2
\$\begingroup\$

I tried to dc sweep V3 in my circuit and measure the gain. But it does not give a correct result.

What I am trying to do is actually recreating some measurements from the datasheet.

Maybe there is a simulation techniques that I do not know yet. Appreciate any suggestion.

enter image description here

enter image description here

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ Your graphs appear to indicate that it does work. \$\endgroup\$
    – Andy aka
    Commented Jul 22, 2020 at 9:10
  • \$\begingroup\$ @Andyaka As I understand it, it should be like the Fig.3 above. it should be linear-in-dB against vgain. \$\endgroup\$ Commented Jul 22, 2020 at 10:04

2 Answers 2

1
\$\begingroup\$

According to the Datasheet of the AD8337 (Rev. D), page 21/28 under section Gain control considerations:

The offset voltage effect of the AD8337, as with all VGAs, can appear as a complex waveform when observed across the range of VGAIN voltage. Generated by multiple sources, each device has a unique offset voltage (VOS) profile while the GAIN input is swept through its voltage range.

At least from what I can understand, you're seeing the influence of \$V_{GAIN}\$ on the offset voltage, which directly affects the output voltage as well.

\$\endgroup\$
1
\$\begingroup\$

I seem to be getting something pretty close to the datasheet:

test

but I had to probe immediately after the pin for the output (V(x)), otherwise there would have been some 18 dB missing (V(o)). The input is also a more meaningful 0.1 V (accounted for in the plot window as *10). Another thing that's corrected from your schematic is the proper use of an input resistance: you placed it across the voltage source, which is useless, instead of in series. I also didn't use LC filtering since the only noise here is the double residuals. The symbol is the generic [Misc]/DIP8, for the lack of one, but the pinout is the same. Also, in your waveform window, did you use log minus log? The correct way is 20 * log10( V(out) / V(in) ).

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.