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Background: I'm a student, working on my final year project. I am designing a front end for a software defined radio. We have an IF frequency at 36Mz (and a BW ~8MHz). We are bandpass sampling this signal at 29MSPs (our signal falls in the 3rd nyquist zone). The ADC takes a differential input and requires a common mode offset. This DC offset is provided by a differential op-amp. We need to filter out the out of band noise from the op-amp and the previous stages before feeding it into the ADC.

In order to achieve this filtering, I have cascaded a bandstop filter (stop bands are at ~1MHz and ~29MHz) with a lowpass filter (cutoff at ~40MHz). This filters out the out of band noise above 40MHz, as well as below 29MHz while still passing the common mode DC offset.

The problem, is that the bandstop filter requires large inductors (6.2uH). These inductors all seem to have very low self-resonant frequencies and are thus basically useless.

The question: Is it possible to design a highpass filter, that also passes DC? Or is there some other strategy to let the common mode offset pass through my filter, but still cut out the noise in the 0-30MHz range? Or, to design a bandstop filter with (much) smaller inductors?

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  • \$\begingroup\$ Not sure if relevant w/o seeing a schematic but is it possible to put a LP filter (for DC) in parallel with a HP filter, and mix the two at the adc input? \$\endgroup\$ – tariksbl Nov 17 '14 at 2:40
  • \$\begingroup\$ Which op-amp and ADC are you using? What do you mean by "The ADC requires a common mode offset"? \$\endgroup\$ – Bruce Abbott Nov 17 '14 at 4:36
  • \$\begingroup\$ If you want to attenuate noise from a DC output use a low pass filter not a bandpass. \$\endgroup\$ – Andy aka Nov 17 '14 at 8:33
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If it passes DC it's not a high pass filter.

What you probably want is a high pass filter (or a bandpass filter centred on 36MHz) and a separate DC source, which are combined at the ADC input.

Your filter is probably designed to deliver signal into a typical load impedance (50 ohms, 330 ohms) which is usually orders of magnitude lower than the ADC input impedance, and that load is typically modelled as a simple resistor from the output to ground. See the first circuit below (replace the capacitor with your actual HPF)

schematic

simulate this circuit – Schematic created using CircuitLab

In the second circuit, you simply return that resistance to your (clean, low noise) DC voltage of choice, that combines your HPF and DC signals correctly. And as your HPF blocks DC, this does not affect the DC voltages in earlier stages of the circuit. (C3 is a decoupling capacitor, to keep the DC supply clean and noise free)

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  • \$\begingroup\$ Can you elaborate upon that last paragraph? I do not understand. \$\endgroup\$ – RJTK Nov 17 '14 at 13:40
  • \$\begingroup\$ Edited, hope it's clear now \$\endgroup\$ – Brian Drummond Nov 17 '14 at 13:57
  • \$\begingroup\$ I see, thanks. should I be terminating the filter directly into the ADC? Or do I need to terminate into some resistance in parallel with the ADC pins? I have a lot of confusion between RF connections where everything needs to be matched, and the sort of typical "analog" circuits where you go from low impedance outputs to high impedance inputs. \$\endgroup\$ – RJTK Nov 18 '14 at 14:49
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You seem to be confusing differential operation vs. non-differential operation and are confusing the input Vref with the signal DC. You must break your Signal side of operation from the bias (Vref) operation.

A case in point. If your Vref is 1.5 volts and you want to pass 0.5 V DC through, then the input to the ADC V+ = (1.5 + 0.5/2) = 1.75 and V- = (1.5 - 0.5/2) = 1.25. You'll notice that for a differential output the DC from the offset is subtracted and does not exist. Vsig = V+ - V- = 1.75 -1.25 = 0.5 (i.e. "DC").

You haven't shown any diagrams or schematics, but I suspect your issue lies in your single ended to differential conversion.

As to your other questions: Yes, you can do band stop and use only capacitors, as inductors and capacitors have reciprocal operations. It just depends on where you place then in the circuit and how many stages you are running.

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  • \$\begingroup\$ Yes I think I used the wrong terminology. I have a differential output from an IC which I need to terminate into the ADC. But the output from the IC is centred about 0v, but the ADC Vref is 1.5V. So I need to lift up the signal with a common mode offset. Is this correct? Makes sense? Or am I mistaken somewhere? This is the first time I'm actually working with these kinds of devices. \$\endgroup\$ – RJTK Nov 18 '14 at 14:46
  • \$\begingroup\$ @Jyan yes you need to provide the conversion to differential and separate out the terms (common mode vs. signal). It can be confusing and hard to get your brain around it sometimes. Often ADC's or single ended to differential convertors feed the middle voltage back precisely so it can be used for this purpose. \$\endgroup\$ – placeholder Nov 18 '14 at 15:46
  • \$\begingroup\$ Yeah okay I think I am on the right track. \$\endgroup\$ – RJTK Nov 19 '14 at 17:37

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