Till now I studied voltage feedback opamps and I know how circuits involving them are solved. What is difference between current feedback and voltage feedback amplifier? What are steps in analysis of current feedback opamps?

 

Interesting question. But I do not know the answer. I never used any CFA op amp. But sometimes I was using CFA with a discrete components and I analysis it with a traditional way, the same as we analysis any transistor circuit.
http://www.ti.com/lit/ml/sloa080/sloa080.pdf

 

Hi,

A short hand view is that while a voltage feedback op amp depends on pushing the input voltage difference toward zero in order to obtain accuracy at the output, the current feedback op amp depends on pushing the input current difference toward zero with the exception that the non inverting input is voltage buffered. So it's not really a true current feedback amplifier it's half current feedback and half voltage feedback ie one input is voltage feedback and the other is current feedback in reality. Internally it is current feedback however.
The other difference is usually current feedback op amp circuit resistances are kept low to keep bandwidth high and usually those circuits work with low impedances anyway.

 

Note that to operate as MrAl states, the inverting input has a very low impedance so that current can readily flow into that input. This of course, is opposite to a voltage feedback op amp where both inputs are very high impedance.
The non-inverting input is voltage controlled (not voltage feedback, as you normally don't connect feedback to the plus input) with a high input impedance, similar to a voltage controlled op amp.

An interesting factoid about current feedback op amps is that they do not have the constant gain-bandwidth product limitation that voltage op amps do.
The bandwidth of a current feedback op amp is not greatly affected by the closed-loop gain as a voltage-feedback amp is, where doubling the gain will typically halve the bandwidth.

 

Hello,

When discussing current feedback vs voltage feedback it is certainly acceptable to call the non inverting input a voltage feedback input. Positive feedback is an acceptable type of feedback which helps certain circuits attain better characteristics overall.

Also, "The bandwidth of a current feedback op amp is not greatly affected by the closed loop gain as a voltage feedback amp is", is not true so i have to wonder who made that up. There are current feedback amps that have half the cutoff frequency for twice the gain, just like any other op amp. They are also affected by the actual value of the feedback resistor where the gain goes down as the resistor value goes up, even with the same gain.
The real value is that they can be so fast to begin with under the right external circuit conditions...like 800MHz.

 

...................

Also, "The bandwidth of a current feedback op amp is not greatly affected by the closed loop gain as a voltage feedback amp is", is not true so i have to wonder who made that up. .................

Wonder no more.
Would you believe Texas Instruments?
Read the last sentence on page 8 of this reference.

 

Wonder no more.
Would you believe Texas Instruments?
Read the last sentence on page 8 of this reference.

Hi,

Yes i would believe that

Here's another little equation i also believe:
TI(t)=NotLivingInTheRealWorld(t)

<chuckle>

Seriously though, it sounds like they are stating that there is never a frequency roll off, but it seems they are just relating theoretical aspects of the design of an op amp. Probably their main assumption is that a capacitor can charge with a current source (originating from feedback) in zero time. If what they say is true across the board, then data sheets would not reflect a decrease in cutoff frequency with gain.
You can look on some data sheets for CF op amps to find out the real life truth
"Look at me Ma, i'm using my op amp at 9.7E+15 Hertz!" ha ha.

 

.......................
Seriously though, it sounds like they are stating that there is never a frequency roll off, but it seems they are just relating theoretical aspects of the design of an op amp. Probably their main assumption is that a capacitor can charge with a current source (originating from feedback) in zero time. If what they say is true across the board, then data sheets would not reflect a decrease in cutoff frequency with gain.
You can look on some data sheets for CF op amps to find out the real life truth
"Look at me Ma, i'm using my op amp at 9.7E+15 Hertz!" ha ha.

Seriously, that's not what they are saying.
Obviously CF op amps have a frequency response rolloff (as stated in their data sheets) it's just that the rolloff point is not as highly affected by the closed loop gain as a VF.
CF amps don't have the strict gain-bandwidth-product limit that VF amps do. That is because CF amps don't require the internal compensation to give a 20dB/decade open-loop rolloff that VF amps have for closed-loop stability.

For example this CF amp data sheet shows that increasing the closed-loop gain by a factor of 10 only reduces the bandwidth by a factor of 2 or less, not a factor of 10 that you would see with a VF amp.

 

Hi,

Yes, but other data sheets do not show that behavior.

But rather than argue this point back and forth to infinity, i will agree that some op amps will be more affected than others, just not all.

The real point i was trying to get at though is that the wording does not sound right. It sounds like they are saying that there is no reduction in response when clearly there is. So really i just though it was misleading, that's all. Your explanation sounds very good however so i think we both got our points across and i am happy you stuck with this one

Maybe we could come up with a better wording...

 

Yes, but other data sheets do not show that behavior.
.......................

Can you post one for a CM op amp? I'm curious.

 

Hi again,

Well, AD8001 for example. This was my choice years ago for a digital scope front end, for better or for worse

I'll save you some trouble:
880MHz gain of 1
440MHz gain of 2

I am pretty sure that 880 divided by 2 equals 440 (little joke there)

 

Hi again,

Well, AD8001 for example. This was my choice years ago for a digital scope front end, for better or for worse

I'll save you some trouble:
880MHz gain of 1
440MHz gain of 2

I am pretty sure that 880 divided by 2 equals 440 (little joke there)

Please trouble me.
Where did those response values come from?
I don't see that on the data sheet.

Page 10 (bottom of first column) of the AD8001 data sheet states "Recognizing that G × RIN << R1 for low gains, it can be seen to the first order that bandwidth for this amplifier is independent of gain (G). This simple analysis in conjunction with Figure 5 can, in fact, predict the behavior of the AD8001 over a wide range of conditions."

 

Please trouble me.
Where did those response values come from?
I don't see that on the data sheet.

Page 10 (bottom of first column) of the AD8001 data sheet states "Recognizing that G × RIN << R1 for low gains, it can be seen to the first order that bandwidth for this amplifier is independent of gain (G). This simple analysis in conjunction with Figure 5 can, in fact, predict the behavior of the AD8001 over a wide range of conditions."

Hi...

 

Hi...

That opening paragraph certainly seems to be in contradiction to the paragraph I quoted.
Perhaps that's related to the way they specify the bandwidth with a very low peaking level.
That certainly is not typical behavior for a CM op amp.

 

Hi again,

Maybe because it is an older model, or perhaps because it is so high in frequency to begin with? Just guessing now as i dont work with them much anymore either. I may have to again soon though.