3
\$\begingroup\$

I've got an oscilloscope that, according to the manual, looks like this (for AC):

---Cac--+-----+--------
        |     |
Uin     Cin   Rin      Udisp
        |     |
--------+-----+--------

Uin is the voltage input, Udisp is the voltage shown by the oscilloscope, Rin is the internal resistance and Cac and Cin are internal capacitances. I'm trying to find |Udisp/Uin| for different frequencies. So I rewrote this as follows with complex impedances:

Uin---1/(j*w*Cac)--+-------------- Udisp
                   |     
             Rin+1/(i*w*Cac)
                   |     
 0 ----------------+--------------

         <=>

Uin---1/(j*w*Cac)---Udisp-----Rin+1/(i*w*Cac)-----0

                             \_______Z1______/
     \______________Z2_______________________/

Using Ohm's law, since the currents over both impedances must be the same, I get:

Udisp / Uin = Z1 / Z2

However, this means that the measured amplitude (|Udisp/Uin|) does not depend on the frequency! I must have made a mistake somewhere, as the oscilloscope obviously can't handle all frequencies equally well. Where is the mistake?

\$\endgroup\$
  • \$\begingroup\$ I just saw that I added Cin and Rin as if they were in series, sorry! But I get the same problem if I do it correctly, namely that the magnitude of Udisp/Uin does not depend on the frequency. \$\endgroup\$ – Andrea Dec 6 '11 at 10:00
6
\$\begingroup\$

I can see two problems here. Firstly, your analysis does depend on frequency. Replacing \$j\omega\$ with 's' for convenience, Z1 is given by :-

\$Z_1 = \frac{R_{IN}}{1+\mathrm sR_{IN}C_{IN}}\$

... from which the transfer function Z1/Z2 can be derived as :-

\$\frac{\mathrm sR_{IN}C_{AC}}{\mathrm sR_{IN}(C_{IN}+C_{AC})+1}\$

This is a first-order high-pass function with a pole at a frequency \$\frac{1}{2\pi R_{IN}(C_{IN}+C_{AC})}\$ so your 'scope will display frequencies higher than this with no loss of amplitude ... according to your model.

The second issue is that your model is incomplete. You are assuming zero source impedance and not taking into account the characteristics of the vertical amplifier which will roll-off at some frequency.

\$\endgroup\$
  • \$\begingroup\$ Thank you. Don't know what I was thinking about that not depending on frequency though :) \$\endgroup\$ – Andrea Dec 6 '11 at 18:31

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.