specific tweeter driver for the K402, in this case it is a Faital HF200 driver.

A few years ago I had a nasty signal from the amp and it destroyed electrolytic capacitors (I think bipolar) that are doing duty on the bass section of the xover in parallel.
A good technician traced the fault and he identified the electrolytic capacitors as broken. The symptom why I brought the xover for repair in the first place was that the channel only had after the blow up a very thin deep bass.

 

Now comes my point. The technician was good to me and he thought I would be happy if he traded the destroyed electrolytic capacitors for MKT. So he had soldered together a package of MKT caps to achieve the same capacity. 
But I was very irritated and disappointed. In the end the K402 tweeter sounds much sharper and more aggressive. And this despite the fact that the replacement was made in the bass range of the xover. And to tie in with the thread here, as I said, these capacitors are parallel to the signal. And yet the sound impression is so different.


 

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1 hour ago, mboxler said:

 

Are you using the component values I stated above to get to these current levels?  I seem to be coming up with something different.

 

No sorry I didn't use the same model you were using.

 

 

Amplitude and frequency are the big variables if we are trying to match each others work. Accuracy wasn't important for the explanation.

 

The values I was using was 2.5mH and 100uF.

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19 hours ago, captainbeefheart said:

 

The capacitor and the woofer coil are in parallel. The current flowing through the capacitor depends on the frequency, the higher the frequency the more current passes through the capacitor. The current from the amp passes through the inductor and through the capacitor right back to the amp, the current through the capacitor does not pass through the woofer voice coil.

 

 

I find these explanations of current flow misleading, or I don't understand how capacitors work.  Current cannot pass through a capacitor...it's an open circuit. Current (or charge) is stored on the capacitor plates, therefore the current can only flow two ways.

 

From the "+" plate of the capacitor, through the inductor, through the amp, to the "-" plate of the capacitor.

 

or

 

From the "-" plate of the capacitor, through the amp, through the inductor, to the "+" plate of the capacitor.

 

The capacitor is storing current that would otherwise pass through the woofer voice coil.  It's not diverting that current back to the amp.

 

Am I correct?

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Let's look at a woofer circuit with 2.5mH series inductor and a 100uF shunt capacitor. Of course we know that adding the capacitor changes the low pass filter to a higher slope, from -6db/octave to -12db/octave. I am using a 1kHz signal with 10v input, of course we are trying to filter out 1kHz so it doesn't pass through the woofer. The capacitor is just 1.59 ohms at 1kHz, it's dominating as far as a load is concerned (compared to the woofer) and taking the majority of the current through it since it's the lowest impedance, current passes back to the source via - ground. So instead of the 1kHz making it to the woofer, the capacitor is giving it a path of lower resistance back to the source.

 

Green is voltage at woofer, it's less than 1v with a 10v input. Blue is current through the 100uF capacitor. Make note current is out of phase with the voltage.

 

 

capcurrent.png

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1 hour ago, captainbeefheart said:

DC current cannot pass through a capacitor.

 

Agree, but as I understand it no current can pass through a capacitor.

 

1 hour ago, captainbeefheart said:

AC current passes through

 

As I stated above, no current can pass through a capacitor.

 

1 hour ago, captainbeefheart said:

but is 90° out of phase with voltage.

 

Since a capacitor is an open circuit, current can only flow from one plate to the other via the circuit itself.

 

Current only flows when there is a change in voltage, and the quicker the voltage change, the greater the current flow.  This current flow is instantaneous, and is 90° ahead of the voltage change.  

 

An example that I would be afraid to try would be a capacitor is series with a driver.  Hook this circuit to a 12 volt battery.  The instant the battery is connected, current will flow through the circuit, including the driver.  The current flow will decrease until the voltage across the capacitor is 12 volts, then will stop.

I guess I wouldn't call this blocking DC.

 

Again, this is how I understand it.

 

Mike

 

 

 

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