Bubba oscillator questions

Do you mean this figure?  It's Figure 18.   It looks to me like maybe you are correct, but I think the phase on either side of the 10kOhm resistor would be the same.  The reason I would take the output at the previous amplifier output (instead of at the amplifier input) is that you don't have a 10 K Ohm resistance limiting the drive capability of that node. 

I can't help you with the 1-5 MHz question.  I have not worked with this circuit.  We do have amplifiers that are fast enough to have plenty of loop gain at 1-5MHz, though I'm not sure that they have open loop bandwidth that high.  You may want to pick a high speed amplifier and try building the circuit in the TINA-TI simulator.  Something like an LMH6609 might work. 

Dear Loren,

Thanks for your reply, but I am honestly extremely surprised by your answers considering you re a TI employee:

(i) "It looks to me like maybe you are correct." Huh? Either I am correct or I am not. Surely you can provide a more definitive answer?

(ii) "I think the phase on either side of the 10kOhm resistor would be the same." Huh? Simple circuit analysis tells you that is clearly impossible. Assuming a high-enough input impedance for the opamp, the current flowing through the 10k and 10nF components is the same; the voltage across the resistor is in phase with the current while that across the capacitor lags by 90 degrees. Therefore the phase on either side of the 10k resistor cannot be the same.

(iii) "We do have amplifiers that are fast enough to have plenty of loop gain at 1-5MHz, though I'm not sure that they have open loop bandwidth that high." Huh? You are a TI employee - if you are not sure about the BW/open loop gain about TI opamps you work with, who is?

May I request a knowledgeable TI engineer to help me with the answer to my original question? Surely there is someone out there who can :-).

Best,

-Ram

I am sorry I can't give you a definitive answer on this circuit.  You are correct, I'm not expert on this, maybe someone who is will chime in. 

As drawn the circuit is very assymentrical, the impedance of the Cosine output is very low while the impedance of the Sine output is very high.  I would expect this to present practical problems unless both points are buffered. 

If you look at figure 19 the "output" of the circuit, it does not look like 90 degrees between the top and the bottom traces, so it looks like there is some room for improvement. 

Bart,

Thank you for confirming this and for the simulation file. However, instead of taking the Cosine output from the left of its 10k resistor, is it not a better idea to take the Sine output from the top of its 10k resistor as I originally suggested? Both strategies would give you quadrature outputs, but with the latter method you would have the benefit of taking both outputs directly from opamp output terminals and therefore reduce or eliminate the need for a buffer stage.

-Ram

I think that's a reasonable assertion for the reason you mention.  The downside of course is one less stage of filtering for each output, and thus more distortion.  The primary concern would be the Sine output, which, with only one stage of filtering, could have a bit of a saw tooth look to the sinusoid in the worst case.  The appropriate choice may ultimately be determined through evaluation, and what performance requirements your application necessitates.

Very interesting and useful insight there - thank you Bart. In my application I will take the outputs as you originally suggested and add buffers.

Best - Ram