27

You can simply forget about impedance matching for home Audio. Impedance matching is needed only where the wavelength of the signal comes close to the length of the cable transporting that signal. Electrical signals travel with almost the speed of light through cables, for the highest audio frequency (giving the shortest wavelength) the wavelength is about ...


26

Probably very little effect at all as long as the dimensions are small. Coming from the left hand side, there will be a reflection from point 'A' followed closely by an (almost) equal and opposite refection from 'B'. As long as the distance from 'A' to 'B' is small, these reflections will effectively cancel-out. As an example, let's say the impedance inside ...


22

I'll try to answer this briefly, but a great resource for this type of question is Eric Bogatin's Signal and Power Integrity -- Simpified. You've listed and described several very high speed protocols that have signal edge rates in the hundreds of picosecond range. What this means is that even traces of just a couple of inches can be considered as ...


20

If your power source is a zero ohm output voltage source, followed by a 50 ohm resistor, then yes, what you think is correct. However, practical RF amplifiers (at least ones designed to be efficient) are never built like that. They tend to have a low impedance common emitter or source stage followed by reactive impedance matching, all designed to operate ...


15

The input impedance of certain devices/circuits (transformers) does not neccessarily need to match their output impedance. Consider a 50Ω (or whatever impedance) antenna as transformer that transforms 50Ω (wire side) to 377Ω (space side). The impedance of the antenna is not (only) given by the impedance of free space but (also) by the way it is constructed....


14

From the looks of the new scope traces added to the question, specifically the Vcc trace, it appears that the ringing is originating in poor regulation of the supply at the point of use - most likely not at the bench supply output. While shorter leads from the bench power supply will certainly help by reducing lead inductance, that won't be enough when the ...


13

Do not confuse power transfer with voltage transfer. If the source is indeed 50 ohms, and your scope is set to 50 ohms, it becomes a voltage divider and what you see is half the voltage that you will see when the scope is set to high impedance. simulate this circuit – Schematic created using CircuitLab Power transfer has to do with the relationship ...


12

The idea is that signals propagate at a finite speed, that is to say a certain signal takes t time to get from one end of the transmission line to the other line. The cable also has some intrinsic capacitance/inductance per unit length, which can be approximated with a characteristic impedance (assuming loss-less): \begin{equation} Z_0 = \sqrt{\frac{L}{C}} \...


12

If you don't have access to a \$50 \Omega\$ feedthrough terminator, a T-connector with a \$50 \Omega\$ termination should work just as well. This was standard practice at my old company when we used external hall-effect current probes with our \$1 M\Omega\$ scopes. (Aside: I personally prefer using external \$50 \Omega\$ terminators even if the scope ...


11

Very few do. If your chip's datasheet does not say it has 50-ohm termination, then it almost certainly does not. Traditional CMOS and TTL logic does not provide matching termination, though a few specialized types (line drivers?) might. Typically drivers are low impedance and receivers are high impedance (with some capacitance). Traditional ECL (emitter-...


10

Soldering directly to the pads of the SMA connector instead of using SMA connectors should work fine electrically. The obvious problem is that the cable is no longer removable. If you're fine with that, then go ahead. It would be good to have some mechanical strain relief if this has a chance of getting flexed. At only 2 inches wavelength, you do need to ...


10

No, the 50-Ohms is not a convention for PCB tracks to carry signals. The 50 Ohms is a standard for coaxial cables and corresponding interconnects - dozens and dozens various SMA/SMB, BNC, type-N, etc. connectors. In fact, typical (thin) PCB traces have 65 - 80 - 100 Ohm characteristic impedance on a typical stack-up (7 mils or 12 mils of FR4 between ground ...


10

how does my signal source see the whole cable? The characteristic impedance (\$Z_0\$) of any transmission line be it coax or twisted pair (screened or unscreened) is determined by: - $$Z_0 = \sqrt{\dfrac{R + j\omega L}{G + j\omega C}}$$ Where R, L, G and C are the resistance, inductance, conductance and capacitance of the actual line per unit length. ...


9

The characteristic impedance of any cable at high frequencies is determined by the inductance per unit length and the capacitance per unit length. It should not to be regarded as a conventional lossy resistor - characteristic impedance is simply the impedance that the cable should ideally be terminated with to prevent reflections. So, reflections happen ...


9

The frequency is low enough for measurements with oscilloscope. Add a small few ohms resistor in series with your antenna. Be sure that the resistor is not a coil. It must be resistive at the operating frequency. Feed with a signal generator a sinewave into the antenna at the right operating frequency. Measure at the same time the voltage and the current ...


9

All the answers name some valid points, but they fail to really answer the question which I want to repeat for clarity: Why is 50 Ω often chosen as the input impedance of antennas, whereas the free space impedance is 377 Ω? The Short & Simple Answer These two impedances have no relation at all. They describe different physical phenomena: the antenna ...


9

RF amplifiers do NOT in general have an output impedance remotely close to 50R..... They are however designed to drive a 50R load! Much like audio amplifiers the source impedance is generally far from the design load impedance, because you DONT want maximum power transfer, you want something closer to maximum efficiency! Depending on the topology, the ...


8

The 50Ω standard is basically just convention. There are various stories about how 50Ω came to be chosen. The article Anindo linked is good. There is also The History of 50 Ω or There’s Nothing Magic About 50 Ohms. But the long and short of it is that it is a compromise between low attenuation and power handling. But it became the standard impedance when ...


8

Forget about clock or signal frequency. Think about edge rise time instead. A perfect square wave of, say, 320 Hz actually contains much higher frequency components: You can see frequency components going right up to 2000 Hz (and they go beyond too). But if you slow the rise time of the square wave, then you actually remove these high frequency components. ...


8

What causes the reflection? Why doesn't it occur with other frequencies? (if it doesn't) Reflections occur at all frequencies when there is a mismatch in impedances. At low frequencies, such as audio, these reflections are difficult to see but they are there all the same. Reflections are generally said to be significant when the frequency is high enough ...


8

Unless you want to do your own S-Parameter de-embedding mathematics, you must fit a 50\$\Omega\$ connector to at least one end of the trace. You can either fit a connector to the other end, or a good quality 50\$\Omega\$ resistor. I tend to use 2 x 100\$\Omega\$ resistors in parallel for lower ground inductance. There are many connector styles to choose ...


8

Impedance matching is tricky, but the role of a quarter wave transmission line is to map from one impedance to another. The actual impedance of the line will not match either the input or the output impedance - this is entirely expected. However at a given frequency, when a correctly designed quarter wave line is inserted with the correct impedance, the ...


8

Option 3 is best It does not matter whether you connect the primaries together or not, you are still stuck with the primary voltage appearing on pins 3 and 4 as a result of transformer action, and still stuck with that interacting with whatever parasitic capacitance exists. At higher frequencies, the leakage inductance will tend to decouple winding 34 from ...


8

The rule of thumb I use is that anything exceeding 1/20 of the wavelength is to be considered a transmission line. And bad terminated transmission lines have reflections that distort the signal. To get a fast approximation of the wavelength, I consider that the speed of a signal is half the speed of light (based on experience with PCBs) and that the speed ...


8

The "terminators" of a CAN bus serve two purposes: They terminate the transmission line. The type of CAN you are referring to is intended to be implemented as a twisted pair. 120 Ω is roughly the impedance of such twisted pair. By terminating the ends with the characteristic impedance of the cable, reflections from the ends are minimized. They ...


8

designed to operate at audio frequency or to match impedance? Both. The specification of a transformer covers both its frequency range, and its turns ratio. It gets more difficult to design a transformer as its frequency range increases. The low frequency end is inductance limited, you need enough turns, and a high permeability core to increase the value ...


7

I'm writing this as an answer because I didn't think there would be enough room in comments. Having said that, it's likely that several of the points I'm making could be the cause of your problems: - Are you using a x10 scope probe? What does the output from pin 2 look like - schmitt triggers will not all trigger at the same point on a badly shaped ...


7

Per other answers and comments, I focused on bringing the overshoot down with some of the suggestions provided. I did the following: shortened the leads going to and from the breadboard, adjusted compensation on the probes (one was slightly under compensated) This reduced measured overshoot from ~2.4V to 1.8V (over 5V). @AndrejaKo's suggestion had the ...


7

Impedance matching is NOT one of the major applications of a common collector circuit. Why should a transistor be able to match an impedance? Impedance matching is done with passive components. A common-collector circuit is good at providing a high input impedance (to a weak signal) and generating a low output impedance (at the emitter) - it is a power ...


7

That sounds like someone gave you a "rule of thumb" for doing series termination. The purpose of a series terminator, placed close to the source or output of the chip, is to match the output impedance of the driver to the impedance of the trace on the board. A lot of times people don't bother / can't figure out the output impedance of the driver so they ...


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