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From the TI application note RF and IF amplifiers with op amps:

schematic: wideband RF amplifier

The source says "The 39-pF capacitor provides peaking to compensate for some high-frequency roll-off, but better IP3 performance can be achieved by removing it and living with the roll-off." Let's just consider it left off.

What function do the resistors between the two op-amp stages serve? The choice of \$50\Omega\$ makes me think of transmission lines, but this amplifier has a usable bandwidth up to around 300 MHz, so wavelength is on the order of 1 meter, significantly more than the distance between the stages (it is a dual-op-amp package), so any reflections here would be fast enough to be negligible.

Additionally, the input and the output are each terminated with \$50\Omega\$ resistors. Here's it's reasonable to assume the attached cable is long enough to be considered a transmission line, and these resistors are providing termination for that line. But, why terminate at both ends? Assuming other circuits are doing the same thing, (terminating input and output), won't this serve to cut the voltage in half? This seems to be rather counterproductive for an amplifier; what's the advantage?

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  • \$\begingroup\$ I think its to minimize/stop reflections. Any discontinuity from the characteristic impedance, will cause a reflection which will distort your signal depending on how 'bad' the reflection is. By adding them to both ends, you have a 50ohm constant impedance (because characteristic impedance at the middle is 50) at every point and so no reflections. Thats my guess here, because it seems to make some sense to me. Its also why this is a comment and not an answer. I'll wait for the big boys to show up lol \$\endgroup\$ – efox29 Jul 10 '13 at 2:46
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    \$\begingroup\$ According to the linked document, "Isolation is accomplished by using interstage termination resistors." But exactly what they're isolating from what isn't immediately obvious to me. \$\endgroup\$ – The Photon Jul 10 '13 at 4:54
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    \$\begingroup\$ I have encountered this App Note before. If I'm not mistaken, the noise figure (12dB per stage) is also counterproductive for this sort of application, which leads me to question the central theme of the document. \$\endgroup\$ – mng Jul 10 '13 at 22:23
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I think you're confusing an aerial and transmission line when you mention wavelength in the order of 1 meter - the transmission line equation is essentially independent of the frequency (and hence wavelength).

You're correct to think in terms of transmission lines . The other thing to realize is that this op amp is a current feedback amp (and not the 'normal' voltage feedback amp).

{see http://en.wikipedia.org/wiki/Current-feedback_operational_amplifier }

The 49.9 ohm terminating the first op amp output sets its output impedance to 50 ohm (nominal). The second 49.9 ohm resistor terminates what is in effect a non resonant 50 ohm transmission line and produces a flatly tuned circuit. The result of this termination is to reduce the gain of the first stage by half which looks rather odd but is necessary to maintain the flat tuning of the stage.

enter image description here

Getting back to the 39pf capacitor. It boosts the signal at the high frequency end and compensates for the roll-off in gain but will mis-match the termination leading to some reflection at high frequency.

The circuit is designed to slot straight into a 50 ohm system transmission line system.

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    \$\begingroup\$ Phil's remark about wavelength seems perfectly valid to me. The author of the linked article shows the same circuit for a 10.7MHz IF amplifier for an FM broadcast receiver. The idea that the connection between the two opamps in the same package should be treated as a transmission line at 10.7MHz is a bit far-fetched - and why 50Ω anyway? \$\endgroup\$ – MikeJ-UK Jul 10 '13 at 8:17
  • \$\begingroup\$ @MikeJ-UK Wavelength is not important in this design as the op amp bandwidth limits the upper usable frequencies resulting in a short, non resonant transmission line (terminating load dominant). The first resonance occurring at quarter wavelength. The designer of the original circuit has decided at some point to keep everything at 50 ohm - probably to aid testing of circuits. Apart from pcb design using 50 ohm strip there is no reason why other values could not have been used. The 10.7MHz circuit shows this by using 332 ohm resistors to match the impedance of the SAW filter. \$\endgroup\$ – JIm Dearden Jul 10 '13 at 8:51
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    \$\begingroup\$ My point is that if the distance between pins 1 and 5 is not large enough to warrant consideration as a transmission line. \$\endgroup\$ – MikeJ-UK Jul 10 '13 at 10:33
  • \$\begingroup\$ @MikeJ-UK I understand your point but I'm only responding to the question. I agree with you that the designer of the original circuit did not treat the two op amps as being on the same chip and simply took an 'off the shelf' application note design for a '50 ohm' amp and put them together. I suspect that it was for testing purposes. It worked so they are happy. Its not my job to optimize the design. It would be interesting to see the effect of totally removing the resistors all together. \$\endgroup\$ – JIm Dearden Jul 10 '13 at 10:55
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    \$\begingroup\$ Its flat because the impedance of the transmission line calculates as pure ohmic resistance.[ Z = (L/C)^0.5] There is nothing (in the first approximation) formula that is frequency dependent. At higher frequencies small parasitic inductances and capacitance in the circuit and dielectric losses in the line start to dominate and the gain falls off. As to why it needs terminating have a look at allaboutcircuits.com/vol_2/chpt_14/5.html In a short transmission line the terminating resistance dominates. \$\endgroup\$ – JIm Dearden Jul 13 '13 at 13:36
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My guess is that the circuit doesn't assume that the op-amps are in the same package (despite the pin-numbers) and does assume that PCB traces are 50Ω. The data sheet says :-

A 50-Ω environment is not necessary onboard, and in fact, a higher impedance environment improves distortion as shown in the distortion versus load plots. With a characteristic board trace impedance based on board material and trace dimensions, a matching series resistor into the trace from the output of the THS320x is used as well as a terminating shunt resistor at the input of the destination device.

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  • \$\begingroup\$ I'd still wonder why a resistor is needed at both ends. By my understanding, it only needs to be at either end (and either will do) to eliminate reflections and standing waves, so why both? \$\endgroup\$ – Phil Frost Jul 10 '13 at 19:55

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