6
\$\begingroup\$

I would like to understand how a standard passive probe is said to have a bandwidth equal to 500 MHz. Actually what is the standard setup to qualify a passive probe?

I found an article on Teledyne LeCroy which explains how to model a passive probe. According to this, the bandwidth of the scope depends on the source resistance and the ground lead inductance, but the source resistance is unknown and the ground lead inductance depends on how you use the probe. I mean the worst case is by using the alligator clip.

enter image description here

When you take a look by doing a simulation, it seems that the bandwidth of its kind of probe can be very low if the source resistance is high.

enter image description here

Does anyone know the standard setup for qualifying a passive probe of 500 MHz?

\$\endgroup\$
7
  • \$\begingroup\$ In a nutshell, does your question boil down to what source impedance is assumed? \$\endgroup\$
    – Andy aka
    Commented Mar 13, 2023 at 16:45
  • \$\begingroup\$ @Andyaka Yes but it also depends on the ground lead inductance \$\endgroup\$
    – Jess
    Commented Mar 13, 2023 at 17:30
  • 1
    \$\begingroup\$ When probing fast-edged signals, you connect to the ground ring on the probe with the shortest conductor possible to circuit ground. Do not use the clip-on ground lead. With high impedance circuits the probe capacitance will reduce your bandwidth and may cause instability in the circuit probed. For fast-edged signals or non-trivial impedance circuits, consider using a low capacitance FET probe. \$\endgroup\$
    – qrk
    Commented Mar 13, 2023 at 17:52
  • 1
    \$\begingroup\$ As said in the earlier comment @jess, the ground lead cannot be used on high frequencies so it's kind of a red-herring to consider it at all. \$\endgroup\$
    – Andy aka
    Commented Mar 13, 2023 at 19:01
  • \$\begingroup\$ Note that the equivalent circuit that you have shown includes the scope. Here is a schematic showing the separate sections. It is from a teardown of a probe, others may vary slightly. electronics-diy.com/schematics/967/Scope-Schematic-1.jpg \$\endgroup\$
    – Mattman944
    Commented Mar 13, 2023 at 20:04

2 Answers 2

2
\$\begingroup\$

A typical x10 probe schematic is shown below, when connected to a typical scope input. The key difference between this schematic, and the schematic of the article mentioned by the OP, is the presence of C1 at the probe tip. C1 compensates for drop-off in frequency response by providing a low-impedance path for high-frequency signals.

schematic

simulate this circuit – Schematic created using CircuitLab

Description of the Schematic:
R1 & R2 form a divide-by-10 voltage divider for DC, but the AC response of is determined by the three capacitors:
C1: This is usually a fixed-value high-voltage capacitor with low inductance and low dielectric loss.
C2: This is an adjustable capacitor, the one you tweak to "compensate a probe".
C4: The capacitance within the scope. (sorry, this should have been "C3".)

If C2 is adjusted so that the value of C2 in parallel with C4 is 9x C1, then we have a properly compensated probe with good high-frequency response. For an intuitive idea about how this probe behaves at high frequency: The load that the signal at the probe tip sees will be 9Mohm in parallel with about 3pF, which gives an approximate RC time constant of ~ 30E-6s == 5.3kHz. So you can see that for the majority of the probe bandwidth the characteristics are determined by the capacitors, not the resistors.

This means that for signal frequencies above 5kHz, the loading effect on that signal is due to the probe capacitance rather than its resistance. At 100MHz, the impedance of a 3pF capacitor is just 530 ohms, and at 500MHz it will be just 106 ohms. Hence the reason why good probes tend to have lower values for C1. If the loading effect of ~500 ohms is excessive for the circuit under observation, then consider using a x100 probe (exchange gain for bandwidth and lower capacitance at the tip), or an active probe (the probe head contains an amplifier to reduce loading effect of the signal, and to improve frequency response).

In regards to test conditions for passive probes: most are specified with a sig-gen having an internal impedance of 50ohm, but it is recommended to read the technical specifications of the particular probe you are using.

For an excellent introduction to scope probes including passive probes, active probes, differential voltage probes, & current probes, may I suggest the following document may be of some value:
https://download.tek.com/document/02_ABCs-of-Probes-Primer.pdf

Here are some other links that may be useful to understand how passive probes trade-off lower gain for higher bandwidth, and practical ideas on implementing passive probes:

Regarding Passive Probes 1:1 and 10:1

Poor man's x100 oscilloscope probe

Why does capacitive loading occur when using passive oscilloscope probes?

\$\endgroup\$
2
  • 1
    \$\begingroup\$ The cable has distributed resistance (as well as L and C), which effectively makes it a "low-Z" probe past some upper cutoff. Which makes sense, you can't make an actual wideband RF probe out of capacitors; the capacitors are just another model approximating the mid-upper range of the probe's response, but the circuit gets more developed as you model it to higher frequencies. \$\endgroup\$ Commented Aug 29, 2023 at 1:47
  • 1
    \$\begingroup\$ @TimWilliams Yes, the concept of a "lossy transmission line" was the key to getting good bandwidth from passive probes. The basic concept was to trade-off gain for bandwidth; this was done decades ago, the patent by John KOBBE filed 1956 is the earliest I am aware of that was specific about using distributed resistance in the co-ax centre wire as the means for this. Refer US2883619. Also this document is very informative, dates back to the 1960s: w140.com/tekwiki/images/6/62/062-1146-00.pdf \$\endgroup\$ Commented Aug 29, 2023 at 2:31
1
\$\begingroup\$

I don't know offhand if there is an actual industry standard, but it won't be anything fancy I can guarantee that. A 50Ω output impedance sine- or square-wave source, with the amplitude perhaps adjusted to compensate for the input capacitance of the probe.

it seems that the bandwidth of its kind of probe can be very low if the source resistance is high

Indeed. That's a given. If you need wide bandwidth and high source resistances, you need FET probes, or you must integrate a FET measurement buffer on the DUT itself to isolate the probe from the high-impedance source. Typically that means matching a pair or RF FETs for the buffer, and following those with a high bandwidth buffer. The matched audio pairs have input capacitances similar to those of 10x probes.

For low-volume devices, integrating lots of performance measurement helpers that reduce the need for expensive specialized test equipment is a big advantage in both validation and future troubleshooting.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.