1
\$\begingroup\$

First schematic below is the initial gain stage of the intermediate signal generated by mixer of radar. It is passed through sallen key high pass filter.

Initial gain and sallen key high pass filter

Second schematic for differential to single ended conversion with some gain. And then another stage with digital potentiometer for adjusting the gain with a microcontroller.

Second part of the circuit

At last stage i need to convert signal back to differential for driving the SAR ADC. I have found 2 reference designs from links below and converted accordingly.

Page 28 of the datasheet for single to differential convertion

Single to differential convertion reference 2

ADC range is 0-3.3V and DC Bias should be around 1.65V.

My questions are:

  1. Is this design correct, especially the second schematic. When you check the stages, is there any problem that i could not see such as wrong conversion, wrong dc bias apply or connection etc. For example can i directly connect pin 14 of the ADA4807 to pin 5.

  2. For DC Bias generation, can i simply use a LDO regulator with high PSRR. I had oscillation problems when i generated DC Bias with op-amp voltage follower.

ADA4807 Datasheet

LT6232 Datasheet

Later Update for further details of the system: For radar, a closest target at 1 meter distance will reflect 1 to 3 kHz 0.05-0.1V IF signal but the signals coming from further targets are a lot weaker down to microvolts. The reason of the high pass sallen key is that distant target will reflect microvolt level signal therefore it is like a noise on closer target. When 40 dB/dec high pass filter is used, the closer signals will become weaker meaning lower amplitudes (fc = 400 kHz, Q = 0.5, damp = 1 for this setup because radar cannot see beyond 200-250 meters but it can be changed for higher cutoff later). This is the end of first schematic. Overall we amplified signal and prepare it for filtering. After the high pass filter section the signal is compensated but still needs a lot more amplification because after first gain stage filtering decreased amplitudes.

Higher sampling rate improves SNR performance along with differential input thats why i kept it differential, but converting single ended would be easier in case of number of components as well. Sampling will be around 3.3 to 5.6 MHz (higher sampling and down sampling is better. It will be certain according to the USB throughput). For now it will be 3.3 MHz sampling rate.

\$\endgroup\$
4
  • \$\begingroup\$ It may be best to convert to single ended at first, before any filtering, then convert back to differential at the end. \$\endgroup\$
    – user69795
    Jan 7 at 23:02
  • \$\begingroup\$ Please update the question with more information. a) What are the high and low corner frequencies of the passband for the overall circuit? b) What is the overall gain in the center of the passband? c) How much of that gain is in the first stage? Based on the answers I might agree with 69795 about making most the circuit single-ended. \$\endgroup\$
    – AnalogKid
    Jan 7 at 23:08
  • 1
    \$\begingroup\$ 2nd schematic, U32A - There is no equivalent to C9 in the non-inverting signal path. \$\endgroup\$
    – AnalogKid
    Jan 7 at 23:10
  • \$\begingroup\$ @AnalogKid Hi, thank you for your answers. I have added more detail. Can you give me feedback again. What do you mean by c9 has no equivalent, is it unnecessary. Should i get rid of it? \$\endgroup\$
    – Cenk
    Jan 10 at 12:44

1 Answer 1

2
\$\begingroup\$

The one thing that I have encountered in the past, is oscillataion do to positive feedback through the common mid-point bias connections. I have shown one such positive feedback path in the diagram as a red line. If the gain around this loop is greater than 1, then it will oscillate.

Look at pin 3 of U32A. A signal <3MHz on this pin will be multiplied by 11. So the rest of the path must attenuate by at least the same amount.

You can find other such positive feedback paths on both schematic diagrams.

If I am not mistaken, there is one long path from DCBIAS on U15A pin 3 through IFF+,through U32A Pin 3, and returning to DCBIAS through U32D Pin 12.

There are 3 non-inverting gain sections for signals on this positive feedpack path totaling about 1700.

This is likely the source of your oscillation.

There is another source for oscillation. As the bias network grows, the capacitive load on an op-amp driver will increase. If the capacitive load is too great for the opamp, it will oscillate. Using a regulator instead may be better. Choose a regulator that can work with 10s of microfarad capacitance on its output.

enter image description here

An op-amp used for mid-point bias must be able to handle all the ac current that must pass through the bias network.

An LDO regulator can sink current only through its output capacitor so may not be appropriate either.

Even if a perfect source is available, the inductance and resistance of the bias distribution network (BDN) will cause the voltage to bounce depending which way the current passes.

To correct this problem, the positive feedback loops must be broken or decoupling applied at each tap from the BDN. Both these methods are shown below.

In Figure 1. R1,C1 and R2,C2 decouple the OA1 and OA2 respectively from the BDN and from each other, But there is still a hardwired connection between each stage.

In Figure 2, The voltage dividers provide a local bias voltage while the capacitors provide an ac ground. The stages are completely separate from each other so cannot interfere with each other. The voltage dividers each can be buffered with an op-amp if necessary, but usually just the bypassed voltage divider is enough. You may not need the capacitor depending on configuration.

I have used both methods successfully and prefer the method in figure 2.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
2
  • 1
    \$\begingroup\$ That red path made wonderful oscillators even back in the vacuum tube days. It's a real menace. Once you see it, though you sure don't forget it :) \$\endgroup\$ Jan 8 at 2:24
  • \$\begingroup\$ Even poor grounding can suffer from this effect. @Kubahasn'tforgottenMonica \$\endgroup\$
    – RussellH
    Jan 8 at 3:00

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.