0.5V in generator become 0.4V in Oscilloscope?

i am using a Digital Data Analyzer to generate 0.5V peak to peak data but i only receive 0.4V peak to peak at my oscilloscope.

could someone tell me how could this happen?

I am thinking if I am using 50Ohm for the data analyzer and the oscilloscope, the value should be 0.25 if divided by half?

Thank you very much?

Added after 4 minutes:

I calculated the input resistance of the oscilloscope is 200Ohm instead of 50Ohm.

anyone know why there is a change in value?

Maybe the signal's frequency is near the bandwidth limit of your signal generator or oscilloscope. Try turning off AC coupling and bandwidth limiting.

Yes, in a 50-ohm system, mixing-up the 50-ohm and high-impedance setting on the scope usually causes a 2X or 0.5X error in amplitude. However, if the cable is improperly terminated and if the frequency is high enough for significant transmission line reflection, then various weird amplitudes are possible.

Maybe the signal generator or oscilloscope is out of adjustment.

Maybe your scope's 50-ohm resistor is damaged. Try measuring it with an ohm-meter.

I change the Oscilloscope from a 500MHz one to a 6GHz one, then the reading is correct, 0.5V to 0.5V.

is it because the 500MHz bandwidth marked on the Oscilloscope is not the actually limit of the Oscilloscope, just an indication?

because I find even the 6GHz Oscilloscope cannot catch a 2.5Gbps clock, the waveform is a bit distorted with too few sampling points.

What frequency is your signal?

The scope's bandwidth spec refers to its -3dB bandwidth. At 500 MHz, the scope could attenuate the signal by as much as -3dB (0.707 times).

A 2.5 GHz square wave consists of a 2.5 GHz fundamental sinewave plus components at 7.5 GHz, 12.5 GHz, 17.5 GHz, and so on. A 6 GHz scope will show the 2.5 GHz fundamental and probably some of the 7.5 GHz component, but very little more. It won't look like a square wave. If the scope also has a relatively low sample rate, then the display will look jagged, unless it uses display interpolation. Bandwidth and sample rate are independent specifications, both should be listed on the instrument's spec sheet.

The scope may also have special sample modes (such as equivalent-time sampling) that eliminates the jagged display when viewing a repetitive waveform such as a clock. Check the operator's manual.

Maybe helpful:
https://www.tek.com/Measurement/App_Notes/RTvET/ap-RTvET.html
**broken link removed**

There is one old rule that can be used in almost any such case: when you have pulses (like clock) and want to get them go through without distortion the required bandwidth can be determined as BW=0.35/tr, where tr is the rise time of the pulse. This means that your oscilloscope must have at least three times more bandwidth than 1/tr value.

Best regards,
Rf-OM

Thank you friends,

echo47 said:

What frequency is your signal?

A 6 GHz scope will show the 2.5 GHz fundamental and probably some of the 7.5 GHz component, but very little more. It won't look like a square wave.

Click to expand...

i made a mistake here. the actually wave is like a triangular or sinewave, not a clock any more.

I input 50MHz signal input into the 500MHz scope and get 0.4V out of 0.5V. then change to a 6GHz scope, can recover the 0.5V data. then I try to use the 6GHz scope to trace a 2.5Gbps clock(I try to test a 2.5Gbps TIA-LA circuit), find the waveform is distorted.

so if following the rule by 3 times 1/tr, let say 2.5Gpbs clock has 100ps tr, then the required bandwidth of the scope should be 30GHz?


I find from the specification of the digital data analyzer MP1632A, it is said that the external clock input's waveform is squared wave when f<0.5GHz and squared wave or sine wave when f>0.5GHz, i guess it is also the generator limiting the output waveform.

According to the theory if you want to get undistorted signal and rise time is 100pS or 1E-10s the required bandwidth should be 0.35/1E-10=3.5E9 or 3.5GHz if 100ps is the rise time of clock signal.

BR,
RF-OM

I assume that your 50 MHz signal is a square wave. Yes? The waveform should look similar on both scopes. The rise/fall time may appear faster on the 6 GHz scope, but the amplitudes should be the same on both scopes. If the 500 MHz scope shows incorrect amplitude, then maybe it needs calibration or repair. Of course, be sure you are operating the scope correctly.

If you tell us what model 6 GHz scope you have, then maybe someone here will know how well it should display a 2.5 GHz clock.

RF-OM's equation BW = 0.35/tr is fine, but his sentence saying "... three times ..." is incorrect.

To echo47,

The coefficient 1/0.35 is roughly equal three times. I believe that such accuracy is okay in this case.

Best regards,
RF-OM

Hi RM-OM, "three times more bandwidth than 1/tr" suggests 3/tr instead of 0.35/tr. That's what confused dog1357.

I understood it now. Excuse my poor language please, I definitely mean right thing but wrote it differently. I apologize for this inconvenience.

Best regards,
RF-OM

I think it's a questionable to some extent to discuss the possible causes of a signal amplitude deviation without knowing how the signal actually looks like. Also, by watching a step response of a longer period square wave (that surely can be output by the said signal generator as well), all speculations regarding bandwith limits, unintended AC coupling etc. could be ended at once.

As echo47 said, a 500 MHz bandwith would always allow complete settling of a 50 MHz square wave. Some modern oscilloscopes have signal scaling options, that aren't recognizable at first look, e. g. Agilent 8000. They are well suited to confuse an operator not familiar with the instrument.