Continue to Site

resistive feedback opamp with tiny loop gain

Status
Not open for further replies.

exp

Full Member level 1
Hi,

In a paper I found the use of an OTA+buffer ("opamp") as a baseband amplifier after the mixer. It uses resistive feedback and the source resistance is given by a series of the antenna and the switch impedance of the mixer. To first order, the amplifier can be descried as the ordinary "inverting opamp amplifier":

https://en.wikipedia.org/wiki/Operational_amplifier#Inverting_amplifier

with typical numbers from the mentioned paper: Rin=55 Ohm, Rf=2k, a0=30 (since it consists of a simple diff pair + source follower in a 65nm process).

Using these numbers I get a Gain of A0 = a0/(a0+1) * 1/(Rin/Rf + 1/(a0+1)) = 16 which is OK.

However, the loop gain is given as T = a0/(1 + Rf/Rin) = 0.8029 < 1 !

The approximation A0 = Rf/Rs is only valid if a0 >> Rf/Rs.

Now I am confused: I always learned the loop gain should be as high as possible for good accuracy, linearity etc.
Does it make sense to use an amplifier like this where the loop gain is even negative? (one advantage is that it is unconditionally stable).

Based on the T expression, it seems like the circuit requires Rin >> Rf (or extremely high a0).

Also, since the distortion coefficients are attenuated by (1+T), wouldn't a configuration like this be a desaster for linearity?

Or do I just miss something?

It is ok to use an amplifier as long as the specs are met. But the big question is that the Gain Ao varies all over the place since the open loop gain of the amp is very low. If Ao=Rf/Ri (i.e very high open loop gain of opamp), it would be fairly constant across PVT changes. But with low open loop gain, it could be difficult to control the overall gain (In case you are planning to implement gain programmeability) of the block across PVT.

Lower loop gain may not give you the advantage of having feed back loop. i.e achieving high linearity numbers could be difficult (You may see the open loop linearity and closed loop linearity numbers to be almost same).

It would be good to increase the open loop gain of the amp to improve the performance.

Ok, thank you, this was a bit what I expected.

First of all, this is not a discrete dircuit but an IC in 65nm so achieving high loop gain is difficult to begin with (even with 2 stages not more than a couple of hundred).

But this is not so much my confusion here: You are talking about the "open loop gain of the amp" which would be the gain of the actual amplifier (a0). I Wouldn't be surprised if increasing a0 would give me better performance.

However, it turns out that the loop gain of the circuit is T0 = Rin*a0/(Rin + Rf)! This is different than the open loop gain of the amplifier (a0) and the effective loop gain T0 is even always much less than a0, specifically if Rf > Rin.

Now I could say: Not a big of a deal, just make Rin >> Rf: done.

This configuration is widely used as a TIA. In this case the source resistance would be high (current source). However, in this case, the source resistance is small (hence it is driven by a voltage source). So even when using moderate Rf values this would imply a low loop gain (for a practical IC OpAmp and not a discrete 120dB opamp). Right so far?

What is more, in the paper, the feedback resistor should be used for matching. The setup is as follows: The antenna (modeled as ideal voltage source in series with Ra=50 Ohm), followed by a switch with on-resistance of Rsw=20 Ohm and then the TIA with feedback resistor Rf.

For matching to occur, the input resistance of the amplifier (i.e., the op-amp only with Rf) should have a certain value which is Ra-Rsw=30 Ohm. Then the required Rf is given by

Rf = (1+a0)*(Ra-Rsw)

The total input resistance of the opamp circuit is given by the series resistance of Ra and Rsw:

Rin = Ra + Rsw

Now I just plug in the definition for Rf into the expression for T0 and I arrive at:

T0 = Rin*a0/(Rin + Rf) = ... = (Ra+Rsw)/(Ra*(1+2/a0)-Rsw) ~ (Ra+Rsw)/(Ra-Rsw)

This shows that T0 is nearly independent of the gain of the amplifier (a0) and very low for typical numbers (e.g. 2.3 for Ra=50 Ohm, Rsw=20 Ohm).

All this is consistent with spice but it still sounds fishy to me ... that the loop gain is to first order independent of the amplifier gain and very, very low.

Last edited:

I believe, you're post is desperately asking for a schematic that clarifies the meaning of the various terms.

Even when recalling commonly used formula signs in analog design text books, they don't make sense at first sight...

No problem:

(this is the simplified single-ended version which has some problems because Rsw=20 only during on-time but the actual circuit is fully differential so there are always 20 Ohm)

Last edited:

For matching to occur, the input resistance of the amplifier (i.e., the op-amp only with Rf) should have a certain value which is Ra-Rsw=30 Ohm. Then the required Rf is given by

Rf = (1+a0)*(Ra-Rsw)

Don't know the said paper, but this approach has nothing in common with regular OP or TIA circuit design. It's the reason for forcing loop gain to unreasonable low values.

There's no point of matching a source impedance by increasing Rf respectively reducing the loop gain. In case of doubt, the closed loop gain would be simply too high, causing output saturation.

Matching input impedance this way can be however done with amplifiers that have a low, well-defined gain, e.g. a LNA. In case an OTA with resistive load is suited for you as amplifier with defined gain, an OTA could be used. But it's neither TIA nor OP design principle.

Don't know the said paper

The said paper(s) are:

"A Passive Mixer-First Receiver With Digitally Controlled and Widely Tunable RF Interface," Caroline Andrews, Student Member, IEEE, and Alyosha C. Molnar, Member, IEEE

"Implications of Passive Mixer Transparency for Impedance Matching and Noise Figure in Passive Mixer-First Receivers," Caroline Andrews, Student Member, IEEE, and Alyosha C. Molnar, Member, IEEE (more details).

It is not perfectly equivalent because the paper discusses a sampling mixer (and hence the op-amp would include a small shunt resistance to ground also) but the general principle incl. my similar numbers fully apply.

, but this approach has nothing in common with regular OP or TIA circuit design. It's the reason for forcing loop gain to unreasonable low values.

There's no point of matching a source impedance by increasing Rf

But this is exactly what the paper proposes.

respectively reducing the loop gain.

Ok, so I make no mistake when I say high Rf vs. low Rs results in low loop gain (as discussed abo

In case of doubt, the closed loop gain would be simply too high, causing output saturation.

I don't get your point here. With the typical numbers I used the closed loop gain A0 = Rf/Rs * T0/(1+T0) is somewhere between 3 and 20 ... very reasonable.

Matching input impedance this way can be however done with amplifiers that have a low, well-defined gain, e.g. a LNA.

But this is exactly the point of the paper: Getting rid of the LNA, connecting the antenna directly to the mixer and using Rf for matching.

In case an OTA with resistive load is suited for you as amplifier with defined gain, an OTA could be used.

I think this is exactly why not an OTA is used but an OP (OTA + Buffer): Without the buffer, the input impedance to the amplifier would be something like (Ro+Rf)/(1+Gm*Ro) = (Ro+Rf)/(1+a0) which includes the Rout of the OTA which is not well controlled either.

Since you are trying to match the input impedance of a TIA to the antenna impedance, you are facing the issue. Typically open loop gain stage with low, defined gain is used to match the impedance as well as to get some gain.

If the switch you shown does Mixing, then the impedance matching equation is not simply what was mentioned earlier? Because the Mixer comes in, changing the entire scenario.

Last edited:

In a paper I found the use of an OTA+buffer ("opamp") as a baseband amplifier after the mixer. It uses resistive feedback and the source resistance is given by a series of the antenna and the switch impedance of the mixer. To first order, the amplifier can be descried as the ordinary "inverting opamp amplifier":

... it consists of a simple diff pair + source follower

I'd say such an amplifier shouldn't be called an OTA, much less an opamp.

Probably the high bandwidth here is more important than the low loop gain (which isn't negative BTW).

If the switch you shown does Mixing, then the impedance matching equation is not simply what was mentioned earlier? Because the Mixer comes in, changing the entire scenario.

No, see the remark under my circuit: For simplicity I use a single ended version here. The actual circuit is fully-differential, having always exactly one closed switch with Ron connected.

If all parasitic capacitances are reasonably small, then the impedance is determined by the indicated resistances up to very high frequencies: The mixing operation does not change any impedances then. This is also consistent with my spice results.

Scenario would change (slightly) if I have a sampling mixer which implies a big load capacitance. In this case the effective resistance after the mixer would change and also a virtual shunt capacitance would be introduced. This is the case for the referenced papers (and explained there). However, as I mentioned already, the effective numbers are quite comparable to my ohmic-only case!

I'd say such an amplifier shouldn't be called an OTA, much less an opamp.

I would so because this is exactly it: a lousy, fully differential op-amp.
When I draw the small signal circuit, this is exactly a Gm, followed by a buffer. It can be analyzed identically to the op-amp circuit (with low open loop gain a0).

Probably the high bandwidth here is more important than the low loop gain (which isn't negative BTW).

(my fault, of course, negative in dB)

Yes, but then I wonder

a) How they achieve the actual matching and precise gain if it's so heavily dependent on a0
b) How they achieve any acceptable linearity. As observed by one other posting (https://www.edaboard.com/threads/344266/) the linearity of a simple diffpair is fairly unacceptable (and also inconsisted with their reported number of -8dBm for low CL).

I agree completely with erikl. An OP is a device that is operated with loop gain > 1 and an overall (closed loop) gain set by the feedback network.

The circuit you are discussing is something different. According to the low loop gain, overall gain and input impedance will strongly depend on a0 and the linearity relies solely on that of the inner amplifier. Consequently there's no satisfying answer to your questions a) and b).

Why dont you try a telescopic structure for TIA? It will mostly work. Or two stage amp as in the second paper you mentioned.

Input impedance of a TIA is 1/Gm. You can use any value of Rf depending on the conversion gain you want. You need to match 1/Gm of the TIA to 50 ohms.

Last edited:

I agree completely with erikl. An OP is a device that is operated with loop gain > 1 and an overall (closed loop) gain set by the feedback network.

Well, ... no. An OP is a device (as you say) and not a circuit which uses an OP! I can happily build the same circuit with a low loop gain using a real, discrete OP. I would agree if you say a circuit which uses an OP the described is not a conventional usage of an OP. But let's stop this discussion, it does not matter.

Why dont you try a telescopic structure for TIA? It will mostly work. Or two stage amp as in the second paper you mentioned.

Input impedance of a TIA is 1/Gm. You can use any value of Rf depending on the conversion gain you want. You need to match 1/Gm of the TIA to 50 ohms.

Why would this matter? A telescopic is nothing more than a Gm cell with a bit higher Gm*Ro=a0 than the one described by me. At the block level this is identical. And, when I add the buffer, then really nothing changes because T0=(Ra+Rsw)/(Ra-Rsw) as shown above.

Even if I omit the buffer, the input impedance is 1/Gm only to first order. As described above:

Rin = (Ro+Rf)/(1+a0) \approx 1/Gm + Rf/(1+a0)

If I did not do a mistake the loop gain is given as:

T0 = a0 * Rin/(Ro + Rin + Rf)

Same as before, this requires a large Rin so that this expression becomes roughly 1.

Even with Rf=0, for a certain loop gain T0:

a0 = Gm*Rin*T0/(Gm*Rin - T0)

If Rin=70 as in my case and we suppose we need Gm=10mS then there is no a0 which would give me a loop gain > 1. Hmm, something looks fishy here ...

- - - Updated - - -

For some reason I cannot edit my posting any more. I went over it a couple of times, I do not find any mistake above. If I replace the Rs with caps and set s=0 I get the familiar OTA T0=beta*a0.
I think this is precicely the reason why you usually use capacitive and not resistive feedback with OTAs. When Rin is large (current source, e.g. 100k) then a large T0 can be achieved.

Just keep us informed if you solve problems a) and b).

Status
Not open for further replies.