Why 50 Ohm Impedance Doesn't Really Matter At Low Frequency

I was watching a video on Youtube on how to measure ESR for capacitors (youtu.be/115erzCCxgE) and it mentioned something interesting.

It mentions the circuit does not have to be terminated with a 50 Ohm termination because of the low frequency.

Can anyone explain why?

Thanks,

Good question. I think it is more useful to think in terms of wavelength instead of frequency. If the size of your circuit is much smaller than the wavelength, than ordinary circuit theory can be used and impedance of the line is not a big factor. When the circuit approaches or exceeds a wavelength in size, then the wave nature of the signal becomes important because voltage can now fluctuate across a line. So impedance at low frequency can be important if you have long wires involved.

Scattering parameters which 50 ohm terminations are relevant to them are used in high frequencies where you cannot measure impedance or admittance parameters accurately due to the difficulty in broadband short and open tests. At low frequencies, where you can measure Z matrix of a network, there is no need for S parameters.

Also, in fact 50 ohm termination is not something special that you must obey. It's related to the average impedance of a coaxial cable for the lowest loss case (77 ohm) and the highest power handling capability case (30 ohm).

Because at that frequency the parallel impedence Xc is very very large and the series impedence Xl is very very small. I.e the inductance and capacitance have next to no effect.

impedance matching is requred to maximize power transfer. in this case, it is not necessary. he is measuring the voltage. also oscillator probe impedance is quite high. he is looking the dc part of the signal on the capacitor. he is not interested in AC voltage.

impedance mathcing is important if you want to transfer power. for example, we do not care impedance match at the opamps, because voltage transfer is important for us.

if the frequency is low, then the length of line between the load and the transmitting device is "electrically short". Under this boundary condition, the entire circuit can be approximated by lumped elements.

In this case you would terminate the circuit in the lumped element load that gave the most transmitted power, which could be any random impedance, such as 27 + J13 ohms, etc., depending on your specific device.

Only IF you have an electrically long line between source and load, then you match your device to 50 ohms, and then at the end of the long transmission line terminate it in 50 ohms resistive load. That way you do not have huge standing waves formed on the transmission line (which cause ripples in output power vs. frequency, and cause inefficient power transfer)