V1=10 V(peak to peak), frequency of voltage source = 1kHz, D1 is a standard 1N4007 Silicon Diode, R1=10 kilo ohm (the load resistor)
In our electrical lab, we were given the circuit shown in the diagram.
I haven't been able to perform this lab due to the pandemic. So I don't have a clear idea of the equipment and what the question means. The question is:
Assume that a BNC cable is connected across the resistor(R1) and fed into the DSO. To get the best possible bandwidth (limited by the scope), we should pick an impedance of 10 k Ohm for the BNC cable?
Yes or No
Could you explain the terms used and what the answer would be?
What I know: In practice, there are 50 ohm and 75 ohm BNC cables.
DSO is an instrument that is used for continuous measurement of properties of signal like voltage and current(unlike the multimeter which reports average values of AC currents). It has a certain range of frequency of input signal that it can measure.
Best Answer
The characteristic impedance of coaxial cable is given by the equation $$ Z =\frac{60 \Omega}{\sqrt{\varepsilon_r}} ln{\frac{D} {d}}$$
Setting $$\varepsilon_r = 1$$
and
$$ Z = 10000\Omega$$ we get D/d = 2.4e72.
So if we set the diameter of the central conductor to 1 mm we get an outer diameter far beyond the diameter of the known universe. I think that such a coaxial cable is not available in your lab.
You also should keep in mind that your instructor is intentionally misleading you by introducing the resistor R1 with the same resistance as characteristic resistance of the coaxial cable he is writing about. The intrinsic resistance of your source is given by your ideal voltagesource (when neglecting the diode) and is therefore zero. It is not 10 kOhm.