For what it is worth I also came up with a relatively low impedance value also.
It was 25 ohms. Details, a 1KVA plitron 55-0-55 with dual primaries. The bridge rectifiers had 0.1uF across all four diodes so I used a 1uF capacitor in the snubber. Sorry no photos, my scope is an old CRT unit and barely visible to the naked eye!!! Time to bite the bullet and buy a new scope.
It was 25 ohms. Details, a 1KVA plitron 55-0-55 with dual primaries. The bridge rectifiers had 0.1uF across all four diodes so I used a 1uF capacitor in the snubber. Sorry no photos, my scope is an old CRT unit and barely visible to the naked eye!!! Time to bite the bullet and buy a new scope.
Yikes! If your injection capacitor Cx = 1 uF then the 4.7k drain pullup resistor "R2" won't be able to charge Cx all the way up to VCC. The available time is half of a 120Hz period, namely 4.16 milliseconds. Your RC timeconstant is 4.7 milliseconds, so your bellringer won't drive full amplitude bings and bongs into your bell (secondary). The damping will be correct and Quasimodo will still give you an optimum Rsnub, but you'll be using runt pulses. Eeeww.
You also won't be able to take advantage of the special feature I put into V4, just to help out folks whose analog scopes produce dim traces in Quasimodo experiments. It's a darn shame.
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I don't think "correct" is the right word, and in fact I didn't use that word in the Quasimodo design note. Search the document; I did.
In my opinion, zeta=1.00 is an appropriate target for beginners, math-averse folks, non-engineers, and people whose attitude is "just tell me what to do, don't ask me to think or to make calculations." A goodly fraction of these folks are quite brave and I admire their Can Do attitude; you'll find several posts on this thread from people whose first-ever use of an oscilloscope was to find the zeta=1.00 point of a snubber! Congratulations and bravo to each of you!
In my opinion, telling a beginner to aim for zeta=1.00 gives a comfortable margin of safety on both sides of the target. If, through operator error or mis-calibrated equipment, they end up with an Rsnub that in real life gives zeta=0.70 that's not the end of the world. The transformer secondary is still damped, and the "oscillatory ringing" waveform is still fairly well behaved; it's plotted in Figure 1 of the design note- take a look. If, through operator error or mis-calibrated equipment, they end up at a real life zeta=1.80 that's also not the end of the world. The transformer secondary is overdamped but not dangerously so, and the "oscillatory ringing" waveform is still well behaved; check out the plot to see.
In fact, I urge curious readers to build a SPICE simulation of a parallel RLC circuit, and to calculate the damping factor zeta of their circuit before running the transient simulation (the equation is in appendix A). Just how good-looking or how bad-looking is the zeta=0.5 waveform? Could you live with it? How about zeta = 0.75? Zeta = 1.50? Zeta = 2.00? Simulate them, look at the waveforms, and look into your heart.
For people much more advanced than beginners (I'll call them brown-belts and black-belts), whose first exposure to damped, second order, linear systems was long before they ever heard of a Quasimodo test-jig: y'all don't really need a recommendation from me. Pick a zeta that makes you happy, using whatever criteria you think are important. I have to imagine that eventually some diyAudio member, whose scope includes FFT capability, will twirl the Rsnub on their Quasimodo to get the deepest null of output harmonics. What zeta that corresponds to, I don't know or won't say. Here's a zany idea for black-belt designers of nosebleed audio products: Since the classic filter alignments each have a different Q (hence a different zeta), it might be excellent salesmanship to brag that your power conditioning subsystem's damping has Bessel alignment (or Cauer, or Butterworth, or ...). None of the reviewers at The Absolute Sound magazine have ever heard that before!
For intermediates, somewhere between beginner and brown-belt, you can shoot for zeta=1.00 and move on with your life; your transformer will work just fine. Or, if you wish, you can dive deeper into the world of snubbers and learn more. A few of the References in the Quasimodo design note, advocate damping your transformer's secondary at a different value of zeta than 1.00. Read them and find out why. Maybe you'll find their reasons irresistibly compelling. You can tinker with SPICE and teach yourself about snubbers; I recommend learning to use the ".MEASURE" feature as it's quite a time-saver. You can visit (The Snubber Design Website) and sign up for their email newsletter, if you so choose. They sell an e-book too; I haven't bought it (yet) so cannot offer an opinion.
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OK Mark,
I'll bite, what do you mean by a "runt pulse"? I'll assume you mean the lack of full charge, or in another words a Quasimodo who is too small to pull the bell rope all the way down. More like a ding-a-ling than a ding-dong Said it before, but I love your approach, extremely helpful.
Mike
I'll bite, what do you mean by a "runt pulse"? I'll assume you mean the lack of full charge, or in another words a Quasimodo who is too small to pull the bell rope all the way down. More like a ding-a-ling than a ding-dong Said it before, but I love your approach, extremely helpful.
Mike
Mike, put the Quasimodo pulse generator's output (node "DRAIN" in the schematic attached to post #1) on your oscilloscope. When Cx = 0.01 microfarads, the 4.7K pullup resistor has plenty of time to pull node DRAIN all the way to Vcc. When Cx = 1.0 microfarads, node DRAIN doesn't get all the way to Vcc and your pulses, they are runts. Photos below.
A quick and ugly kludge workaround is to reduce the pullup resistance, by temporarily tack-soldering a 1K resistor in parallel with the 4.7K R2. Leave this temporary resistor's leads at least 2cm long, so you'll have an easy time removing it later. OF COURSE reducing this resistance increases the power consumption of Quasimodo --- so if you're running Quasimodo off a 9V battery, be 5X more careful than before. You may find it easier to attach this resistor to diode D4 instead of R2. Consult the schematic; both are equally good.
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When I replicate your first image I get this:
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The three shots below are infinity, 22.6R and 15.6R
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Actually I owe Andrew a lot more. This little sub-discussion led to one of these classic "penny-drop" moments where suddenly my understanding of basic Ohm's law, generalised Ohms law, capacitor reactance, frequency filtering via capacitors and more suddenly suddenly took a big leap forward with a big "ah".
But, can I check the following? For a simple RC component, isn't the capacitor's reactance in series with its ESR and the resistance of the resistor and so the total impedance is simply their sum(Z= Zc + ESRc + R)? With Zc = 1/(j.2.Pi.f.C) a very small cap (0.15uF) means the reactance of the cap dominates total Z at 50Hz which is large and I=V/Z implies very little current. Is this correct? If the ESR=0 and R=0, the current across the cap (ignoring phase) would be V/Zc = 2.Pi.f.C.V = 1mA in amplitude (2x3.14x50x0.15nx15x1.414). This matches the Spice model. I need to refresh my memory on how to add the complex Zc to ESRc and R.
But, can I check the following? For a simple RC component, isn't the capacitor's reactance in series with its ESR and the resistance of the resistor and so the total impedance is simply their sum(Z= Zc + ESRc + R)? With Zc = 1/(j.2.Pi.f.C) a very small cap (0.15uF) means the reactance of the cap dominates total Z at 50Hz which is large and I=V/Z implies very little current. Is this correct? If the ESR=0 and R=0, the current across the cap (ignoring phase) would be V/Zc = 2.Pi.f.C.V = 1mA in amplitude (2x3.14x50x0.15nx15x1.414). This matches the Spice model. I need to refresh my memory on how to add the complex Zc to ESRc and R.
The (complex) current through a (complex) impedance is the (complex) voltage, divided by the (complex) impedance. Ohm's Law in complex variables.
In this case (resistor in series with nonideal capacitor), the impedance's real_part is the resistance of the snubber resistor plus the real_part of the ESR of the capacitor. The impedance's imaginary_part is (1 / (2 * pi * freq * Capacitance_in_farads)).
So it's a complex division: I = V / Z where the boldface vectors I, V, and Z are all complex numbers. The imaginary_part of the numerator is zero but the imaginary_part of the denominator is not zero. It's your decision whether to perform the calculation in polar coordinates, or in Cartesian coordinates. In this problem the polar representation requires 2 coordinate_transforms but the division is trivial. In this problem the Cartesian representation requires 1 coordinate transform but the division is quite messy. Take your choice.
(reference 1)
(reference 2)
(reference 3)
(reference 4)
(reference 5)
In this case (resistor in series with nonideal capacitor), the impedance's real_part is the resistance of the snubber resistor plus the real_part of the ESR of the capacitor. The impedance's imaginary_part is (1 / (2 * pi * freq * Capacitance_in_farads)).
So it's a complex division: I = V / Z where the boldface vectors I, V, and Z are all complex numbers. The imaginary_part of the numerator is zero but the imaginary_part of the denominator is not zero. It's your decision whether to perform the calculation in polar coordinates, or in Cartesian coordinates. In this problem the polar representation requires 2 coordinate_transforms but the division is trivial. In this problem the Cartesian representation requires 1 coordinate transform but the division is quite messy. Take your choice.
(reference 1)
(reference 2)
(reference 3)
(reference 4)
(reference 5)
Or, more simply, take the tip on pg 33 of Horowitz's The Art of Electronics and ignore phase in which case the division becomes easy.
Thanks for the references. I was just wanting to confirm that the total impedance components (cap reactance, ESR of cap and resistor) were indeed all in series and hence additive to get total impedance. I would have thought the addition of the three impedance components easier than the formula presented by Andrew but I will first go through the references posted by Mark.
(In any event, as Stormsonic noted, in this example the ESR of the cap and R are trivial and can be ignored for comprehension of the bigger picture.)
Anyway, apologies for the diversion. I am having to teach myself this stuff and I was getting horribly confused while reading (and re-reading) Horowitz's book. (I won't embarrass myself further by giving an example of just how confused!) Then, for some reason, when I read Andrew's straightforward first statement in post 271 it was as if all the tumblers clicked into place and the door opened. Thanks again!
Thanks for the references. I was just wanting to confirm that the total impedance components (cap reactance, ESR of cap and resistor) were indeed all in series and hence additive to get total impedance. I would have thought the addition of the three impedance components easier than the formula presented by Andrew but I will first go through the references posted by Mark.
(In any event, as Stormsonic noted, in this example the ESR of the cap and R are trivial and can be ignored for comprehension of the bigger picture.)
Anyway, apologies for the diversion. I am having to teach myself this stuff and I was getting horribly confused while reading (and re-reading) Horowitz's book. (I won't embarrass myself further by giving an example of just how confused!) Then, for some reason, when I read Andrew's straightforward first statement in post 271 it was as if all the tumblers clicked into place and the door opened. Thanks again!