# I am having trouble measuring the amplitude of a 97 MHz signal with an oscilloscope with a bandwidth of 100 MHz

First, I designed a Colpitts oscillator with a frequency of around 94-97 MHz. I added a common emitter amplifier to the design to increase the output amplitude of the oscillator, then I connected the probe of the oscilloscope to the output and measured the amplitude of the signal with the oscilloscope.

When the probe is at 1x, the amplitude of the signal is 633 mV.

Here is the screenshot of the signal while the probe is at 1x:

When the probe is at 10x, the amplitude of the signal is around 380 mV.

Here is the screenshot of the signal while the probe is at 10x:

I'm not sure if the signal is measured correctly since the bandwidth of the oscilloscope and probes is 100MHz.

What is the actual amplitude of the signal?

If the measurement is correct when the probe is 10x, the amplitude of the signal should be 380mVx10=3.8V.

I am attaching the circuit diagram.

I recorded the operation of the circuit on video.I am testing the modulation in the video via a radio. There are also FFT and oscilloscope measurements of the modulation in the video.Please warn me if I have done something wrong.link is here https://drive.google.com/file/d/1YryFCI-n6-iACd4vVtxXLwiRebVg5j8W/view?usp=sharing

• The bandwidth of an oscilloscope is typically the frequency at which the amplitude is down -3 dB. Thus, if the the bandwidth is 100 MHz and your signal source is 97 MHz, it is very likely your amplitude measurement could be off by 20% or or more. Commented Mar 22, 2022 at 23:55
• Some thoughts. X10 lightens the load on the circuit by about a factor of 10 (which may increase the signal as it is loaded less), but there's a compensation cap and you need to go through a procedure to make sure it is set correctly. The X1 doesn't need that procedure. The X10 also diminishes the signal to the scope. The scope may, or may not, automatically adjust things on the display, accordingly. Also, the scope limit typically means that the signal is already reduced by that point. So it won't be an accurate reflection when you are that close to the limit, either.
– jonk
Commented Mar 22, 2022 at 23:56
• @jonk Is the amplitude of the signal 3.8V or 633mV ? Which one is right? Commented Mar 23, 2022 at 0:01
• @MathKeepsMeBusy Is the amplitude of the signal 3.8V or 633mV ? Which one is right? Commented Mar 23, 2022 at 0:02
• In the X1 position the scope input capacitance is probably about 100pF or more. Also note that the probe bandwidth is only 6Mhz in the X1 setting. That will load the circuit a great deal. Even in the x10 position the input capacitance will probably be 10pF or so that is a significant load at ~100MHz. Commented Mar 23, 2022 at 0:13

These are a few observations with assumptions due to lack of design details.

• 3dB DSO @ 100 MHz = 70.7% of signal or 30% attenuation

• 1:1 Probes can never be use >20 MHz due to cable capacitance 100 pF /m and ground inductance <10 nH/cm

• A 10: 1 probe calibrated, is always your best bet for RF for signals >= 20 MHz BUT the ground clip must be very short (<1") or never used and only with spring coil differential sig/gnd. on tip/ring

• output impedance = Rc is unknown but assumed low enough

• Most likely your signal is 3.8V/0.7 ~ 5.3Vpp with a 10V supply @ 30mA implies Rc < = 330 ohms

Although my design is simple, by changing your design specs, for Vpp, Zo, Imax and THD, you can get a decent square or sine wave by changing R ratios and bias for symmetry.

This one is > 9V square wave. with the right transistor and probe methods.

• I have attached the circuit diagram. Commented Mar 23, 2022 at 1:20
• Should I buy a larger bandwidth oscilloscope or buy a larger bandwidth probe to measure the signal accurately? Do I have any other choice? My budget is not enough. Commented Mar 23, 2022 at 7:23
• It depends why you need this. Learn to simulate tinyurl.com/ycbtqsgk and search for 10:1 probe solutions with coil spring on this site and also make 10:1 attenuators to reduce load and lower source impedance tinyurl.com/ycbtqsgk Commented Mar 23, 2022 at 12:46
• tinyurl.com/y6wzwjnc interactive sim with 10: 1 divider to emitter to sample collector Commented Mar 23, 2022 at 12:59
• You can't measure a 100 MHz 9Vpp square wave accurately with a 10:1 probe due to C and 1GHz BW. You need a FET Diff Probe  tinyurl.com/y7watbba Commented Mar 23, 2022 at 13:32

As people stated in the comments, the oscilloscope plus probe (x10 setting) -3dB bandwidth is 100MHz. Thus, at 97 MHz, which is very close to 100MHz, the amplitude could be down by 0.71 times (-3dB point).

At x1 setting, the input capacitance of your probe is likely to be in the 100pF range when coupled to an oscilloscope. Depending on your amplifier circuit (you don't show a schematic), you could be severely loading the amplifier output and causing extra loading on the oscillator circuit due to Miller capacitance of your output amplifier. Plus, at x1 setting, the bandwidth of the probe is likely to be around 10MHz (couldn't find any data on your probe, so numbers are from a Tektronix P6129B probe manual).

At x10 setting, the probe capacitance is like to be in the 18pF range which will provide less loading on the output of your amplifier. The x10 setting on your probe is likely to be more accurate, but since you are so close to the bandwidth of the measurement system, the output voltage is likely to be higher than 3.8V.

To make things more complicated, a x10 probe's impedance has a complex nature. The following image is the input impedance of a Tektronix P6105A 100MHz, x10 probe. You can see that the impedance at 100MHz is Rp=400 ohms in parallel with XCp=200 ohms (8pF). Thus, the x10 probe may have significant loading on your amplifier at 100MHz (depends on your amplifier design).

Tektronix P6105A probe impedance taken from the manual.

• @grk The last three photos in the post contain prop and oscillator input resistor and capacitor information. Commented Mar 23, 2022 at 0:51
• @OzGtZt A schematic of your circuit would be useful as the output structure of your amplifier is unknown. The whole point of my post is an oscilloscope probe has a complex impedance which changes over frequency. Note at 100MHz, the example graph shows that the input impedance of a x10 probe is much lower than 10Mohm. Just noticed the y-axis of the graph is inconsistent. I'll replace that with a scan from a paper copy.
– qrk
Commented Mar 23, 2022 at 1:14
• @grk I have attached the circuit diagram. Commented Mar 23, 2022 at 1:21
• Should I buy a larger bandwidth oscilloscope or buy a larger bandwidth probe to measure the signal accurately? Do I have any other choice? My budget is not enough. Commented Mar 23, 2022 at 7:23

You are falling into the same trap I did: Help with 144Mhz small signal amplifier not amplifying

Main issue is the scope probe loads down the circuit. To avoid this I used a resistor divider and put the 10x probe on that and used that to measure the circuit, both input and output. This gave me a relative value that could be compared, and it was more accurate since the probe wasn't loading down the circuit.

• You made an explanation but I did not understand it. Could you please explain your solution method by drawing a circuit diagram? It would be nice if it was detailed. Commented Mar 23, 2022 at 20:21
• electronics.stackexchange.com/questions/591298/… use this or 75 ohm to 1:1 probe with 750 ohm on DSO using BNC T for 20 dB pad or whatever probe coax Zo is. Commented Mar 23, 2022 at 22:37