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Capacitors in parallel effects on frequency

Discussion in 'General Electronics Discussion' started by Agriias, Oct 29, 2014.

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  1. Agriias

    Agriias

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    Oct 16, 2014
    Hi all,

    I have a question on capacitors and signal chains. If I have a signal that goes through two capacitors in parallel of different values such as .01uf o and .005nf what would be the result?

    I understand that the lower .005nf capacitor would act as the path of least resistance to the higher frequencies, but does it short out all frequencies above its cutoff leaving only the remaining frequencies below that to go through the other capacitor while blocking anything below it as well?

    Another way of stating it is does the frequency essentially split off in both directions with redundancy through both capacitors?

    Also another question: if they were equal in value they would add together and have a mutual lower frequency cut off correct? with the same frequencies going through both capacitors?

    Thanks,
    Agriias
     
  2. Gryd3

    Gryd3

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    My understanding of Capacitors in parallel, is that their capacitance is added together. Although I am curious to see other responses, as I am under the impression the new capacitance value would be used to determine frequency response. I'm here to learn though, so don't take that as the correct answer until we get more info!
     
  3. hevans1944

    hevans1944 Hop - AC8NS

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    Capacitors, by themselves, don't have a lower cutoff frequency. The reactance of a capacitor varies continuously as a function of frequency and can be found, at any given frequency, by means of this formula: Xc = 1 / (2 pi f C), where pi is about 3.1416, f is the frequency in hertz, and C is the capacitance in farads. Two capacitors, of any particular values, connected in parallel are the same as a single capacitor whose value is the sum of the two capacitor's individual values. Makes no difference if one is much larger or smaller than the other: the two in parallel act as one. So if C1 = 0.01 μF (10 nF) and C2 = 0.005 nF, then in parallel they are the same as a single 10.005 nF capacitor.

    There is no such thing as "the path of least resistance" at any frequency. Circuits obey Ohm's Law and Kirchoff's Law and current and voltage are what they are.

    Of course with real capacitors other factors may come into play as a function of frequency. Real capacitors have inductance, an effective series resistance (ESR), and a dissipation factor or loss tangent that is of particular importance at high frequencies. These must all be taken into consideration before making statements about "cutoff frequency". Capacitors are often used in filters to create low-pass, high-pass, band-pass, and band-reject (notch) frequency characteristics. In none of these filters is there a specific "cutoff frequency" but rather a continuous change in what frequencies are passed and what frequencies are attenuated. It is quite common to parallel a "trim" capacitor with a specific calculated (or measured) value to make the total value of the sum more closely match what the circuit needs to perform to specification.

    Sometimes a lower-valued capacitor of different construction is paralleled with a higher valued capacitor, typically an electrolytic, to take advantage of a lower ESR that the lower-value capacitor has over the higher-value capacitor. Therefore you see electrolytic capacitors, which have a relatively high inductance and ESR, used in power circuits but "bypassed" with ceramic or metal film capacitors of lower values and lower inductance and ESR.
     
  4. shrtrnd

    shrtrnd

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    Yeah, I don't know if this is a classroom theoretical question or a practical application question, but hevans1944 said what matters in practical application.
    The composition of the capacitor at higher frequencies is what matters.
     
  5. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    I don't know if the question about the splitting of current had been answered. But if you have say two capacitors (ideal) one valued at 100 uF and the other at 10 uF connected in parallel and a 1 Volt 1 KHz sine wave is applied across them .The circuit current is split between the two at a certain ratio dependant on their values. So in this case there will be a 10:1 current ratio. The 100 uF displace 10 x the current of the 10 uF.So the max approx. current displaced in the 100 uF is 633 mA and in the 10 uF 63 mA.
    Thanks
    Adam
     
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  6. Agriias

    Agriias

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    Oct 16, 2014
    Let me clarify with a further example:

    I have a input signal coming off the plate of a tube amplifier and it travels through two capacitors joined together at a node before splitting off. One capacitor leads to a tone control/ pot and the thev capacitor leads to the volume control.

    Each capacitor is of a different value with the higher .02 UF only blocking low frequencies and the .02nf blocking a higher level of frequencies. I am to lazy to figure out what the cutoff is..

    Anyways. the signal goes through both tone and volume here it is recombined before heading to the next output stage.

    Lets say both capacitors can allow frequencies over 10k hertz to pass easily. when the signals recombine are they essentially doubling up into 2 -10k frequencies? OR back to my originl question does the capacitor that is .02nf (the lower capacitor essentially short circuit all the frequencies over its cutoff and leave only what is left to go through the .02uf capacitor?

    The end result in the first scenario where the frequences are redundant would seem to be one where the higher frequencies are essentially doubled or boosted, in the second scenario we are simply segregating the higher frequencies in a filter before recombining with the remaindr of the signal that wasnt passed.

    On another note, i get that the current will be calculated by combining the total capacitance of tresistors to determine voltage and current passing through each, my question is more about filtering and frequency.
     
  7. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Let me clarify, it's not just the capacitor that determines this as Hop (Hevans1944) mentioned above, it's the resistance also that needs to be included. To have a defined cut-off you would normally have a resistor also in the circuit. The cut-off frequency, because I can be bothered is 1/(2*pi*R*C) in Hertz or 1/(R*C) in Radians per second. Or if you have only one RC you can use the single pole transfer function if you find it easier.

    [​IMG] .

    [​IMG] .

    [​IMG] .

    Thanks
    Adam
     
  8. Agriias

    Agriias

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    Oct 16, 2014
    I get how to calculate the cutoff frequency 1/ (2 pi f c) but that gives me the reactance in ohms for a given frequency.

    for instance if I put 100hz through a 1uf capacitor the resistance is 1.59ohms. this isn't very much so it will pass all audible frequencies.

    Why do you need a resistor to have a defined cutoff? If you lower the capacitance the reactance increases versus that original 100hz. If the reactance is high enough versus a given frequency wouldn't the ohms inherent in that reactance be enough to outright block the frequency?

    Also in the example in my last post lets say that the pots are 250k ohm of variable resistance.


    "Capacitors are often used in filters to create low-pass, high-pass, band-pass, and band-reject (notch) frequency characteristics. In none of these filters is there a specific "cutoff frequency" but rather a continuous change in what frequencies are passed and what frequencies are attenuated. It is quite common to parallel a "trim" capacitor with a specific calculated (or measured) value to make the total value of the sum more closely match what the circuit needs to perform to specification."

    I think this is getting at what I am trying to comprehend. So does a filter section with low pass and high pass filters essentially just drop the voltage/current of the passed frequencies in relation to the original signal before they merge again to go to the final output?

    Thanks for all the answers guys

    Agriias
     
  9. Gryd3

    Gryd3

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    Jun 25, 2014
    I think you need to post a sample circuit / diagram.

    There is many different types of filters that can be built, and many different ways to build each one.
    There are some common traits, but rather than guessing what you are imagining, show us a picture.

    There is truth to what you say. A capacitor will attenuate lower frequencies, but there is usually other components in play when this behavior is wanted.
    They also have a common use as coupling capacitors to join or pass an AC signal from one portion of a circuit to another.

    I am also unsure by your initial post, if the two capacitors are connected completely in parallel to each other... or if you have two parallel circuits, each with it's own coupling capacitor.
     
  10. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    You Said
    "Why do you need a resistor to have a defined cutoff? If you lower the capacitance the reactance increases versus that original 100hz. If the reactance is high enough versus a given frequency wouldn't the ohms inherent in that reactance be enough to outright block the frequency?"

    Here's why, with no other load you won't get any attenuation. Here is 1KHz 1 Volt sine wave. You work out the XC.
    Adam

    caps2.JPG

    caps.jpg
     
  11. Agriias

    Agriias

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    Oct 16, 2014
    ok.

    I am getting for C2 = Xc = .16 Ohms C1 = XC = 159K Ohms

    Thus C1 would totally block the 1KHertz frequency.

    I find this interesting if I use ohms law and put 1v/.16 it is giving a current value of 6.25?????? This makes no sense. I can figure the current by putting C1/C1+C2 and then multiplying the sum by 6.28E6 which comes to 1.00531-E6... essentially nothing?

    the current is 1v/159,154 = 6.28-E6

    This example is extreme in the values range 1uf-1pico in that one side is essentially blocking all flow and the other side is passing it all. so there is no real filtering I think.

    I'll work on the picture and post what I am talking about.

    Thanks
     
  12. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    Try 159 MOhm and not 159 KOhm. But as you can see both are passing this frequency without any attenuation.
    Adam
     
  13. Agriias

    Agriias

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    Oct 16, 2014
    I don't follow. How could the 159M ohm pass the frequency when the reactance is so high? Wouldn't only the 1uf capacitor pass the frequency given its small reactance?
     
  14. Gryd3

    Gryd3

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    Jun 25, 2014
    Because we have no 'other' resistance to account for.

    There will be no voltage drop without current flow. This voltage drop is what attenuates the signal.
    You need additional components, you need a circuit.
    There are exceptions, but in this case, you should draw up a diagram so we can provide specifics.
    Otherwise we will spin our wheels and smoke may emerge from your ears.
     
  15. Agriias

    Agriias

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    Oct 16, 2014
    Ok... so this was my first time using this editor that I found. I am a novice with schematics so bear with me. So assume this is an old school tube amplifier for a guitar. And the 500hertz frequency is coming off the plate into the filter section. In this case are we simply attenuating the signals amplitude in relation to each other?

    If I used a more traditional filter you could shunt the unwanted frequencies to ground or return and hence bypass the speaker and get rid of them that way?

    schemeit-project (1).png


    Sorry for all the confusion I am new to all of this.
     
    Last edited by a moderator: Oct 30, 2014
  16. Gryd3

    Gryd3

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    Jun 25, 2014
    Not a problem, thank you very much for producing a diagram for us.
    This looks like two high pass filters in parallel to me. C1/R1 and C2/R2.
    In this example, the capacitors themselves aren't in parallel with each other, but the High-pass filters are ;) This changes the question a bit.
    You have already demonstrated that you have the skill to determine the impedance for any given frequency for a capacitor (even though you missed a few 0s ;))
    Why not work out the cutoff frequency for each highpass.
    At first glance it appears to simply mix the output signals from each filter to the speaker.

    Of course, there is always more for me to learn so I could have missed something.
     
  17. BobK

    BobK

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    Say you connect a 1M resistor to the + terminal of a 1V battery. What is the voltage between the other end of the resistor and the - terminal?
    How about a 1K resistor? A 1Ω resistor?

    Answer to all: 1V.

    You need two resistors to make a voltage divider.

    An RC filter is voltage divider in which the capacitor acts like one of the resistors and a resistor is used for the other. The capacitor has a reactance which is dependent on frequency. The resistor has a reactance that is not dependent on frequency. You cannot determine a "cutoff" frequency without knowing both the resistance and the capacitance.

    And, by the way, cutoff frequency is not where the signal is fully blocked (it never is.) The cutoff frequency is where it is down by 3dB meaning it is still 70.7% of the original signal. I.e. for a high pass filter, you can think of frequencies above the cutoff as being barely attenuated and anything lower will be increasingly attenuated, going down by half for each halving of the frequency.

    Bob
     
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  18. hevans1944

    hevans1944 Hop - AC8NS

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    I really do not like, and mostly avoid, circuits where potentiometers are used "backwards" and behave mostly like rheostats. I like to take the signal from the wiper of a potentiometer, not inject a signal into the wiper. But that's just the way I learned to use potentiometers. Your Mileage May Differ (YMMD).

    In the circuit @Agriias posted (which must be just a fragment of the actual real circuit), the so-called tone control will "short out" the audio signal from the plate of the amplifier tube through C1 when the wiper is rotated to the "ground" end of the potentiometer. Depending on the impedance of the plate circuit this may or may not have much effect on the speaker output volume because the reactance of C1 is fairly high at audio frequencies. The volume control does the same thing, shorting out the plate circuit of the amplifier tube when the potentiometer wiper is rotated to the "ground" position, but I guess this is okay if you like doing things that way. And of course you can't drive a speaker with both of its terminals shorted together. Yeah, I know... picky, picky, picky... and just when things were going along so well. I suggest Googleing (is that a word?) "audio tone control circuits" to get some idea of the many different ways people think tone controls should be implemented. Unfortunately, you will not find anything there on digital signal processing (DSP), which is the modern way to do it, but all of the analog circuits involve capacitor and resistor networks in various interesting arrangements, some quite clever.

    But getting back to the OP's original question:
    You must discard your idea that capacitors are somehow "frequency selective" with upper or lower cutoff frequencies. They are frequency selective only in the sense that their capacitive reactance is an inverse function of frequency: higher frequencies yield lower reactance and lower frequencies yield higher reactance. But reactance is a continuous function of frequency. There is no particular frequency that would result in an abrupt "cutoff" in frequency response.

    Consider the following circuit: a capacitor connected in series with a resistor with an AC excitation voltage applied to this series combination. Measure the signal across the resistor (using whatever means at your disposal) and you will find that the signal increases with increasing frequency and decreases with decreasing frequency. At no frequency (except DC, which is not a frequency) does the signal totally disappear. At sufficiently high frequencies the signal will essentially be the same as the excitation signal because the capacitive reactance becomes a vanishingly small value compared to the resistance, which of course does not vary with frequency. At sufficiently low frequencies, there is a vanishingly small signal because the capacitive reactance becomes very large compared to the fixed resistance. Nowhere in between does the signal across the resistor vanish because a cutoff frequency is reached.

    Then consider this: you can get the same frequency response from this resistor-capacitor circuit with ANY value of capacitor as long as you make the product of resistance (in ohms) and capacitance (in farads) the same. So, for example, a 1 μF capacitor and a 1 MegΩ resistor will yield the same frequency response, measured across the resistor, as a 10 μF capacitor and a 100 kΩ resistor, or an 0.1 μF capacitor and a 10 MegΩ resistor. The particular values you might select would depend on the characteristics of the following circuitry. The impedance of a 1 μF capacitor in series with a 1 MegΩ resistor is much lower than the impedance of an 0.1 μF capacitor in series with a 10 MegΩ resistor. You can verify this using an on-line calculator. Just set the value for L to zero and plug in values for R and C.

    73 de AC8NS
    Hop
     
    Last edited: Oct 30, 2014
    Arouse1973 likes this.
  19. Agriias

    Agriias

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    Oct 16, 2014

    Thanks Hevans! I get it now. Yes, I was under the assumption that the frequencies vanish or are totally removed as opposed to attenuated to have less volts/current.

    Why is it that in diagrams of a high pass filter they often show the capacitor before the resistor? wouldn't it make sense for it to be before the capacitor so that the DC energy that gets blocked has a path to return? Also, When they display low pass they always have the capacitor passing the higher frequencies out of the signal chain before the resistor..
     
  20. Arouse1973

    Arouse1973 Adam

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    Dec 18, 2013
    The resistor after the capacitor is what is causing the attenuation. Think of it as being a frequency sensitive voltage divider. In a high-pass filter you are not interested in DC because this is blocked by the capacitor apart from a small leakage current. The resistor the other side of the capacitor is what is supplying the return path for the displaced current.
    Adam
     
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