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Determine the cut off frequency of any filter formed from output impedance and input impedance

Discussion in 'Electronic Repair Modifications DIY' started by Guitarfreak, Aug 10, 2011.

  1. Guitarfreak

    Guitarfreak Well-Known Member

    Say you have a piece of electronic equipment which has an input impedance of 1Megohms, and an input capacitance of 10 picofarads as determined by a ceramic capacitor across the input resistor. This obviously creates an RC impedance filter which will act on any AC entering the device. Now enter into the equation a line capacitance (cable and/or source capacitance) of whatever is entering said device. Is the line capacitance ignored until it goes higher than 10pF? Or does it at all times sum with the input capacitance of the device?
  2. MrEase

    MrEase Active Member

    You seem to have a few misconceptions here.

    First, an input capacitance of just 10pF is very unlikely to be created by a discrete 10 pF capacitor. Rather it will arise from stray capacitance of the input wiring and of the input device itself. In reality, you may well find input capacitances much greater than this as capacitors are often added (normally 100 pF up to 1000 pF) to prevent RF interference.

    Second, the 1MOhm and 10pF do NOT form an RC filter, obviously or otherwise. What DOES create a low pass filter is the output impedance of whatever is driving the input and the cut off frequency will be determined by the output impedance of the driver, the input impedance of the input and the impedance of the cable connecting them. Hoping to explain this a bit better, consider the case where the driver output impedance is actually 0 Ohms. It does not matter what load you put on this as there will be no loss of signal, hence no filter action. Of course in this case, input currents could become very large but who cares if the source can drive it.

    So on to the real answer to your question. To determine the the cut off frequency of any filter formed from output impedance and input impedance you must do the proper calculations. There is no way around this if, for instance, your input is from a passive guitar. The output from this itself will be a reactive impedance and the calculations depend on good measurement of the guitar. Sadly, as you move the guitar volume and tone controls, the output impedance will also vary. Altogether a quite complex scenario.

    If we consider the output from a pre-amp though, certain approximations can be made. Output impedance can usually be assumed to be purely resistive in the audio band and this can be quite easily measured and is unlikely to vary. Hence we would expect a figure of a few 10's or 100's of Ohms. Then add the input capacitance of the input stage to the capacitance of the cable (Yes, they sum at all times). The frequency where capacitive impedance equals the source impedance is your 3dB cut off frequency. This approximation assumes that the input resistance (1MOhm in your example) is very much greater than the source impedance.

    Hope that helps.
    audiokid likes this.
  3. Guitarfreak

    Guitarfreak Well-Known Member

    I appreciate the advice, thank you.

    Yes, the circuit in question is intended to be used with an electric guitar, which for the sake of this conversation can be assumed to be using a passive humbucking pickup with a coil impedance of 15k and a volume pot of 500k, no tones. The actual capacitor in use in the input of the floor device is larger than 10pF, I was just using it as an example. The proposed effect is low pass not high pass, which I believe will still be achieved by a parallel capacitor on the input. I do yet have a few questions.

    1. Now I can calculate the impedance Z seen at the input of the device as given by the network of volume pot resistance, input resistor, and input stage (gain device) all in parallel in order to find the ultimate Z load on input. From there I can calculate what value capacitor to use in order to gain a certain level of high end rolloff correct? (I know that this is simplified, but is this at least somewhat correct and applicable?) Should I include in this calculation of parallel resistances the impedance of the pickup coil as well? Or should that be left out of the equation?

    2. How do I go about calculating my guitar's reactance in order to make the same calculations? I have a standard voltage/current meter, but I have no way of measuring capacitance nor inductance. I have some electronics classes lined up for the next few semesters, but I'd like to jump right in and push further with a current project of mine, so I apologize for misconceptions based on personal experience and not textbook fact.

    3. I've received advice long long ago that it is unwise to put a capacitor directly across a guitar pickup because various and damaging things may occur due to parasitic oscillations. Does this apply in this case as well when we place a capacitor in parallel with the input resistor? I'd think not because input capacitance always exists even regardless of a capacitor being there or not, but I ask just to make absolutely sure.
  4. MrEase

    MrEase Active Member

    No problem.

    With a resistive source impedance and capacitance to ground you will always have a first order low pass function. When inductances are involved to ground though YMMV!

    You can't calculate the output impedance of the guitar from the DC resistance of the pickups and the pot. The pickup is a really a small AC generator and it's output impedance will not be equivalent to the DC resistance. Nor are different pickups all that similar (as you can tell by the variation of their DC resistance). A normal multimeter is not going to be able to measure the source impedance of the output voltage.

    The absence of tone controls helps somewhat but you will be unable to produce a consistent low pass filter as changing the volume control will vary the output impedance considerably. At full volume, the pot is a shunt load on the pickup, so the output impedance will be the pickup output impedance loaded by 500k (in your case). At half level, the wiper will have 250k to ground and 250k in series with the pickup. The output impedance will be considerably increased and this will cause a large reduction in the low pass cut off frequency and thus change the tone considerably. Let's say, for the sake of argument, that the pickup output impedance is 25k resistive. At full volume, the output impedance will be close to 25k. At half volume, the output impedance will be much nearer to 125k. This will give you a 5:1 variation in cutoff frequency. This is without taking in account changing you guitar lead from a 5 metre to a 10 metre... All told, trying to use a load capacitor in your amp to produce a low pass filter is a complete waste of time. This is why guitar tone controls are ahead of the output volume and we already know that people spend countless time changing pickups, caps and pots in their guitars in search of a sound they like.

    This advice makes no sense to me. To create parasitic oscillations (or any oscillations) you always need a feedback path with a gain of EXACTLY one at the oscillation frequency. This is simply not possible with passive electronics (there are ALWAYS losses), thus there is simply no possibility of oscillation. In the amplifier it could be a different question, however unlikely, as there IS gain there. Changing the value of a capacitor where parasitic oscillations do occur can sometimes be effective although intuitively you would only expect it to change the frequency of oscillation. There are though many other factors that can result in the new frequency not having enough gain to oscillate.

    This is getting far too advanced for you at the moment though. Keep up the learning, all your experimentation will stand you in good stead for the future and you are guaranteed to have moments when that lightbulb above your head suddenly lights up brilliantly!
  5. Boswell

    Boswell Moderator Distinguished Member

    Lots of points to you (GF) for going though this thought process, but no useful low-pass result is possible when you have a simple volume pot on the guitar. The output impedance of your guitar varies from zero (volume pot at minimum) up to a maximum of about 130K Ohm (volume pot around half way up). You can't usefully use capacitance loading to achieve a consistent low-pass filter when your source impedance varies like that. In fact, the opposite normally applies: you try to minimize the capacitance loading on the guitar output in order to avoid the tone altering as you change the volume pot.

    Through examining simple setups of this sort it becomes clear why pickup amplifiers mounted in the guitar are often used. The alternative is a short guitar lead to a stomp box that has a buffer on its input. After the buffer, all things are possible.
  6. MrEase

    MrEase Active Member

    Thanks Boswell. That's a much more succinct explanation than mine!

    I keep getting bogged down in trying to add some detail that might help. Too often I just bury exactly what I'm trying to explain with too many words... :<)
  7. audiokid

    audiokid Staff

    I'm the infamous king of that. You explained this better than I could as well. I would have probably taken a whole page to explain why I can't explain what I'm trying to explain lol.
  8. Guitarfreak

    Guitarfreak Well-Known Member

    I feel that I must thank you gentlemen, for although I cannot yet describe in my own words what makes it different, I now see (or hear) an improvement. I have moved my tone shaping cap from input to directly after the buffer, but keeping the same calculated cutoff frequency, and I have achieved better sound from it. Before it worked really well, I got a little bit of a darker tone, but it got spongy as well. Which is not a bad thing in some instances... Now with the cap post-buffer but still pre-gain, I get a tighter more direct sound, but rounded and shaped the way that I wanted it to be, and I think it's closer to my intended sound for having done so.

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