Advice on BNC cables for beginners electronic lab
Do I need to worry about impedance when connecting the equipment together and buy (for example) 50 ohm cables or would straight through work OK?
You seem to have the false impression (see my highlight in the above quote) that a 50 ohm cable will incorporate 50 ohm resistors. A 50 ohm cable (coax or otherwise) is telling you about the impedance it will present when using RF signals (greater than 1 MHz) and has nothing to do with real resistance values measured using a multimeter (for example).
All cables have a characteristic impedance (including "straight through" cables) and its impedance is represented by the formula: -
$$Z_0 = \sqrt{\dfrac{L}{C}}$$
Where L is the distributed inductance per metre and C is the distributed capacitance per metre. If you looked at some cables in their data sheets you might see typical values such as: -
- L = 0.25 μH per metre
- C = 100 pF per metre
Take the square root of the ratio and you get 50 ohms (nothing to do with actual resistive components).
Standard lab equipment such as signal generators usually use 50 ohm impedance so BNC cables between equipment should also have 50 ohm impedance. However, not all scopes have built-in 50 ohm termination so external BNC T junction and 50 ohm terminator may be useful.
A cable with a 50-ohm impedance will respond to transients that are fed to it the way a 50-ohm resistor would during the time it takes a signal to pass down the cable and back. If whatever's connected to the far end doesn't behave like a 50-ohm resistor, the cable will behave as though the cable injected as a transient the difference between the voltage change produced at the far end, versus the voltage change that would have appeared with a 50 ohm resistor. There are three general ways one may use such a cable:
Drive a load that behaves like a 50-ohm resistor, with minimal impedance at the source end. The far end will absorb all of the energy sent down the cable, resulting in a clean signal. If the far end impedance isn't exactly 50 ohms, some of the signal will get reflected, and most of that will get re-reflected at the source.
Drive a load that behaves like a 50-ohm resistor, but also have a 50-ohm resistor at the source. This will cause the signal to be attenuated by 50%, but even if some signal gets reflected because the load impedance isn't exactly 50 ohms, most of that signal will get absorbed by the source rather than being re-reflected.
Ignore the cable impedance. This will result in signals getting reflected at both ends of the cable, but if the cable is short and the signals aren't too fast, the reflections may die down fast enough that they don't pose a real problem.
The second approach gives the cleanest signals, at the cost of 50% attenuation. The last approach is simplest, and is often "good enough". For many purposes, the first ends up being a good choice.