Ask The Applications Engineer22
by Erik Barnes
CURRENT FEEDBACK AMPLIFIERSI
Q.
I'm not sure I understand how currentfeedback amplifiers work as compared with regular op amps. I've heard that their bandwidth is constant regardless of gain. How does that work? Are they the same as transimpedance amplifiers?
A.
Before looking at any circuits, let's define voltage feedback, current feedback, and transimpedance amplifier. Voltage feedback, as the name implies, refers to a closedloop configuration in which the error signal is in the form of a voltage. Traditional op amps use voltage feedback, that is, their inputs will respond to voltage changes and produce a corresponding output voltage. Current feedback refers to any closedloop configuration in which the error signal used for feedback is in the form of a current. A current feedback op amp responds to an error current at one of its input terminals, rather than an error voltage, and produces a corresponding output voltage. Notice that both openloop architectures achieve the same closedloop result: zero differential input voltage, and zero input current. The ideal voltage feedback amplifier has highimpedance inputs, resulting in zero input current, and uses voltage feedback to maintain zero input voltage. Conversely, the current feedback op amp has a low impedance input, resulting in zero input voltage, and uses current feedback to maintain zero input current.
The transfer function of a transimpedance amplifier is expressed as a voltage output with respect to a current input. As the function implies, the openloop "gain", v_{O}/i_{IN}, is expressed in ohms. Hence a currentfeedback op amp can be referred to as a transimpedance amplifier. It's interesting to note that the closedloop relationship of a voltagefeedback op amp circuit can also be configured as a transimpedance, by driving its dynamically lowimpedance summing node with current (e.g., from a photodiode), and thus generating a voltage output equal to that input current multiplied by the feedback resistance. Even more interesting, since ideally any op amp application can be implemented with either voltage or current feedback, this same IV converter can be implemented with a current feedback op amp. When using the term transimpedance amplifier, understand the difference between the specific currentfeedback op amp architecture, and any closedloop IV converter circuit that acts like transimpedance.
Let's take a look at the simplified model of a voltage feedback amplifier. The noninverting gain configuration amplifies the difference voltage, (V_{IN+} V_{IN}), by the open loop gain A(s) and feeds a portion of the output back to the inverting input through the voltage divider consisting of R_{F} and R_{G}. To derive the closedloop transfer function of this circuit, V_{o}/V_{IN+}, assume that no current flows into the op amp (infinite input impedance); both inputs will be at about the same potential (negative feedback and high openloop gain)).
With
and
substitute and simplify to get:
The closedloop bandwidth is the frequency at which the loop gain, LG, magnitude drops to unity (0 dB). The term,
1 + R_{F}/R_{G}, is called the noise gain of the circuit; for the noninverting case, it is also the signal gain. Graphically, the closedloop bandwidth is found at the intersection of the openloop gain, A(s), and the noise gain, NG, in the Bodé plot. High noise gains will reduce the loop gain, and thereby the closedloop bandwidth. If A(s) rolls off at 20 dB/decade, the gainbandwidth product of the amplifier will be constant. Thus, an increase in closedloop gain of 20 dB will reduce the closedloop bandwidth by one decade.
Consider now a simplified model for a currentfeedback amplifier. The noninverting input is the highimpedance input of a unity gain buffer, and the inverting input is its lowimpedance output terminal. The buffer allows an error current to flow in or out of the inverting input, and the unity gain forces the inverting input to track the noninverting input. The error current is mirrored to a high impedance node, where it is converted to a voltage and buffered at the output. The highimpedance node is a frequencydependent impedance, _{Z(s)}, analogous to the openloop gain of a voltage feedback amplifier; it has a high dc value and rolls off at 20 dB/decade.
The closedloop transfer function is found by summing the currents at the V_{IN} node, while the buffer maintains
V_{IN+} = V_{IN}. If we assume, for the moment, that the buffer has zero output resistance, then R_{o} = 0
Substituting, and solving for V_{o}/V_{IN+}
The closedloop transfer function for the current feedback amplifier is the same as for the voltage feedback amplifier, but the loop gain (1/LG) expression now depends only on R_{F}, the feedback transresistanceand not (1 + R_{F}/R_{G}). Thus, the closedloop bandwidth of a current feedback amplifier will vary with the value of R_{F}, but not with the noise gain, 1 + R_{F}/R_{G}. The intersection of R_{F} and Z(s) determines the loop gain, and thus the closedloop bandwidth of the circuit (see Bodé plot). Clearly the gainbandwidth product is not constantan advantage of current feedback.
In practice, the input buffer's nonideal output resistance will be typically about 20 to 40 Ohms, which will modify the feedback transresistance. The two input voltages will not be exactly equal. Making the substitution into the previous equations with
V_{IN} = V_{IN+} I_{err}R_{o}, and solving for V_{o}/V_{IN+} yields:
The additional term in the feedback transresistance means that the loop gain will actually depend somewhat on the closedloop gain of the circuit. At low gains, R_{F} dominates, but at higher gains, the second term will increase and reduce the loop gain, thus reducing the closedloop bandwidth.
It should be clear that shorting the output back to the inverting input with R_{G} open (as in a voltage follower) will force the loop gain to get very large. With a voltage feedback amplifier, maximum feedback occurs when feeding back the entire output voltage, but the current feedback's limit is a shortcircuit current. The lower the resistance, the higher the current will be. Graphically, R_{F} = 0 will give a higherfrequency intersection of Z(s) and the feedback transresistancein the region of higherorder poles. As with a voltage feedback amplifier, higherorder poles of Z(s) will cause greater phase shift at higher frequencies, resulting in instability with phase shifts > 180 degrees. Because the optimum value of R_{F} will vary with closedloop gain, the Bode plot is useful in determining the bandwidth and phase margin for various gains. A higher closedloop bandwidth can be obtained at the expense of a lower phase margin, resulting in peaking in the frequency domain, and overshoot and ringing in the time domain. Currentfeedback device data sheets will list specific optimum values of R_{F} for various gain settings.
Current feedback amplifiers have excellent slewrate capabilities. While it is possible to design a voltagefeedback amplifier with high slew rate, the currentfeedback architecture is inherently faster. A traditional voltagefeedback amplifier, lightly loaded, has a slew rate limited by the current available to charge and discharge the internal compensation capacitance. When the input is subjected to a large transient, the input stage will saturate and only its tail current is available to charge or discharge the compensation node. With a currentfeedback amplifier, the lowimpedance input allows higher transient currents to flow into the amplifier as needed. The internal current mirrors convey this input current to the compensation node, allowing fast charging and dischargingtheoretically, in proportion to input step size. A faster slew rate will result in a quicker rise time, lower slewinduced distortion and nonlinearity, and a wider largesignal frequency response. The actual slew rate will be limited by saturation of the current mirrors, which can occur at 10 to 15 mA, and the slewrate limit of the input and output buffers.
Q.
What about dc accuracy?
A.
The dc gain accuracy of a current feedback amplifier can be calculated from its transfer function, just as with a voltage feedback amplifier; it is essentially the ratio of the internal transresistance to the feedback transresistance. Using a typical transresistance of 1 MOhms, a feedback resistor of 1 k Ohms, and an R_{o} of 40 ohms, the gain error at unity gain is about 0.1%. At higher gains, it degrades significantly. Currentfeedback amplifiers are rarely used for high gains, particularly when absolute gain accuracy is required.
For many applications, though, the settling characteristics are of more importance than gain accuracy. Although current feedback amplifiers have very fast rise times, many data sheets will only show settling times to 0.1%, because of thermal settling tails a major contributor to lack of settling precision. Consider the complementary input buffer above, in which the V_{IN} terminal is offset from the V_{IN+} terminal by the difference in V_{BE} between Q1 and Q3. When the input is at zero, the two V_{BE}s should be matched, and the offset will be small from V_{IN+} to V_{IN}. A positive step input applied to V_{IN+} will cause a reduction in the V_{CE} of Q3, decreasing its power dissipation, thus increasing its V_{BE}. Diodeconnected Q1 does not exhibit a V_{CE} change, so its V_{BE} will not change. Now a different offset exists between the two inputs, reducing the accuracy. The same effect can occur in the current mirror, where a step change at the highimpedance node changes the V_{CE}, and thus the V_{BE}, of Q6, but not of Q5. The change in VBE causes a current error referred back to VIN, whichmultiplied by RFwill result in an output offset error. Power dissipation of each transistor occurs in an area that is too small to achieve thermal coupling between devices. Thermal errors in the input stage can be reduced in applications that use the amplifier in the inverting configuration, eliminating the commonmode input voltage.
Q.
In what conditions are thermal tails a problem?
A.
It depends on the frequencies and waveforms involved. Thermal tails do not occur instantaneously; the thermal coefficient of the transistors (which is process dependent) will determine the time it takes for the temperature change to occur and alter parametersand then recover. Amplifiers fabricated on the Analog Devices highspeed complementary bipolar (CB) process, for example, don't exhibit significant thermal tails for input frequencies above a few kHz, because the input signal is changing too fast. Communications systems are generally more concerned with spectral performance, so additional gain errors that might be introduced by thermal tails are not important. Step waveforms, such as those found in imaging applications, can be adversely affected by thermal tails when dc levels change. For these applications, currentfeedback amplifiers may not offer adequate settling accuracy.
Part II will consider common application circuits using currentfeedback amplifiers and view their operation in more detail.
