Why the parasitic inductance matters

So, what this parasitic inductance of 812 nH is going to mean for our resistance box? In simple terms, an inductor behaves like a frequency dependent resistor with AC signals, and the value of that "resistor" can be calculated using the following formula.

We call this value inductive reactance, but really is just a fancy name for the frequency dependent resistance that inductors have on AC signals. We just call it inductive reactance instead of resistance to prevent confusion.

So, with an AC signal of 10 Khz for example, the parasitic inductance will behave pretty much like a resistor in series with our resistance box with a value of 51 mΩ.

Which may not sound like a lot, but as we increase the frequency the inductive reactance will also be increased. For example, with an 1 Mhz signal it's going to be around 5.1 Ω, with a 10 Mhz one it will be around 51 Ω and with a 20 Mhz one it will be around 102 Ω! That means at high frequencies on the lowest ranges the inductive reactance is going to dominate, thus limiting the usability of our resistance box on the lower ranges.

Testing the impact of the parasitic inductance

To demonstrate that issue I set my resistance box to 50 Ω, while I had it in series with my signal generator which also has a 50 Ω output impedance, and I measured the peak to peak voltage while the frequency was at 1 Khz, 10 Khz, 100 Khz, 1 Mhz, 10 Mhz and 20 Mhz.

On my signal generator I also set the amplitude of the signal to be exactly 2 Vp-p, so, since we basically have two 50 Ω "resistors" in series forming a voltage divider, we would normally expect to see 1 Vp-p across the resistance box regardless the frequency of the signal.

But, since our resistance box is not perfect and it has an amount of parasitic inductance as we are going to see that's not actually going to be the case.

On the image below you can see the signal across the resistance box while its frequency is at 1 Khz:

Here while it is at 10 Khz:

Here while it is at 100 Khz:

And finally, here while it is at 1 Mhz:

As we see, the signal on all those frequencies is about 1 Vp-p, exactly as we expected it to be. But see what happens when we increase the frequency to 10 Mhz:

As you can see, the amplitude of the signal has started to increase, let's increase the frequency a little more, to 20 Mhz:

As you can see, the amplitude is now about 1.5 Vp-p instead of the theoretical 1 Vp-p. Since the output impedance of the signal generator is fixed at 50 Ω, this can only mean that the resistance of the resistance box has been increased by roughly 100 Ω!

And that makes sense, because if we calculate the inductive reactance of the parasitic inductance, which we have already measured to be about 812 nH, at 20 Mhz we get,

that it is about 102 Ω.

That means instead of just the 50 Ω resistance of the resistance box we also have an additional 102 Ω in series which comes from the parasitic inductance. So, using the voltage divider formula we can calculate that the voltage across our resistance box is going to be,

roughly 1.5 Vp-p, which gets validated from our measurement.