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I plan to make a high-speed resistive probe according to the book "High-speed digital design: A Handbook of Black Magic". There is one thing I don't understand there:

Why this probe (unlike standard 10:1 passive probes), does not need to be AC compensated by adding a capacitor in parallel to the 950Ω resistor?

According to my simulations, not adding this capacitor reduces the bandwidth to only a dozen MHz (screen below). I understand that from some frequency the wire stops looking like 100pF capacitance and starts looking like 50Ω resistance. From what frequency does this happen, and is there no signal attenuation below this frequency as in the simulation below?

enter image description here Top trace is Impedance, bottom is output signal

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4 Answers 4

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Cables are only capacitive (or inductive) equivalent at low frequencies. Your model must be extended to include the (new) frequency of interest. In other words: the cable will be some higher-order LC lumped equivalent -- or a transmission line most generally.

The frequency is determined by the electrical length. A 1m cable at 67%c is 0.2 m/ns, or for 1m, 5ns. So the λ/4 cutoff is a 20ns period, or 50MHz. Inductance or capacitance will be a good approximation at frequencies much lower than this, say λ/10 or below (<20MHz). Inductance manifests for Z < Zo, and capacitance for Z > Zo (where Z is the system impedance, and Zo the characteristic impedance of the cable).

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As discussed by Tim, you have to model your coaxial cable as a 50 Ohm transmission line (at least I hope you are using 50 Ohm coax). There's a bit more to the story however.

The question is what the input to the scope looks like. If you are using a scope that has no internal 50 Ohm termination and you add your own terminator externally, then what you have is more or less correct (we are ignoring the extra transmission line parasitics from having the terminator external to the scope). Also, some lower end scopes with internal 50 Ohm terminators also simply put 50 Ohms in shunt with the original 1Meg/16pF input.

But this arrangement causes an inherent bandwidth limitation. Suppose you considered the ideal scenario in which you drive the scope input with a 50 Ohm source impedance but without any cabling between the source and the scope. You would essentially form an RC network with an RC time constant at RC/2, where R is 50 Ohms and C is 16 pF. The pole sets an inherent bandwidth limitation at around 400 MHz for these numbers.

For a well designed high frequency scope, the 50 Ohm internal termination will not have 16 pF in parallel with it. It is perfectly reasonable to expect well less than a few pF of capacitance with modern input stage amplifiers, but mileage will vary depending on the scope you are using.

Taking all this into account, you have to revise your circuit model. The 1 Meg resistor is inconsequential, the 16 pF capacitor should be an order of magnitude smaller, and, as per Tim's answer, the coax should be modeled as a 50 Ohm transmission line (NOT a lumped capacitance) with length corresponding to your situation.

Regarding your question about transmission lines, an ideal lossless 50 Ohm transmission line looks like 50 Ohm all the way to DC. In practice, the finite loss of 50 Ohm line results in a slightly higher impedance at low frequencies (and it actually becomes dispersive), but those effects are fairly negligible in your situation.

50 Ohm line can look capacitive or inductive if the electrical length of the line is shorter than roughly a quarter wavelength, as mentioned by Tim. This property is a manifestation of multiple reflections and used to construct stepped impedance filters. If you properly terminate a 50 Ohm line in 50 Ohms and you drive the coax with a step, you will not see an RC exponential settling at the load -- you will see a step.

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Here is what I get with these values. Note the behavior of the line.
T1 is 1 m classical coaxial line (101 pF/m) 50 Ohm / 5 ns delay.

enter image description here

And here, I add some "parasitics" with 950 Ohm.

enter image description here

Update:

Here is an example of a probe -30 dB.

enter image description here

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  • \$\begingroup\$ Thank for simulation, what software do you use for such simulations? \$\endgroup\$
    – piotr
    Commented Oct 9, 2023 at 10:28
  • \$\begingroup\$ microcap v12 free archive.org/details/mc12cd_202110 \$\endgroup\$
    – Antonio51
    Commented Oct 9, 2023 at 10:30
  • \$\begingroup\$ Improve it radically by stopping the reflection at the left end of the coax,too. C1 causes a substantial reflection at 100 MHz and higher. Insert a resistor in parallel with the coax input. Of course, it increases the attenuation , if you cannot reduce the resistance of R1 \$\endgroup\$ Commented Oct 9, 2023 at 11:23
  • \$\begingroup\$ Ok. Will add a 50 Ohm accross C3 and recheck when I am home. \$\endgroup\$
    – Antonio51
    Commented Oct 9, 2023 at 11:28
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    \$\begingroup\$ This does still seem low-frequency for this probe design. The commercial LeCroy PP066 manages 7.5 GHz with this type of probe. \$\endgroup\$
    – Hearth
    Commented Oct 9, 2023 at 12:52
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Forget the simulations for building a resistor scope probe. With fast edges, the SPICE model is very difficult to create.

I've had success using brass tubing with a RN55 resistor inside the tubing. Adjust how much the tubing covers the resistor (acts like a capacitor) to adjust the high-frequency response using a pulse with a very fast rise time. Once the position is set, tack solder the coax braid to the inner wall of the brass tubing.

The unit shown below uses a RN55 499 ohm resistor (11x probe) and a pin from a machined IC socket for the probe tip. Brass tubing is 1/8 inch OD with 0.014 inch wall thickness which fits the RN55 resistor body quite nicely. Cable is 1.2 metres of RG316 (Teflon insulated 50 ohm coax). This particular probe's 3dB corner frequency is around 1 GHz. The coil of wire is the ground point.

enter image description here

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