I am designing the front end(single ended to differential) for the ADC using LTC6406 to interface with the 20MSPS ADC. My input frequency is 140 MHz. Simulation Sch is shown below:sch

L1, L2 and C1 are used to make the filter of 140MHz. When I fed input of 140MHz 2Vpp with DC offset of zero, it works fine. Waveform shown below:wav1

But when I fed this circuit with 140MHz, 2Vpp with DC offset of 1V, I get the waveform shown below:wav2

Can the output waveforms shown in both the figures be fed to ADC and ADC will give same response.

I am unable to make out the modification in circuit for 140MHz, 2Vpp, 1V DC offset. And looking into part numbers of L1, L2 for 140MHz filter, the datasheet of the inductor talks about self resonant frequency and Q factor of inductor. How Q and resonant frequency of inductor will affect my filter.

Final Edit: Here is the rectified Sch with waveform: sch wav4


1 Answer 1


Well you haven't said what your supply voltage is or what is wrong with the output but this might be a clue: -

enter image description here

For the inductor, the higher the Q the less damping there will be in the L-C filter. This can be a good thing (and a bad thing i.e. it may create unwanted overshoot with transients).

The self resonant frequency is specified because the inductor has parasitic parallel capacitance and this will form a resonant circuit with the inductance - this area generally is to be avoided but in your circuit (low pass filter) I don't think a SRF above a GHz will be much trouble. Whatever the SRF is, it will form a parallel blocking impedance at SRF.

  • \$\begingroup\$ Will my second waveform, be okay for ADC to work fine.??? \$\endgroup\$
    – AKR
    Commented Aug 27, 2013 at 17:56
  • \$\begingroup\$ If you look at V(out+) - V(out-), does it look like V(in)? \$\endgroup\$
    – The Photon
    Commented Aug 27, 2013 at 18:00
  • \$\begingroup\$ @Andy. No Andy it doesn't look same, \$\endgroup\$
    – AKR
    Commented Aug 27, 2013 at 18:04
  • \$\begingroup\$ @AnandKumarRai I think you meant that comment for ThePhoton. \$\endgroup\$
    – Andy aka
    Commented Aug 27, 2013 at 18:07
  • 1
    \$\begingroup\$ @AKR - in your question, directly above the 2nd picture you say your signal has a 1V dc offset and this will push-apart the two waveforms by 1V - it looks like it does too. Were you trying to design an AC coupled amplifer? Or maybe just having a bad day (it happens!) \$\endgroup\$
    – Andy aka
    Commented Aug 28, 2013 at 17:18

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