# Trying to get the full potential of my 100 MHz oscilloscope

I posted a question yesterday in which I realized a 100 MHz oscilloscope with 100 MHz x10 probe will have hard time capturing high frequency signals such as the ringing on rising/falling edge, and using a higher bandwidth probe allows the whole system to capture more precise samples.

After some searching I found the following formula (source):

With the formula, I can calculate that:

• 100 MHz probe + 100 MHz scope ~= 70 MHz total bandwidth
• 200 MHz probe + 100 MHz scope ~= 80 MHz total bandwidth
• 500 MHz probe + 100 MHz scope ~= 98 MHz total bandwidth

Does this mean that, in order to get the full potential out of my 100 MHz scope, I should invest in a 500MHz probe? Or is there other formula that I need to take into consider, which may somehow "bottleneck" the total bandwidth of the system?

• Which exact scope and which exact probes you have? Their data sheets must contain information about their bandwidth. It would be poor marketing if your formula is true, because it would mean you can never utilize a 100 MHz scope to 100 MHz bandwidth unless you have a probe with infinite bandwidth. Commented Mar 19 at 20:14
• If you want to show "100 MHz", try bringing the signal in through a coax, straight out of a signal source like a function generator. (and add a coax termination to the scope input, if not built into the scope, as typically not for 100MHz ones) Commented Mar 19 at 21:32
• Interesting question. I always assumed that if I bought a 100 MHz scope that was bundled with 100 MHz probes, the combined bandwidth would be 100 MHz. That is, each would have a little excess bandwidth to arrive at the target bandwidth. If I was still working, I would have access to equipment to test my theory. Commented Mar 19 at 23:42
• I don't think the 100 MHz refers to the 3dB bandwidth but rather the passband of the probe. The 3dB bandwidth of a quality probe is probably a lot higher than the width of the passband. Commented Mar 20 at 1:27
• @user1850479 The manual of that specific 100 MHz Tektronix P2100 says : Bandwidth DC to 100 MHz. No other mentions about bandwidth. What are your definitions for the passband then for a 100 MHz probe? Commented Mar 20 at 12:49

## 2 Answers

If you purchase a $$\100 \space \mathrm{MHz}\$$ oscilloscope with the intention of observing a $$\100 \space \mathrm{MHz}\$$ signal, you will encounter significant measurement error. The bandwidth $$\BW\$$, in $$\\mathrm{Hz}\$$, of your oscilloscope determines the Amplitude Error, as there will be attenuation as the frequency increases, which becomes more significant at the point of $$\-3 \space \mathrm{dB}\$$ (corner frequency $$\\omega_c\$$ in $$\\mathrm{rad/s}\$$). If you are interested only in in the the high frequency side, you can model it rougly as a simple first order system (remebering that the bandwidth stuff refers to analog part of scope):

$$\frac{V_o(\omega)}{V_i(\omega)}= \frac{1/j \omega C}{1/j\omega C+R}= \frac{1}{1+j\omega RC}$$

As

$$RC=\frac{1}{\omega_c}=\frac{1}{2 \pi BW} \qquad (1)$$

the magnitude of that transfer function, as function of frequency $$\f = \frac{\omega}{2\pi}\$$, in $$\\mathrm{Hz}\$$, is given by:

$$\left | \frac{V_o(f)}{V_i(f)} \right | = \frac{1}{\sqrt{1 + \left ( \frac{f}{BW} \right )^2}} \qquad \lt 1$$

Also:

$$\left | V_o(f) \right | = \frac{1}{\sqrt{1 + \left ( \frac{f}{BW} \right )^2}} \space \left | V_i(f) \right | \qquad (2)$$

The definition of Amplitude Error is:

$$\text{Amplitude Error (%)}= 100 \left (1- \left | \frac{V_o(f)} {V_i(f} \right | \right )$$

Or:

$$\boxed { \text{Amplitude Error (%)}= 100 \left (1- \frac{1}{\sqrt{1 + \left ( \frac{f}{BW} \right )^2}} \right ) }$$

If you prefer in a plot:

So, if you put sinewave with amplitude $$\ 1 \space \mathrm{V}\$$ and frequency $$\100 \space \mathrm{MHz}\$$ as input in a $$\100 \space \mathrm{MHz}\$$ ($$\BW\$$) oscilloscope, the Amplitude Error error will be $$\29.3 \space \% \$$. Also, using $$\(2)\$$, the voltage $$\V_o\$$ will be read as aprox. $$\0.707 \space \mathrm{V}\$$. On the other hand, with a $$\500 \space \mathrm{MHz}\$$ scope, the Amplitude Error will be aprox. $$\ 1.94 \%\$$ (or $$\ V_o= 0.98 \space \mathrm{V}\$$). In summary: A $$\100 \space \mathrm{MHz}\$$ scope will have up aprox. $$\30 \%\$$ error at $$\100 \space \mathrm{MHz}\$$. To keep errors below the $$\3 \%\$$, the maximum signal bandwidth that can be measured is about $$\30 \%\$$ of $$\100 \space \mathrm{MHz}\$$, or $$\30 \space \mathrm{ MHz}\$$. In a different way: In order to accurately measure a $$\100 \space \mathrm{MHz}\$$ signal (below $$\3 \%\$$ error), you need at least $$\300 \space \mathrm{MHz}\$$ of bandwidth.

Another consideration is with respect to the rise time of the signal to be measured (which is the time taken by a signal to change from a specified low value to a specified high value, typically from $$\10 \%\$$ to $$\90 \%\$$). You can have a rise time of $$\10 \space \mathrm{ns}\$$ on a "square" signal with a frequency of either $$\10 \space \mathrm{Hz}\$$ or $$\1 \space \mathrm{MHz}\$$. In other words, it is the ability to follow this rise time that identifies the quality of the measurement system, not the frequency alone.

A common expression relating the rise time $$\tr\$$, in seconds, with the bandwith $$\BW\$$ in $$\\mathrm{Hz}\$$, is:

$$BW = \frac{0.35}{tr}$$

That expression can be easily obtained considering that circuit in the beginning of this post, along $$\(1)\$$ and the equation for capacitor charging (from $$\10 \%\$$ to $$\90 \%\$$). So, a $$\100 \space \mathrm{MHz}\$$ oscilloscope has a rise time of $$\3.5 \space \mathrm{ns}\$$ and a $$\500 \space \mathrm{MHz}\$$ one, a rise time of $$\1.75 \space \mathrm{ns}\$$.

Through the Central Limit Theorem, it follows that whenever you have a measurement system formed by others non interacting cascaded sub-systems (one after the other), free from overshoot in their step-responses, each with its own rise time $$\tr_1\$$, $$\tr_2\$$, ... and $$\tr_n\$$, the rise observed time from input to output $$\t_r\$$ is given by:

$$tr = \sqrt{tr_1^2 + tr_2^2 + \text{...} + tr_n^2}$$

If $$\tr_s\$$ and $$\tr_p\$$ are the rise time for the scope and the probe, respectively, then the equation for System Bandwidth in your reference can be obtained as follows:

$$tr = \sqrt{tr_s^2 + tr_p^2}$$

$$\frac{1}{tr} = \frac{1}{\sqrt{tr_s^2+tr_p^2}}$$

$$\frac{BW}{0.35}=\frac{1}{\sqrt{tr_s^2+tr_p^2}}$$

$$BW = \frac{1}{\frac{1}{0.35}\sqrt{tr_s^2+tr_p^2} }$$

$$BW=\frac{1}{\sqrt{\left ( \frac{tr_s}{0.35} \right )^2+ \left ( \frac{tr_p}{0.35} \right )^2}}$$

$$BW=\frac{1}{\sqrt{\left ( \frac{1}{BW_s} \right )^2+ \left ( \frac{1}{BW_p} \right )^2}}$$

Where $$\BW\$$, $$\BW_s\$$ and $$\BW_p\$$ are the system, scope and probe bandwidths, respectively.

Therefore, the response is yes, that expression you have mentioned and associated calculations make sense. But, I believe that the most important is to keep your error lower as possible (remember the rule of $$\3 \%\$$).

Finally, in a note from Fluke, there is an alternative approach considering the harmonic content in your signal:

Perhaps an even better, and much more conservative, approach is to consider the stated bandwidth to be that of a fifth harmonic of the frequency you want to measure. A fifth harmonic, likely present in a typical pulse, might begin to become attenuated at the stated bandwidth. That would indicate that we can rely on a 500 MHz bandwidth oscilloscope to show a full and undistorted picture of the input at about 100 MHz, maintaining good fidelity of the signal you're trying to measure.

Using this criteria, a measurement system with $$\500 \space \mathrm{MHz}\$$ bandwidth is suggested to capture the fifth harmonic of a $$\100 \space \mathrm{MHz}\$$ square wave.

• IMHO, using Mathjax for values in text blocks looks a bit strange - the SI units display weirdly... Commented Mar 20 at 20:36
• In line with what @Greenonline says, if you need to, you can use \mathrm{MHz} to produce $\mathrm{MHz}$ in the middle of a mathjax/TeX block. The standard way to present units is always in upright lettering, whereas italics are reserved for variables. Commented Mar 20 at 21:34
• It was applied. Commented Mar 21 at 17:19

The meaning of bandwidth here is the -3dB frequency. This is basically the frequency at which the output amplitude is ~71% of the input signal's amplitude (-3dB power loss corresponds to half the power across the same load. You can work out the rest from logarithms.).

ImgSrc

The diagram above is just an example – doesn't represent the actual response of either the probe or the scope.

If the probe and the scope have the same bandwidth, the resultant diagram will show a "sharper" response. See the brown lines below:

As you can see, the -3dB frequency is now less than $$\f_H\$$. This is basically where your formula comes from.

Does this mean that, in order to get the full potential out of my 100 MHz scope, I should invest in a 500MHz probe?

Well, I can't give exact numbers as it must have already been discussed in the previous question you linked. But technically, in any case, the probe's BW should be higher enough than the scope's.

and using a higher bandwidth probe allows the whole system to capture more precise samples.

You need to define precise or precision here. An effective BW of 71 MHz (100 MHz scope and probe) corresponds to ~5 ns rise time. If you invest in a 500 MHz or a 1 GHz probe, you'll be able to measure rise times of 3.5 nanoseconds. If that's really what you need then invest in 500 MHz (or even higher) probes.

A real example from our lab: We normally require to be able to measure pulses having widths of as low as 20 nanoseconds (e.g. dead times between pulses), so the maximum bandwidth we require is about 200 MHz (from $$\0.35/t_R\$$ where the $$\t_R\$$ is rise time, in our case it's one-tenth of the minimum pulse width = 2 ns). In our lab, we use 400 MHz scopes and they are shipped with 500 MHz probes which is sufficient for us. We also have 1 GHz and 1.5 GHz active probes in case we need to measure faster rise times.

• It can also be hard to find a 500 MHz passive probes with a suitable capacitance for a 100 MHz scope I guess.
– pipe
Commented Mar 19 at 21:35
• I really appreciate the real life example you provided at the end of your answer, it's actually way easier for an electronics amateur like me to understand. Both answers are really good and helpful so it's really hard to decide which one to accept. I chose to accept the other answer because I think the more detailed explanation of how the formula came from will help more for the professional users of this stack exchange. Commented Mar 27 at 16:40