Demodulating differential signal.

[Alex_Bam]

Hello,
I am demodulating differential signal using LPF based demodulator. The schematic of the entire system is attached which includes diff modulator, coupling capacitors, and demodulator.

A modulated signal of 17 Mhz is demodulated. For the demodulator, I designed a 2-stage differential amplifier to amplify the diff signal then LPF is utilized to filter high frequencies, finally, the output signal is retrieved using Schmitt trigger. For LPF I used a single capacitor, the output impedance of 2-stage op-amp and capacitor constitute LPF.

Can anyone comment Why the designed 2 stage op-amp and LPF is not amplifying and filtering the modulated signal? According to my
understanding, the design op-amp parameters are adequate to demodulate the signal.

The design parameters of 2-stage op=amp are: ICMR+= 4V ICMR-=1.5V , GBW=40Mhz, Gain=68dB, Vdd=5V, C_load= 2pF.

Thanks.

 

[WHONOES]

I see that your own design amplifier uses no feedback, as such it is operating open loop. Your gain bandwidth product would then be divided by the gain factor of 68dB which is approx' 2511. Therefore the available bandwidth will be 17MHz / 2511 which yields 6.7KHz.
Could that be part of your problem?