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# LMH6618: Differential amplifier DC shift remover circuit

Part Number: LMH6618

Hi,

I have implemented a differential amplifier circuitry using LMH6618 op-amp (as shown in the attached file). At the inverting input I have applied a DC voltage through a pot and at the non-inverting pin I have applied a sine wave with DC shift of 2.5V. My intention is to remove the DC voltage of sine wave input at the output. From theoretical analysis I guess the output should be Vout = Vdc*(-Rf/R1)  +  Vin*(1 + Rf/R1). According to this formula, I applied 5V DC at the inverting input. But the waveform suggests that the sine wave at the output is still DC shifted. Moreover I was expecting the sine wave to be twice the amplitude at the non-inverting input. But the waveform gain is less than this. Can anyone help me analyse the circuit and what should I do to remove DC shift of sine wave? This circuit is part of my project where I'm trying to remove DC shift of a video signal.

Thanks,

Sandeep M.

LMH6618_sch.TSC

• Well Sandeep, that pot you have in the inverting side is part of the gain equation for the non-inverting input. So, if for instance it is midscale, it adds 5kohm to your 549 ohm R to give a total gain R of 5.549kohm, should show a gain of 1+(549/5549) = 1.1V/V