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The advice when dealing with amplifying high-frequency signals always seems to be "use a current-feedback amplifier, not an op amp". (see 1, comments on 2, 3 and comments on its question) But why is this? What about current-feedback amplifiers makes them inherently faster than conventional voltage-feedback amplifiers?

I've also seen it said that their gain is largely independent of frequency, as compared to op amps which have a gain-bandwidth product that is relatively constant, meaning their gain is limited at high frequencies. Is this related to the high bandwidth, or is it a separate advantage of the CFA?

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    \$\begingroup\$ At the most fundamental because in a bipolar circuit, we can switch currents faster than voltages (all other things being equal). \$\endgroup\$ – Peter Smith Apr 8 '20 at 14:14
  • \$\begingroup\$ analog.com/media/en/training-seminars/tutorials/MT-034.pdf \$\endgroup\$ – Andy aka Apr 8 '20 at 14:22
  • \$\begingroup\$ Voltage feedback opamps have high impedance inputs, CFAs have low impedance inputs, at least the one taking feedback, equivalent to the top stage of a cascode, with similar speed advantages \$\endgroup\$ – Neil_UK Apr 8 '20 at 14:44
  • \$\begingroup\$ Where did you read this? Currently there are voltage feedback OP-amps available with GBW in the multiple gigahertz. 20 years ago you might have needed a CFA for the applications where you'd now use those op-amps, though. \$\endgroup\$ – The Photon Apr 8 '20 at 15:03
  • \$\begingroup\$ @ThePhoton Multiple different answers on this website, in fact. \$\endgroup\$ – Hearth Apr 8 '20 at 15:09
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I think, the explanation is as follows:

  • Voltage opamps (if they are unity-gain stable) are internally compensated, which means: Their open-loop gain has a pretty small 3-dB cutoff frequency (20...200 Hz). As a consequence (with 20dB drop per decade) and - let`s say Aoo=100 dB - the transit frequency is app. (2...20) Mhz.

This is necessary because the resistor ratio in the feedback loop determines (a) the desired closed-loop gain as well as (b) the loop gain which is responsible for stability properties. Hence, both gains (closed-loop gain and loop gain) are directly coupled and cannot be set independent to each other.

  • Current-Feedback-Amplifiers (CFA): In contrast to the voltage opamps, the loop gain is determined not by the gain-setting ratio of the feedback resistors, but by the value of the feedback resistor (between output and inv. input) only.

Therefore, the loop gain can be set independent on the closed-loop gain at a value that allows good and stable operation. Hence, the amplifier is not required to be fully compensated. The 3dB cutoff of the open-loop gain and, with it, the transit frequency can be designed much larger than for an opamp.

Here are the closed-loop gain expressions:

  • Opamp: Acl=[Ai]*[1/(1+Ai/Ao(jw))] with A(ideal)=Ai=(1+R2/R1)

  • CFA : Acl=[Ai]*[1/(1+R2/Ztr(jw))] with A(ideal)=Ai=(1+R2/R1) and Ztr(jw)=transfer impedanz

Comment to CFA: Because the stability can be ensured by proper selection of the feedback resistor R2, in the data sheets an "optimum" value for R2 is specified (recommended). The closed-loop gain can be set with R1.

Answer to the last question: The bandwidth of any amplifier with feedback is always set by the loop gain (which, for opamps, is closely related to the closed-loop gain). For CFAs however, the loop gain is constant (set by the fedback resistor R2). Therefore, the closed-loop bandwidth is also constant and there is no dependence on the closed-loop gain which is set by the resistor ratio.

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If a circuit can avoid Miller Effect, so the precious signal charge is not wasted in fighting internal junction (collector-base, drain-gate) charge demands, the bandwidth soars.

For example, examine the schematic of the UA715 operational amplifier, designed in the late 1960s.

The input differential pair is CASCODED, minimizing the Miller Effect, and producing a fast-settling circuit.

Notice the input capacitance is seldom specified on opamps.

And when spec'd, the operating conditions are not made clear.

And when slewing, the Cin can be much smaller than when linearly responding and settling.

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Consider a non-inverting amplifier configuration using a voltage feedback amplifier. The gain can be shown to be

$$\frac{1 + R_1 / R_2}{1 + (1 + R_1 / R_2)/A_\mathrm{OL}} = A_\mathrm{CL} \frac{1}{1 + \frac{A_\mathrm{CL}}{A_\mathrm{OL}}}.$$

Hence the attenuation factor is dependent on the closed loop gain.

For a non-inverting amplifier using a current feedback amplifier with an open loop transconductance \$T(f)\$, the closed loop gain can be shown to be approximately

$$\frac{1 + R_1/R_2}{1 + R_1/T} = A_\mathrm{CL}\frac{1}{1 + R_1/T}.$$

Hence the attenuation factor is dependent on \$R_1\$. However, the closed loop gain can be controlled only with \$R_2\$ while keeping \$R_1\$ constant. Thus the bandwidth can be independent of the closed loop gain.

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  • \$\begingroup\$ user110971...could it be that you have mixed 6dB with 3dB? \$\endgroup\$ – LvW Apr 10 '20 at 9:17
  • \$\begingroup\$ @LvW Remember it is voltage. An attenuation of 0.5 in voltage is 6 dB in power. \$\endgroup\$ – user110971 Apr 10 '20 at 9:21
  • \$\begingroup\$ user110971...so you consider the ratio Acl/Aol as real (and not as imaginary)? To me Aol should be considered with a -20dB/dec slope (and 90deg phase shift) \$\endgroup\$ – LvW Apr 10 '20 at 9:34
  • \$\begingroup\$ @LvW Good point. Yes, I agree. I edited out the 6 dB part. You shouldn’t look at a single point of the frequency response in isolation anyway. You should consider the entire thing. I don’t know why I put it there. \$\endgroup\$ – user110971 Apr 10 '20 at 9:47

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