Analysis of Active Filter

Discussion in 'Homework Help' started by BenBa, Dec 10, 2013.

  1. BenBa

    Thread Starter New Member

    Dec 7, 2013
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    [​IMG]

    I am unsure how this filter works (i am in an intro to analog circuits class and we barely covered active filters).

    Here is my attempt at drawing the input voltage:

    [​IMG]

    you can see I used the 20KHz wave as an envelope for the 100Khz signal, drawing 5 full waves of the 100KHz within one half wave of the 20KHz, but i am not fully sure if this is the correct thing to draw as i have never worked with overlapping signals...

    As for the voltage at the point specified, i believe that all that does is cut off the bottom portion of what I drew as well as decrease the voltage by 0.7 volts (The standard voltage drop of a diode). But i am not sure if this is correct.

    For part B I have no clue how to approach this. I didn't know this filter had gain, i thought active filters just filtered...

    EDIT: For part a i just realized that there is a capacitor acting as ripple control, so the output should not just be the top half of the input, but how do i calculate this ripple control? I know that delta V = I_load/fC for this half wave recitifer, but is Iload the entire rest of the circuit or is it just the next resistor?

    Still super confused about part b, i know that DC level frequencies get eliminated because of the capacitor in the middle, but how do i determine the cutoff frequency? The op amps seems like all they are doing is buffering the signal and not actually filtering it.

    EDIT 2: I just realized my scaling is off, i should have 2.5 waves of the 100KHz within on half wave of the 20KHz, so 5 waves total over the full length of the 20KHz.
     
    Last edited: Dec 10, 2013
  2. #12

    Expert

    Nov 30, 2010
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    Remember your virtual ground principle. One end of R1 is connected to an op-amp input that is at________________________.

    I_load must be E/R. What is your rectified voltage going to be if you assume no impedance at Vin (the signal source)? Run the math for ripple voltage now.

    and, yes. It doesn't look much like a filter with the frequencies given.
     
  3. BenBa

    Thread Starter New Member

    Dec 7, 2013
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    the op amp input is at ground, so all the voltage drops across R1.

    I_load must be E over R, what is E here? My rectified voltage is going to be Vin-0.7 because 0.7 is the diode voltage drop and Vin is 2Volts peak to peak, so after the diode drop it is 1.3 volts peak to peak so Vmax is 1.3/2 = 0.65V. I dont know what you mean by "no impredance at Vin"

    assuming I have made the right decisions then I_load = 0.65V/1,000 = 65 microAmps?

    So our ripple voltage drop delta V = 65microAmps/(100,000 Hz * 0.015 microHenris) = 0.0433 Volts.

    So does that mean that at the peak of each of the rectified waves it will fall off 0.433 volts until it hits the next wave and gets brought back up to Vmax?
     
  4. #12

    Expert

    Nov 30, 2010
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    Part 1: true.

    Part 2: 2 volts (peak to peak), rectified is not 1.3 volts or 1.3 volts divided by 2. After the diode, the entire negative half is missing and you have some .3 volts of the positive half remaining. and you typed the resistor value wrong. R1 is 10,000 not 1,000 The math is right, the typing is off. Then the .015 is micro farads, not micro henris.

    all the rest: true.

    As for source impedance: If the 2 volts (peak to peak) was arriving through a one million ohm resistor, it would make a lot of difference in the answer, so we assume the 2 volts is arriving from a source with no resistance in the path.

    And, finally, if your teacher decides the diode only uses up .6 volts, he will think you are wrong, but you seem right to me.
     
  5. BenBa

    Thread Starter New Member

    Dec 7, 2013
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    So let me run these numbers through.

    after the diode rectifies the signal it is .3 volts at its peak.
    So the Iload is 0.3/10,000 = 0.3mA. Then the delta V =0.3mA/(100,000*0.015microFarads) = 0.02Volts.

    So if i drew it it would look like this


    [​IMG]
     
  6. #12

    Expert

    Nov 30, 2010
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    Your numbers are right. You need to get rid of that outer envelope line and remember that there are only 2 1/2 of the fast waves per half cycle. Then remember that there is a time gap where the 2 1/2 waves going in the negative direction don't exist.
     
  7. BenBa

    Thread Starter New Member

    Dec 7, 2013
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    So would this be a good image of the final wave form?

    [​IMG]
     
  8. tbinder3

    Member

    Jun 30, 2013
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    Between the two "humps" you will have a time gap where the negative part of your sinusoidal wave would've been.

    So draw the wave like it should, erase the negative part, your good to go. I thought your other graph looked better than this more recent blob of nothing.
     
  9. crutschow

    Expert

    Mar 14, 2008
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    Here is an LTspice simulation of the input circuit. The DC voltage goes to a little more than 0.4V, indicating about a 0.6V diode drop, and the sawtooth ripple is about 30mVpp.

    Edit: I don't know why this is called an active filter. :confused: All the op amps do are buffer the signal from a simple RC passive filter.

    Rectifed Wave.gif
     
    Last edited: Dec 14, 2013
  10. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    It uses active element, the op amp. Why the op amp is there, is probably to provide high input resistance or low output resistance so that not to load the next circuit.
     
  11. tbinder3

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    Jun 30, 2013
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    Just to go along a little with what shteii01 started. High input impedance, low output impedance, characteristics of an ideal op amp. This so-called "active" filter is classified as such because of the 2 active components, the op-amps. If you were to use say an RLC circuit, you could have a very nice "passive" filter, but of course, the amplification wouldn't be there, in fact an Ao of just 1, in just a basic RLC filter.

    Earlier you stated you were drawing the waveform for the input voltage, why thats important and all, comparing that with the output voltage is what you should be trying to do, to understand whats going on. Not only that, but draw waveforms after each "stage" or even after every element if necessary. Like I would draw the input waveform, a waveform after the first op-amp, and etc… Wherever you think it's important, but comparing those results I think is what's important here.
     
  12. crutschow

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    True. But that doesn't make it an "active filter". It's just a passive filter with buffering. For an active filter you need feedback in the frequency shaping elements.
     
  13. tbinder3

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    Jun 30, 2013
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    I agree with what your trying to say, but for educational purposes, it's not a horrible example. Intro to AC courses like to be specific, it's either this or that, with no debate (although I think debate is wonderful in open room class discussions). Since there is an op-amp, more than likely it is a passive filter to the textbook. And like I was saying in the post above yours, probably just used as an example for students to understand concepts there already working with, feedback, virtual ground, and filters in general.
     
  14. crutschow

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    Well, perhaps I'm being unduly pedantic but if you are being educated, then I would think the proper terms should be used. Otherwise it can lead to confusion later. The term "Active Filter" normally refers a circuit that uses feedback to generate higher order complex pole and zero filters without using inductors, not a first order passive RC filter with an active buffer. :)
     
    tbinder3, LvW and JoeJester like this.
  15. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    What book had this question?
     
  16. LvW

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    Jun 13, 2013
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    Full agreement. Knowing the terms is important for a good understanding.
     
  17. tbinder3

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    Jun 30, 2013
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    Like I said I agree with what your saying. Issue that rises in my head: What if they have no clue what a higher order complex pole is? Or even a buffer… Yes, they should know the correct terms and what those terms mean, but breaking that down to a simplistic definition that the reader can understand. And of course everyone's own definition is going to be different and reflects how that person understands it best.
     
  18. LvW

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    Jun 13, 2013
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    tbinder3 - OK, agreed.
    Thus, we can use the following description/definition:
    * In a passive filter the frequency-dependence is caused by passive elements only (R,L,C). Amplifiers - if any - are used for decoupling or amplification purposes only.
    * In active filters there are amplifiers with frequency-dependent feedback, which mainly determine the filter characteristic. It is the main advantage of these filters that inductors can be avoided.
     
  19. crutschow

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    If you agree with me then there should be no "buts". Whether "they" understand what an active filter is or not is beside the point. If they don't know the meaning then they need to learn it. It's fine to break it down to simpler terms to explain it but you don't change the meaning. Technical terms do not have arbitrary meanings as determined by someones "own definition". That's the road to technical anarchy. :eek: Technical terms have precise definitions that are generally agreed to by all to minimize ambiguity in conveying technical information.
     
  20. JoeJester

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    Apr 26, 2005
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    How about sticking with something that is published rather than your own.

    From Chapter 16, Op Amps for Everyone
    And the active filter ...
    Low frequencies were < 1 MHz according to the publication.
     
    Last edited: Dec 17, 2013
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