-2
\$\begingroup\$

I am modeling a simple transimpedance amplifier with a current source input and some extra components for the photodiode model. For my AC plot the Y-axis says dB, not dbv (or dbV) so what am I referencing? I'm not sure how that works here (current source input, plus some negative bias on the opposite side). In the meantime I switched to absolute values and mentally compensate for what seems like a 1000x value of the real value. thanks

\$\endgroup\$
  • 1
    \$\begingroup\$ You might get better answers by adding a screenshot \$\endgroup\$ – Christian May 24 '17 at 19:31
  • 2
    \$\begingroup\$ electronics.stackexchange.com/questions/212779/… \$\endgroup\$ – jonk May 24 '17 at 19:52
  • 1
    \$\begingroup\$ IIRC, it's dBV (for a voltage probe). If you want to check, just add a 1 V AC source to your design, ground one pin, and put a probe on the other one. \$\endgroup\$ – The Photon May 24 '17 at 20:36
  • \$\begingroup\$ It's dBW, dBV or dBI, depending on whether you're measuring power, voltage, or current. What is your problem? \$\endgroup\$ – Bruce Abbott May 26 '17 at 8:44
  • \$\begingroup\$ thank you all, you've been very helpful. it's amazing how much this program can do that is not documented very well, or at all. \$\endgroup\$ – Matthew Carson May 27 '17 at 17:56
0
\$\begingroup\$

Remember dB is a ratio, in this case vin to vout if you divide V/V you get a unitless value which dB is. You can always change the axis to linear, log or whatever you want if you need further clarification by right clicking on the axis and selecting a different mode.

| improve this answer | |
\$\endgroup\$

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.