Electronic – Why do current-feedback amplifiers have a higher bandwidth

amplifierhigh frequency

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?

Best Answer

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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