6 ms·
The bucket brigade device: An analog shift register
- throw_m239339 2y agoPopular way to make analogue delay effect pedals, in fact, that's how I heard about the expression first, while wondering how analogue effect pedals worked without some sort of tape system.
- brudgers 2y agoI found Moritz Klein's video on bucket brigade (audio) delays informative: https://youtu.be/4LjP5Y1yxXs?si=l2_dLDy6Cyf6_sGh https://youtu.be/4LjP5Y1yxXs?si=l2_dLDy6Cyf6_sGh
- rwmj 2y agoBurying the lede there! Can someone expand on the bit about how CCDs are like bucket brigade delaylines?
- equestria 2y agoIt's basically the same thing, except the initial charges in each cell are created through the photoelectric effect when the sensor is exposed to light. After that, it's the same conveyor-belt-style readout proces as for the BBD. It's also why CCDs tended to have problem horizontal or vertical halos around bright lights. Nowadays, most cameras use CMOS sensors, where the amplification and readout circuitry is integrated directly into each photosensitive cell.
- Cerium 2y agoCheck out the datasheet for this single row CCD sensor[1]. Look at pages 2 and 6, which show the block diagram and control signal waveforms. There is a single analog output pin. To read the device the "analog shift register" is used, which is a bucket brigade device to move the charge from the photo diode to the output buffer. [1] https://www.spectralproducts.com/pdf/ToshibaTCD1304.pdf https://www.spectralproducts.com/pdf/ToshibaTCD1304.pdf
- kazinator 2y agoA bucket brigade delay obtains a sample of a voltage on one end, and then shifts copy of that charge among the cells to the other end. The delay line as a whole contains a window of samples at any point, but that window is such is not accessed. A CCD line obtains a window of samples directly: the cells are charged according to light sensors. Then this window of samples is shifted out in the same way as a bucket brigade delay line. So the main differences are that the bucket brigade is clocked continuously, running nonstop and that only one of the cells obtains the sample. The CCD runs only briefly after an image has been exposed and capture has been triggered, in order to shift out the samples, and sampling takes place at every cell. A bucket brigade can be slowed down or sped up to change the delay. That affects the sampling rate and therefore frequency resolution. That's how we can create a chorus effect for musical instruments. So the clock speed is a direct functional parameter. Slowing down or speeding up the CCD doesn't make a difference to the result, except that there's likely an ideal range of rates, balancing between the time needed to accurately charge one cell to be equal to its neighbor, while being fast enough to avoid leakage.
- fecal_henge 2y agoA modern application of simmilar concept: https://www.psi.ch/en/drs https://www.psi.ch/en/drs
- deleted 2y ago[deleted]
- ggeorgovassilis 2y agoCan someone explain why this doesn't distort the signal? Eg. with all these capacitors one might think it would filter out higher frequencies?
- duped 2y agoIt does. It also obeys the Shannon-Nyquist sampling theorem when you take into account the clock frequency (you can't delay signals above half the clock frequency). It also exhibits nonlinearities that will create harmonic distortion. At least in the pro audio space, those qualities are actually desirable.
- atoav 2y agoThey can be desirable. If I want to normalize the group delay of a bunch of drum mics, I sure as hell not gonna use a BBD-Delay for it. But if I want a flavourful analog delay onna guitar track, why not.
- pierrec 2y ago>a nunch ofndrum mkcs "Am I having a stroke" comes to mind :) A bunch of drum mics? I'm no recording engineer, but I guess this is about mixing different signals from the same setup, with the signals having different delays (possibly due to filters, cables, or different types of mics)
- brudgers 2y agoDistance. Microphones at different distances receive a signal at different times. When combined, the signals are not phase aligned. This causes comb filtering where the combined waves reinforce and cancel each other. Though that is sometimes desirable (e.g. comb filtering is important for spatial ambience such as reverb). Some audio engineers find careful microphone placement solves many potential problems. Drums are notoriously hard because modern recording techniques use many microphones on a kit.
- atoav 2y ago
- anonymousiam 2y agoThere was also this popular (in the day) successor (sad-1024 dual analog delay line) https://www.alldatasheet.com/datasheet-pdf/pdf/1161976/ETC2/SAD-1024.html https://www.alldatasheet.com/datasheet-pdf/pdf/1161976/ETC2/...
- Eduard 2y agodo I interpret the screenshots correctly: that's a whopping 2.5 millisecond delay?
- alfiedotwtf 2y agoWhich is within the range needed for things like guitar pedals (flangers and chorus)
- nothacking_ 2y agoThis is just a 185 stage device, but later ones would have thousands of stages, and could be chained together for even longer delays. Sample rates could also be much lower then 40 kHz as low as 6 kHz was used for echo effects.
- amelius 2y agoIf you momentarily put two capacitors in parallel, the charge will divide between the two, right? So how does all the charge end up on the other side?
- rcxdude 2y agoThe other side of the 'input' capacitor is pulled down to a much lower voltage, which means the charge prefers to flow into it. (And then this alternates so the next time the one it is transferring charge to is pulled lower)
- amelius 2y agoI still wonder if all the charge moves from left to right, or if some of the charge stays behind. And how would that work in a CCD chip where thousands of positions are shifted through?
- rcxdude 2y agoFrom what I can find, it's something like 99.999% of charge moves, but it varies from CCD to CCD and can cause artifacts in bigger ones if it's too low. I couldn't find a great indication of what affects it, other than defects in the crystal lattice causing charge trapping, which apparently tends to be an issue in aging telescopes, especially space ones like Hubble.
- amelius 2y agoOK but what I mean is that if you tie two capacitors together, and one of them has a terminal at a much lower potential, then unless that voltage is minus infinity, then the other one will not be completely discharged.
- magicalhippo 2y agoCapacitors store a certain amount of charge, in that regard like a small battery. The voltage on the positive side is relative to the negative side. If you have a 5V supply with respect to ground (your reference voltage), and you connect a capacitor between it and ground, it will charge to 5V. If you then disconnect the capacitor and connect the negative side to the 5V supply, then the positive side will still be 5V relative to the negative side of the capacitor, but the positive side will be 10V with respect to ground. Now take a second capacitor. Let's call the positive sides of the two capacitors as p1 and p2 respectively, and the negative sides for n1 and n2 respectively. If you then connect a second capacitor in parallel with the first, so both n1 and n2 are connected to 5V, the charge will equalize between them. If you then connect n1 to ground, then p1 will be negative relative to p2 and charge will start flowing into capacitor 1 from capacitor 2. Using transistors as digital switches, you can do this "dance" of alternating where you connect the negative sides to and thus move the charge along.