A current-feedback amplifier is a form of op-amp which most commonly serves as the error amplifier of a negative-feedback circuit where the input error is a current.

In this article, we'll talk about the step responses, slew rates, input buffers of CFAs.

#### CFA Series

This article is part of a series. Though this article is designed to be useful on its own, you may consider reading the preceding part before continuing.

The rest of the series can be found here:

- Part I: Introduction to the CFA: Current-Feedback Amplifiers vs. Voltage-Feedback Amplifiers
- Part II: Characteristics of Current-Feedback Op-Amps: Benefits of CFA Design vs. VFAs
- Part III: Applications and Limitations of the Current-Feedback Amplifier: Dual CFAs and Composite Amplifiers

In the previous article, we highlighted some unique features of the current-feedback amplifier (CFA) using the idealized model (presented as Figure *1b* in the previous article):

**Figure 1.** A noninverting amplifier using the idealized model of the CFA

**Figure 1.**A noninverting amplifier using the idealized model of the CFA

We now wish to investigate how a practical CFA lives up to the above ideal expectations.

It turns out that a primary source of deviation from the ideal is the fact that the input buffer exhibits a *non-zero output impedance *r_{n}, so we shall now work with the more realistic model of Figure 2*a.*

*(a) (b)*

**Figure 2**. Investigating the effect of the input buffer’s output resistance r_{n} upon the feedback factor ß.

**Figure 2**. Investigating the effect of the input buffer’s output resistance r

_{n}upon the feedback factor ß.

We note that the following equations still hold from when we first introduced them in the previous article (where they were presented as Equations (8) and (9). We'll repeat them here for convenience:

*Equation (1)*

*Equation (2)*

However, we now have* **V _{n}*

*≠ V*

_{p }because of the voltage drop across

*r*. In fact, Ohm’s law gives

_{n}*V*, or, if we recall that

_{p}– V_{n}= r_{n}I_{n}*V*

_{p}= V_{i},

*Equation (3)*

Substituting Equation (3) into Equation (2), we eliminate *V _{n}*. Collecting for

*I*and then substituting into Equation (1), we eliminate

_{n}*I*to obtain an expression involving only

_{n}*V*and

_{o}*V*Collecting once again and solving for the ratio

_{i}.*V*gives the

_{o}/V_{i}*closed-loop gain A*, which we express as

*Equation (4)*

where

*Equation (5)*

and

*Equation (6)*

The presence of *r _{n}* has no effect upon

*A*. However, it

_{ideal}*raises*the

*1/ß*curve, and in so doing it

*reduces*both the loop gain

*T*and the crossover frequency

*f*, compared to the idealized case

_{x}*r*

_{n}*= 0*, for which

*T = z/R*and

_{F}*f*(Henceforth, we visualize

_{x}= f_{t}.*f*

_{t}_{ }as the crossover frequency extrapolated to the limit

*r*.)

_{n}→ 0

#### Example: The Effect of r_{n} in a DC Circuit

To get an idea, let us rework the DC example from the first article (presented there as Figure *2b*)—but with *r*_{n }= 40 Ω in place.

*(a) (b)*

**Figure 3**. Visualizing DC voltages and currents for the cases (a) r_{n} = 40 Ω and (b) r_{n} = 0.

**Figure 3**. Visualizing DC voltages and currents for the cases (a) r

_{n}= 40 Ω and (b) r

_{n}= 0.

Using Equation (6), we find* T* = 303.03 (a drop from T = 400 from the circuit in the previous article). Then we find *A* via Equation (4), *I*_{n }via Equation (1), *V _{p} – V_{n}* via Ohm’s law, and

*V*

*via Equation (3).*

_{n}Moreover, we calculate the currents through *R _{G}* and

*R*via Ohm’s law. The results are visualized in Figure 3a (for easy comparison, the example from the previous article is repeated in Figure 3

_{F}*b*).

It is apparent that the most noticeable effects of *r _{n}* are to cause

*V*

_{n}*≠ V*, and also to contribute a slight reduction in

_{p}*V*due to the decrease of

_{o}*T*from 400 to 333.

### The Closed-Loop AC Response

We now wish to take a closer look at the crossover frequency *f _{x}* of Figure 2

*b*.

As discussed in our comparison of CFAs and VFAs, this frequency approximates the *–3-dB frequency* of the closed-loop response *A(**jf**)*.

Since *z(**jf**)* is of the dominant-pole type, the GBP = |z| × f is constant on the sloped portion of the curve, so we can write

*Equation (7)*

Solving for *f _{x}* gives, after simplification,

*Equation (8)*

This expression is reminiscent of the VFA equation we presented in the previous article as Equation (16), except for the presence of the term *r _{n}/R_{F}*, which is a well-designed CFA is such that

*r*<< 1. Consequently, there is a bit of gain-bandwidth tradeoff also in CFAs, but this is much lighter than in VFAs. For instance, in the circuit with

_{n}/R_{F}*r*

_{n }= 40 Ω and

*R*= 1.25 kΩ of Figure 3

_{F}*a*, adjusting

*R*for (1 +

_{G}*R*) = 1, 10, and 100 V/V, we get

_{F}/R_{G}*f*=

_{x}*f*/1.032,

_{t}*f*/1.32, and

_{t}*f*/4.2, as shown in Figure 4.

_{t}

**Figure 4**. Illustrating the effect of r_{n} = 40 Ω upon the closed loop bandwidth for A_{ideal} = 0, 20, and 40 dB.

**Figure 4**. Illustrating the effect of r

_{n}= 40 Ω upon the closed loop bandwidth for A

_{ideal}= 0, 20, and 40 dB.

These compare quite favorably with the VFA values of *f _{x}* =

*f*/10, and

_{t}, f_{t}*f*/100.

_{t}

### Step Responses

We conclude our VFA–CFA comparison by taking a look at the step responses. Turning first to the VFA, we use the simplified model of Figure 3*a*, which is typical of many VFAs. We shall focus on the input stage (*Q _{1}* through

*Q*), so we lump together all subsequent stages into a single inverting stage assumed to have gain –

_{4}*a*. (The purpose of C

_{2}_{c}is to provide an open-loop response of the dominant-pole type.)

The bias current *I _{E}* splits between

*Q*and

_{1}*Q*to form

_{2}*i*and

_{1}*i*(for simplicity, we are assuming negligible BJT base currents). Now,

_{2}*i*flows into the diode-connected transistor

_{1}*Q*, and since

_{3}*Q*, its matched twin, is subjected to the

_{4}*same base-emitter voltage drop*as

*Q*, then

_{3}*Q*will just mirror

_{4}*Q*current

_{3}’s*i*, this being the reason why the

_{1}*Q*pair is said to form a

_{3}-Q_{4}*current mirror*. By KCL, the capacitor current is

*i*

_{1}– i_{2}.

*(a) (b)*

**Figure 5**. (a) Simplified VFA model connected as a voltage follower, and step responses: small-signal exponential transient (top), and slew-rate-limited ramp (bottom).

**Figure 5**. (a) Simplified VFA model connected as a voltage follower, and step responses: small-signal exponential transient (top), and slew-rate-limited ramp (bottom).

When *v _{O}*

*= v*, the circuit is said to be

_{I}*balanced*because

*i*, so the capacitor current is

_{1}= i_{2}*i*= 0. If we now apply an input step of suitably small magnitude

_{1}– i_{2}*V*, we are making

_{m}*Q*less conductive and Q1 more conductive, thus unbalancing the circuit to give

_{2}*i*> 0. This will cause

_{1}– i_{2}*C*to charge up and produce a transient at the output, which in turn will gradually reduce the initial imbalance via negative feedback. Just as in the case of a plain

_{c}*R-C*network, we get an

*exponential transient*, and its time constant is τ = 1/(2πf

_{t}), as shown in Figure 5

*b*, top. (For the 741 op-amp, which has

*f*= 1 MHz, we have τ = 159 ns.)

_{t}Now, if we increase *V _{m}*, we’ll reach a point at which

*all*of

*I*will be diverted to

_{E}*Q*, causing

_{1}*C*to charge at a

_{c}*constant rate*, as shown in Figure 5

*b*, bottom. This is the

*maximum rate*at which

*v*can swing. Aptly called the

_{O}*slew rate*(SR), it is easily found via the capacitance law to be

*Equation (9)*

For the 741 op-amp, which uses *I _{E}* = 20 µA and

*C*= 30 pF, we get SR = 0.67 V/µs. The maximum amplitude V

_{c}_{m}before the onset of slew-rate limiting is obtained by equating the initial slope of the exponential transient to SR, or

*V*

_{m}*/τ*= SR, which gives

*V*. For the 741 op-amp, we get

_{m}= I_{E}/(C_{c}×2πf_{t})*V*= 116 mV. Under slew-rate limiting,

_{m}*v*ramps up at a constant rate until it comes within 116 mV of the final value, after which it completes the rest of the transient in an exponential fashion.

_{O}

#### Step Responses of a CFA

Let us now turn to the step response of the CFA, but using the circuit schematic of Figure 6, which will also help us better understand the CFA’s overall internal workings.

We make the following observations:

- The function of
*Q*and_{1}*Q*is to form a low-output-impedance push-pull stage, while_{2}*Q*and_{3}*Q*provide a Darlington-type function to increase the input impedance. Also, the base-emitter voltage drops of_{4}*Q*and_{3}*Q*are meant to compensate in a complementary fashion for those of_{4}*Q*and_{1}*Q*so as to eliminate any DC offset between nodes_{2}*V*_{n}*V*. This is the_{p}*input buffer*, represented at the bottom of Figure 6 as a unity-gain buffer with output resistance*r*._{n} - In negative-feedback operation, the current
*I*drawn by the external feedback network splits between_{n}*Q*and_{1}*Q*such that_{2}*I*, by KCL (again, we are assuming negligible BJT base currents)._{n}= I_{1}– I_{2} - The trios
*Q*and_{5}-Q_{6}-Q_{7}*Q*form high-quality current mirrors of the so-called Wilson type, for the purpose of mirroring_{8}-Q_{9}-Q_{10}*I*and_{1}*I*, respectively, in and out of an important node called the_{2}*gain node*(GN). Denoting GN’s equivalent impedance towards the ground as z, it is apparent that the net current into z is a replica of the inverting-input current*I*itself, so we model the entire action of the mirrors by means of a_{n}*single*unity-gain current-controlled current source (CCCS), as shown at the bottom. - BJTs
*Q*through_{11}*Q*form the output buffer, identical in topology to the input buffer_{14}*Q*through_{1}*Q*(for simplicity, its output resistance_{4}*r*is usually ignored because it plays a much lesser role than_{o}*r*). Thanks to this buffering action, we write_{n}*V*and we model the CFA with a CCVS as in Figure_{o}= zI_{n}*1*. - The impedance is modeled as the parallel combination of equivalent resistance
*R*and capacitance_{gn}*C*, or_{gn}*z = R*. Expanding,_{gn}||[1/(j2πfC_{gn})]

*Equation (10)*

With reference to Figure 2b, we can easily find an expression for f_{t} by imposing *z _{0 }× *

*f*

_{p}*= R*, obtaining

_{F}× f_{t}

*Equation (11)*

In a well-designed CFA, *R _{gn}* is on the order of 10

^{6}Ω,

*C*is in the picofarad range, and

_{gn}*R*is on the order of 10

_{F}^{3}Ω. The time constant

*τ = (*

*R*

_{F}C_{gn}*)*is then in the nanosecond range.

**Figure 6**. Circuit schematic of a CFA (top) and its constituent blocks (bottom).

**Figure 6**. Circuit schematic of a CFA (top) and its constituent blocks (bottom).

We are now in the position to state that there is *no slew-rate limiting* in CFAs because the input buffer will deliver to GN *whatever current it takes* to ensure an exponential transient at the output, regardless of the input step size *V _{m}*. For instance, the circuit of Figure

*3*

*a*, in response to an input step with

*V*

_{m}*= 1 V, will provide an initial current of*

*I*= 6.06 mA.

_{n}= V_{p}/[r_{n}+ (R_{G}||R_{F})]Were we to remove *R _{G}* in order to operate the CFA as a unity-gain voltage follower with a 10 V input step, the initial current would be

*I*= 7.75 mA, much larger than the maximum current

_{n}= V_{p}/(r_{n}+ R_{F})*I*= 20 µA available in the 741 op-amp!

_{E}CFAs are well suited to applications that might suffer from slew-rate-induced distortion and nonlinearity.

Continue reading in Part II: Benefits of CFA Design vs. VFAs.

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