Our TDS2014 o-scope is measuring signal from a 33210A waveform generator. The signal out of the generator is a 1VPk-Pk sine wave. But the o-scope is reading the signal as 2VPK-PK. Is there something wrong with this?

We checked the multiplication of the probe and o-scope . Their multiplication were the same.

Has anyone encounter this problem?


The signal generator is expecting to be run into a 50 ohm impedance - if you did that the voltage would drop to 1 Vp-p: -

enter image description here

See the little red square I've added - inside that square is written 50 ohm.

From the manual it says this: -

The Agilent 33210A has a fixed series output impedance of 50 ohms

See also this page for confirmation. BTW you can rely on the 2Vp-p providing you run the output into a high impedance. With a load of (say) 1 kohm, the output will be 1.905 Vp-p. With a 10 k load the output will be 1.990 Vp-p and, all the time the display will indicate 1 Vp-p.


When the output is set to 50 ohm mode, it expects a 50 ohm termination, which would cause 50/50 voltage divider. So, it outputs double in order to give you the expected signal at your board. In high impedance mode, the impedance of the function generator >> 50 ohms, so it outputs the normal amount of voltage.

  • \$\begingroup\$ The o-scope has a 50 ohm input impedance. See page 1-2: atecorp.com/ATECorp/media/pdfs/data-sheets/… When I change the generator's impedance to HI-Z, the o-scope measures the pk-pk voltage correctly. I don't understand why there is an impedance mismatch. \$\endgroup\$ – e.engineer Apr 28 '17 at 15:17
  • \$\begingroup\$ When the generator (Hi-Z) outputs a 1V pk-pk to a Fluke 115 DMM, the Fluke measured 0.353VRMS. Is this right? \$\endgroup\$ – e.engineer Apr 28 '17 at 17:59
  • \$\begingroup\$ For the generator, in "50 ohm mode" it outputs double the signal. Because it has an internal 50 ohms and expects an external 50 ohms, yielding 1x the signal (resistive divider). But, if the external impedance is 1M instead of 50 ohms, the divider is than 1M:50 instead of 50:50. The 1M then shows the majority of the voltage drop. Since the 1M is the scope input, the scope will see the 2x signal coming out of the generator \$\endgroup\$ – Daniel Bogdanoff - Keysight May 1 '17 at 18:59

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.