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I have a DDS chip with complementary current outputs that I need to filter. I'm using a design like this:

enter image description here

I need a different cutoff frequency, however - 2 MHz instead of 50 kHz. Note that I'm not using the particular op amps specified in that schematic - I'm using the LT6207 instead.

I understand the basic operation - TIAs followed by a differential input amplifier - and after reading the active filter design techniques chapter for Op amps for Everyone, I think I understand the basics of active filters, but I'm stumped as to how to apply those principles to this design.

Unless I'm mistaken, it consists of a first order filter (the TIAs) followed by a third order filter, which violates Op amps for Everyone's prescription that only odd order filters contain order-1 stages. Thus, their calculations and coefficient tables don't seem to be terribly useful.

How can I calculate the component values to adjust my filter for a new cutoff frequency?

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  • \$\begingroup\$ I usually think of filter orders in terms of capacitors. Your input stages are first order (C1, C2). Your output stage is second order (C4, C5/C6). In between, there is C3: a first order PASSIVE filter stage. \$\endgroup\$ Oct 23, 2014 at 9:45
  • \$\begingroup\$ I afraid you will have some problems for a cut-off at 2 MHz because the GBW of the opamps is only 10 MHz. \$\endgroup\$
    – LvW
    Oct 23, 2014 at 9:55
  • \$\begingroup\$ Try using a simulator like LTSpice. \$\endgroup\$
    – Andy aka
    Oct 23, 2014 at 10:01
  • \$\begingroup\$ @AlanCampbell That helps - but how do I then actually figure out the filter values? \$\endgroup\$ Oct 23, 2014 at 11:12
  • \$\begingroup\$ @LvW Sorry, I should have said that I'm not using that particular opamp. I'll update the question. \$\endgroup\$ Oct 23, 2014 at 11:12

1 Answer 1

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First you will need to find a suitable opamp which, at 2MHz, is not the 5534. GBW should ideally be 2 orders of magnitude more than your cutoff, or at least 100MHz (or close) and keeping it stable may be an issue.

Second, you can scale R-C networks to change the frequency. So for example, R1 sets the gain of the input stage and R1C1 set the frequency. You may be able to simply scale C1 as 50/2000 * 2200pf, or 56pf. (Ditto C2)

(You may have to reduce the stage gain, in which case reduce R1. 82 ohms and 560pf would work or some intermediate value keeping R1C1 constant.)

Ditto the passive 1st order stage R3,R4,C3 scale together so you could scale C3 to ... 560pf. The relatively low resistor values here should be fine at 2MHz.

The last stage (2nd order) is a little more complex because if you scale the time constants differently you will also affect the Q, or peakiness of the stage. But again the resistor values look fine so I would simply scale the capacitors, C4=680pf, C5,6=200pf (ideally 205pf).

And simulate as Andy says.

If it doesn't behave as expected, compare the original unscaled simulation with the scaled version. Look at each stage e.g. U1 output, separately.

The opamp characteristics will interact with these ideally scaled component values and the response may not be quite as expected especially if GBW or the output slew rate is too low.. A breadboard will introduce further stray capacitances and inductances, and the PCB layout will be slightly different again...

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  • \$\begingroup\$ So, in a nutshell, scale all the capacitors by 2000/50, except the first stage, which I may also want to alter the capacitor/resistor ratio in order to adjust the gain? That seems like a good empirical solution, though I'd still like to understand how to do it from a theory POV. \$\endgroup\$ Oct 23, 2014 at 11:57
  • \$\begingroup\$ 50/2000, but yes. There's a basic "principle of similarity" involved in the scaling : in filter textbooks (search Amazon for "Zverev"!) you often find filter designs normalised to impedance Z=1 and omega=1, frequency f=1/(2*pi), you get to do the scaling yourself! \$\endgroup\$
    – user16324
    Oct 23, 2014 at 12:02
  • \$\begingroup\$ I've simulated this, and I get reasonable results visually - about 49dB/decade, with minimal peaking - pretty much the same as the original filter. \$\endgroup\$ Oct 23, 2014 at 12:11
  • \$\begingroup\$ Thinking more closely about that, 49db/decade doesn't seem like a great figure for a 4th order filter. Is there a reason it wouldn't be higher in this design? \$\endgroup\$ Oct 23, 2014 at 12:17
  • \$\begingroup\$ Two first-order sections probably leave it underdamped. Which would make the first decade fairly poor. Second decade ought to be better. Except with real opamps... No, this is probably as expected. Above 1 MHz I'd be inclined to use passive L-C filters instead. \$\endgroup\$
    – user16324
    Oct 23, 2014 at 12:24

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