2
\$\begingroup\$

For example, when I have the P2221 ( http://www.farnell.com/datasheets/1808826.pdf ),The bandwidth when setting it to 1x, would only be 6Mhz. Looking at the input resistance and input capacitance: \$ \frac{1}{(2*\pi*RC)}= \frac{1}{(2*\pi*110*10^{-12}*1*10^6)} \approx 1447 \$ and not 6 million. How do you get to 6 million?

Does this bandwidth mean there's an attenuating for signal higher than 6Mhz or does it mean it will filter out all signals above 6Mhz or something else? How can I calculate the frequency when attenuating starts? When setting it to 10x, the bandwidth is now 200Mhz. Why this difference, when the capacitor is roughly 10 smaller and the resistor is roughly 10 times bigger, so RC does not change that much.

\$\endgroup\$
1
\$\begingroup\$

It means pretty much the same as any other bandwidth: It tells you where the response of the circuit is down 3dB. In practice, if you're looking at signals with frequencies more than about 1/10 of the bandwidth, you can expect significant distortion because of the loss of the higher frequencies.

The tricky thing here is that the circuit in question has three parts: lumped resistance and capacitance in the probe; distributed resistance, capacitance, and inductance in the cable; and lumped resistance and capacitance in the oscilloscope's input circuit. So you can't look just at the input resistance and capacitance of the probe; that's only part of the overall circuit. That's useful information to figure out how the probe loads the circuit being measured, but it's not all there is to calculating the input bandwidth.

\$\endgroup\$
  • \$\begingroup\$ 'It tells you where the response of the circuit is down 3dB'. So when I use a 100Mhz, does the signal at the 1x setting gets attenuated, primarily because of the bandwidth or because of the input resistance or is it both? 'if you're looking at signals with frequencies more than about 1/10 of the bandwidth, you can expect significant distortion' So I should expected distortion with the 100Mhz signal, despite that the bandwidth is 200Mhz at the 10x? \$\endgroup\$ – Dorka Apr 18 '15 at 19:00
  • \$\begingroup\$ @Dorka - the input resistance is part of the overall attenuation; it's allowed for in calculating the bandwidth. And, yes, your 200MHz probe will start to mislead you at around 20MHz. Note, however, that how badly things get distorted depends on the nature of the signal. For a pure 100MHz sine wave, you'll see a sine wave with a slight reduction in amplitude. For a 100MHz square wave you'll see something that looks more like a sine wave than a square wave because of the loss of the high frequencies. \$\endgroup\$ – Pete Becker Apr 18 '15 at 19:05
  • \$\begingroup\$ How do you get to 1/10 of the bandwidth? If the cut-off frequency is 200Mhz, wouldn't 20Mhz be still at the straight horizontal line part of the bodeplot of a low-pass filter? \$\endgroup\$ – Dorka Apr 18 '15 at 19:26
  • \$\begingroup\$ @Dorka - 1/10 is a rule of thumb for bandwidth issues. Once you get above that you start getting into noticeable losses. \$\endgroup\$ – Pete Becker Apr 18 '15 at 19:52
  • \$\begingroup\$ @Dorka - The bandwidth of the probe is due to a combination of scope input resistance (typically 1 Mohm) and input capacitance (typically ~12 - 15 pF) \$\endgroup\$ – WhatRoughBeast Apr 18 '15 at 20:14
1
\$\begingroup\$

The answer lies in the fact that oscilloscope probes are not simple RC circuits. Instead, the cable from the tip to the scope is a lossy coax, and its impedance interacts with the scope input resistance/capacitance to provide a low-pass filter with a cutoff frequency of ~6 MHz for your particular probe and a normal scope input. See https://www.youtube.com/watch?v=OiAmER1OJh4 for an entertaining video, with the important stuff starting about 10 minutes in. Note that this is a function of the particular coax the probe manufacturers chose. You can get dedicated 1x probes with bandwidths up to 15 MHz.

\$\endgroup\$
0
\$\begingroup\$

Something to think about: high speed digital probes have a super short ground lead to prevent ringing in the scope waveform. Maybe this might be another reason why it is spec'ed at 6MHz.

\$\endgroup\$
  • 1
    \$\begingroup\$ I'm sorry, I did not understand what you meant. Could you explain in in more detail? \$\endgroup\$ – Dorka Apr 18 '15 at 20:09
  • \$\begingroup\$ @Dorka - a typical probe has a clip-lead that's about 6 inches long that you connect to circuit ground. That 6-inch wire can pick up stray signals, and for really high-speed circuits, you need to worry about that, and maybe use something shorter. But that's probably not the issue here. \$\endgroup\$ – Pete Becker Apr 19 '15 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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