High bandwidth in CFA(current feedback amplifier)

[engrvip]

what is the exact reasone for such high bandwidths in CFAs.Although i know that BW is independent of gain in CFAs but that to me seem to be a good way to justify high BW using mathematical analysis.......can somebody explain what are the physical or functional reasons which cause such high BW in CFAs???????

 

[timo67]

I advice you tou read
**broken link removed**
or any Application Note on the current feedback amplifiers
Comlinear CLC400 for rxample

 

[engrvip]

well i have gone thru lot of these ANs but they try to explain this high BW phenomena thru mathematical analysis ......what i want to know is what is thr physical phenomena which causes this high BW.

 

[timo67]

In reality CFAs do not have higher bandwidths than traditionnal Op amps
The particularity of CFAs is that their bandwidth does not decrease when they are used with with a resistive feed back to make an amplifier.
On the other hand traditionnal Op Amps have a constant Gain Bandwidth product.
If you multiply the gain by 10 your banwidth is automatically divided by 10.
CFA's are current mode circuits and I advise you to read :
Analogue IC design: The current mode approach
By Toumazou, Lidgey and Haig
Chapter 16 for example

 

[wpchan05]

There are two kinds of bandwidth you need to be aware: large signal BW and small signal BW. Large signal BW is usually related to slew rate perfomance. Since CFA uses currents as variable for for signal processing, there is ideally no limit on slew rate (remember current see capacitor as a short). So many ads on CFA specify their SR over 1000V/us, not tens of volts per micro sec as common in traditional voltage opamp.
small signal BW relates closed loop unity gain BW. Since CFA can be modelled very good as single pole device, a large GBW can be achieved. Of course, if you want to prove the gain bandwidth independence as found in CFA, you need math!

 

[engrvip]

Large signal BW being high can be explained due to high slew rate in CFAs.But still not able to fathom the reasons for small signal BW being so large.

What i can understand of ur explaination is that CFAs have got a very good single pole response which means that it's first pole is also very high in value to give it a very good 3db bandwidth and can this be related to the fact that parasitic values in CFAs are smaller then VOAs and also compensation cap. Cc at end of first stage is also small contributing to this large figure for small signal BW.

Plz correct me if my understanding of this fact is worng?????????????????

And one more thing what is generally the difference between these two values of large and small signal BWs if we talk in general for any ckt and also in this particular case of CFAs.because i generally dont find the spec. of large signal BW in data sheets what we see is only 3dB BW spec.

thanks

 

[wpchan05]

Large signal BW being high can be explained due to high slew rate in CFAs.But still not able to fathom the reasons for small signal BW being so large.

What i can understand of ur explaination is that CFAs have got a very good single pole response which means that it's first pole is also very high in value to give it a very good 3db bandwidth and can this be related to the fact that parasitic values in CFAs are smaller then VOAs and also compensation cap. Cc at end of first stage is also small contributing to this large figure for small signal BW.

Plz correct me if my understanding of this fact is worng?????????????????

And one more thing what is generally the difference between these two values of large and small signal BWs if we talk in general for any ckt and also in this particular case of CFAs.because i generally dont find the spec. of large signal BW in data sheets what we see is only 3dB BW spec.

thanks

most CFA are bipolar devices, so they have a very large gm as compared to CMOS. Non-dominant poles in CFA are roughly equal to gm/C; this gives CFA a relatively large GBW.

However, the unique properties of CFA is that the GBW increases as you increase the closed loop gain. This is the result of using current feedback, which you can understand more CFA if you do a small signal analysis of it.

 

[engrvip]

Basically the first pole in CFA transfer function response ( which also determines it s 3-dB BW) depends only on the value of feedback resistor. (This can be seen in various AN's).

And since this feedback resistor (Rfb) value is always kept constant to the value given in the data sheet , hence the closed loop gain ( 1+ Rfb/Rff) is varied only by changing value of feed-forward resistor(Rff).

Hence the 3-db BW of CFA does not change even though gain is changed by varying feed-forward resistor value.

 

[crutschow]

Another factor that helps the high frequency operation of a CFA is that, since the inverting input is a low impedance current input rather than the high impedance voltage input of a standard op amp, the CFA is much less affected by the unavoidable stray capacitance at the inverting input. This stray capacitance can cause significant overshoot and ringing with a standard op amp, much more so than with a CFA.

To see this, note that the standard op amp inverting voltage input must change its voltage slightly due to the signal voltage change, which requires the charging and discharging of the stray capacitance. This charging current must be provided from the input and feedback resistors which distorts the frequency response and adds phase shifts. A CFA inverting low-impedance current input has very little voltage change due to signal voltage change, thus the stray capacitance requires very little charge current and thus has little effect on the frequency response and phase shift.

 

[LvW]

The explanations as given in post#8 can be easily verified by inspecting the corresponding real closed-loop gain function (for finite and frequency-depending open-loop gain properties):

Example: Non-inverting amplifier (negative feedback via R2-R1).
It is easy to compare both cases if we write the closed-loop gain function as a product (gain ideal)*(error function) which turns out to be V2/V1=(1+R2/R1)/D(s)

1.) Voltage opamp: D(s)=[1+(1+R2/R1)/Aol(s)]
2.) Current-feedback amp CFA: D(s)=[1+R2/Ztr(s)]

As can be seen, in both cases the closed-loop gain approaches (1+R2/R1) for ideal devices (opamp open-loop gain (Aol) and CFA transfer resistance (Ztr) approaching infinite, resulting in D(s)=1).
However, there is one important difference:

1.) Voltage opamp: In D(s) we see that (1+R2/R1)/Aol(s)=1/[(Aol*R1/(R1+R2)]=1/LG(s) ;( LG=Loop gain)
2.) Cuurent-feedback In D(s) we have R2/Ztr(s)=1/(Ztr(s)/R2)=1/LG(s)

In both cases the error function contains the finite loop gain LG(s).
However, in case 1 (voltage opamp) the loop gain cannot be selected independent on the ratio R2/R1 (that means: not independent on the desired gain), whereas in case 2 (CFA) it is only R2 that controls the loop gain.
As a result, voltage opamps mus be frequency-compensated in order to operate for all closed-loop gains.
In contrast, CFA units can be made stable by proper selection of the feedback resistor R2 - independent on the desired gain (1+R2/R1). For this reason, CFA units do NOT require any frequency compensation and, hence, provide a larger bandwidth.