Q. Why would I want to drive a capacitive load?
A. It's usually not a matter of choice. In most cases, the load capacitance is not from a capacitor you've added intentionally; most often it's an unwanted parasitic, such as the capacitance of a length of coaxial cable. However, situations do arise where it's desirable to decouple a dc voltage at the output of an op ampfor example,when an op amp is used to invert a reference voltage and drive a dynamic load. In this case, you might want to place bypass capacitors directly on the output of an op amp. Either way, a capacitive load affects the op amp's performance.
Q. How does capacitive loading affect op amp performance?
A. To put it simply, it can turn your amplifier into an oscillator. Here's how:
Op amps have an inherent output resistance, R_{o}, which, in conjunction with a capacitive load, forms an additional pole in the amplifier's transfer function. As the Bode plot shows, at each pole the amplitude slope becomes more negative by 20 dB/ decade. Notice how each pole adds as much as 90° of phase shift. We can view instability from either of two perspectives. Looking at amplitude response on the log plot,circuit instability occurs when the sum of openloop gain and feedback attenuation is greater than unity. Similarly, looking at phase response, an op amp will tend to oscillate at a frequency where loop phase shift exceeds 180°, if this frequency is below the closedloop bandwidth. The closedloop bandwidth of a voltagefeedback op amp circuit is equal to the op amp's bandwidth product (GBP, or unitygain frequency), divided by the circuit's closed loop gain (A_{CL}).
Phase margin of an op amp circuit can be thought of as the amount of additional phase shift at the closed loop bandwidth required to make the circuit unstable (i.e., phase shift + phase margin = 180°). As phase margin approaches zero, the loop phase shift approaches 180° and the op amp circuit approaches instability. Typically, values of phase margin much less than 45° can cause problems such as "peaking" in frequency response, and overshoot or "ringing" in step response. In order to maintain conservative phase margin, the pole generated by capacitive loading should be at least a decade above the circuit's closed loop bandwidth.When it is not, consider the possibility of instability.
Q. So how do I deal with a capacitive load?
A. First of all you should determine whether the op amp can safely drive the load on its own. Many op amp data sheets specify a "capacitive load drive capability". Others provide typical data on "smallsignal overshoot vs. capacitive load". In looking at these figures, you'll see that the overshoot increases exponentially with added load capacitance. As it approaches 100%, the op amp approaches instability. If possible, keep it well away from this limit. Also notice that this graph is for a specified gain. For a voltage feedback op amp, capacitive load drive capability increases proportionally with gain. So aVF op amp that can safely drive a 100pF capacitance at unity gain should be able to drive a 1000pF capacitance at a gain of 10.
A few op amp data sheets specify the open loop output resistance (R_{o}), from which you can calculate the frequency of gainthe added pole as described above.The circuit will be stable if the frequency of the added pole (f_{P}) is more than a decade above the circuit's bandwidth.
If the op amp's data sheet doesn't specify capacitive load drive or open loop output resistance, and has no graph of overshoot versus capacitive load, then to assure stability you must assume that any load capacitance will require some sort of compensation technique.There are many approaches to stabilizing standard op amp circuits to drive capacitive loads. Here are a few:
Noisegain manipulation: A powerful way to maintain stability in lowfrequency applicationsoften overlooked by designersinvolves increasing the circuit's closedloop gain (a/k/a "noise gain") without changing signal gain,thus reducing the frequency at which the product of openloop gain and feedback attenuation goes to unity. Some circuits to achieve this, by connecting RD between the op amp inputs, are shown below. The "noise gain" of these circuits can be arrived at by the given equation.
Since stability is governed by noise gain rather than by signal gain, the above circuits allow increased stability without affecting signal gain. Simply keep the "noise bandwidth" (GBP/A_{NOISE}) at least a decade below the load generated pole to guarantee stability.
One disadvantage of this method of stabilization is the additional output noise and offset voltage caused by increased amplification of inputreferred voltage noise and input offset voltage. The added dc offset can be eliminated by including C_{D} in series with R_{D}, but the added noise is inherent with this technique. The effective noise gain of these circuits with and without C_{D} are shown in the figure.
C_{D}, when used, should be as large as feasible; its minimum value should be 10 A_{NOISE}/(2 pR_{D}GBP) to keep the "noise pole" at least a decade below the "noise bandwidth".
Outofloop compensation: Another way to stabilize an op amp for capacitive load drive is by adding a resistor, RX, between the op amp's output terminal and the load capacitance, as shown below.Though apparently outside the feedback loop, it acts with the load capacitor to introduce a zero into the transfer function of the feedback network, thereby reducing the loop phase shift at high frequencies.
To ensure stability, the value of R_{X} should be such that the added zero (f_{Z}) is at least a decade below the closed loop bandwidth of the op amp circuit.With the addition of R_{X},circuit performance will not suffer the increased output noise of the first method, but the output impedance as seen by the load will increase. This can decrease signal gain, due to the resistor divider formed by R_{X} and R_{L}. If R_{L} is known and reasonably constant, the results of gain loss can be offset by increasing the gain of the op amp circuit.
This method is very effective in driving transmission lines.The values of R_{L} and R_{X} must equal the characteristic impedance of the cable (often 50ohms or 75ohms) in order to avoid standing waves. So R_{X} is predetermined, and all that remains is to double the gain of the amplifier in order to offset the signal loss from the resistor divider. Problem solved.
Inloop compensation: If R_{L} is either unknown or dynamic, the effective output resistance of the gain stage must be kept low. In this circumstance, it may be useful to connect R_{X} inside the overall feedback loop, as shown below. With this configuration, dc and lowfrequency feedback comes from the load itself, allowing the signal gain from input to load to remain unaffected by the voltage divider, R_{X} and R_{L}.
The added capacitor, C_{F}, in this circuit allows cancellation of the pole and zero contributed by C_{L}.To put it simply, the zero from C_{F} is coincident with the pole from C_{L}, and the pole from C_{F} with the zero from C_{L}.Therefore, the overall transfer function and phase response are exactly as if there were no capacitance at all. In order to assure cancellation of both pole/ zero combinations, the above equations must be solved accurately. Also note the conditions; they are easily met if the load resistance is relatively large.
Calculation is difficult when R_{O} is unknown. In this case, the design procedure turns into a guessing gameand a prototyping nightmare. A word of caution about SPICE: SPICE models of op amps don't accurately model openloop output resistance (RO); so they cannot fully replace empirical design of the compensation network.
It is also important to note that C_{L} must be of a known (and constant) value in order for this technique to be applicable. In many applications, the amplifier is driving a load "outside the box," and C_{L} can vary significantly from one load to the next. It is best to use the above circuit only when C_{L} is part of a closed system.
One such application involves the buffering or inverting of a reference voltage, driving a large decoupling capacitor. Here, C_{L} is a fixed value, allowing accurate cancellation of pole/zero combinations. The low dc output impedance and low noise of this method (compared to the previous two) can be very beneficial. Furthermore, the large amount of capacitance likely to decouple a reference voltage (often many microfarads) is impractical to compensate by any other method.
All three of the above compensation techniques have advantages and disadvantages. You should know enough by now to decide which is best for your application. All three are intended to be applied to "standard", unity gain stable, voltage feedback op amps. Read on to find out about some techniques using special purpose amplifiers.
Q. My op amp has a "compensation" pin. Can I overcompensate the op amp so that it will remain stable when driving a capacitive load?
A. Yes. This is the easiest way of all to compensate for load capacitance. Most op amps today are internally compensated for unitygain stability and therefore do not offer the option to "overcompensate". But many devices still exist with inherent stability only at very high noise gains. These op amps have a pin to which an external capacitor can be connected in order to reduce the frequency of the dominant pole. To operate stably at lower gains, increased capacitance must be tied to this pin to reduce the gainbandwidth product. When a capacitive load must be driven, a further increase (overcompensation) can increase stability–but at the expense of bandwidth.
Q. So far you've only discussed voltage feedback op amps exclusively, right? Do current feedback (CF) op amps behave similarly with capacitive loading? Can I use any of the compensation techniques discussed here?
A. Some characteristics of current feedback architectures require special attention when driving capacitive loads, but the overall effect on the circuit is the same. The added pole, in conjunction with opamp output resistance, increases phase shift and reduces phase margin, potentially causing peaking, ringing, or even oscillation. However, since a CF op amp can't be said to have a "gainbandwidth product" (bandwidth is much less dependent on gain), stability can't be substantially increased simply by increasing the noise gain. This makes the first method impractical. Also, a capacitor (C_{F}) should NEVER be put in the feedback loop of a CF op amp, nullifying the third method. The most direct way to compensate a current feedback op amp to drive a capacitive load is the addition of an "out of loop" series resistor at the amplifier output as in method 2.
Part Number  Ch  BW MHz 
SR V/ms 
v_{n} nV/ Hz 
i_{n} fA/ Hz 
V_{OS} mV 
I_{b} nA 
Supply Voltage Range [V] 
I_{Q} mA 
R_{O} ohms 
Cap Load Drive [pF] 
Notes 
AD817  1  50  350 
15  1500  0.5  3000  536  7  8  unlim  
AD826  2  50 
350 
15  1500  0.5  3000 
536  6.8  8  unlim  
AD827  2  50 
300  15  1500  0.5  3000 
936  5.25  15  unlim  
AD847  1  50 
300 
15  1500  0.5  3000 
936  4.8  15  unlim  
AD848  1  35  200  5  1500  0.5  3000 
936  5.1  15  unlim  G_{MIN}=5 
AD849  1  29  200  3  1500  0.3  3000 
936  5.1  15  unlim  G_{MIN}=25 
AD704  4  0.8  0.15  15  50  0.03  0.1  436  0.375  10000  
AD705  1  0.8  0.15  15  50 
0.03  0.06  436  0.38  10000  
AD706  2  0.8  0.15  15  50 
0.03  0.05  436  0.375  10000  
OP97  1  0.9  0.2  14  20  0.03  0.03  440 
0.38  10000  
OP279  2  5  3  22  1000  4  300  4.512 
2  22  10000  
OP400  4  0.5  0.15  11  600  0.08  0.75  640 
0.6  10000  
AD549  1  1  3  35  0.22  0.5  0.00015  1036  0.6  4000  
OP200  2  0.5  0.15  11  400  0.08  0.1  640 
0.57  2000  
OP467  4  28  170  6  8000  0.2  150  936  2  1600  
AD744  1  13  75  16  10  0.3  0.03  936  3.5  1000  comp.term  
AD8013  3  140  1000  3.5  12000  2  3000  4.513  3.4  1000  current fb  
AD8532  2  3  5  30  50  25  0.005  3.6  1.4  1000  
AD8534  4  3  5  30  50  25  0.005  3.6  1.4  1000  
OP27  1  8  2.8  3.2 
1700  0.03  15  844  6.7 
70  1000 

OP37  1  12  17  3.2 
1700  0.03  15  844  6.7  70  1000 
G_{MIN}=5 
OP270  2  5  2.4  3.2 
1100  0.05  15  936  2  1000 

OP470  4  6  2  3.2  1700  0.4  25  936  2.25  1000 

OP275  2  9  22  6  1500  1  100  944  2  1000 

OP184  1  4.25  4  3.9 
400  0.18  80  436  2  1000 

OP284  2  4.25  4  3.9 
400 
0.18  80  436  2  1000 

OP484  4  4.25  4  3.9  400 
0.25  80  436  2  1000 

OP193  1  0.04  15  65  50  0.15  20  336  0.03  1000 

OP293  2  0.04  15  65  50  0.25  20  336  0.03  1000 
Q. This has been informative, but I'd rather not deal with any of these equations. Besides, my board is already laid out, and I don't want to scrap this production run. Are there any op amps that are inherently stable when driving capacitive loads?
A. Yes. Analog Devices makes a handful of op amps that drive "unlimited" load capacitance while retaining excellent phase margin. They are listed in the table, along with some other op amps that can drive capacitive loads up to specified values. About the "unlimited" cap load drive devices: don't expect to get the same slew rate when driving 10 µF as you do when driving purely resistive loads. Read the data sheets for details.
References
Practical Analog Design Techniques, Analog Devices 1995 seminar notes. Cap load drive information can be found in section 2, "Highspeed op amps" (Walt Jung and Walt Kester).
Application Note AN257: "Careful design tames highspeed op amps," by Joe Buxton, in ADI's Applications Reference Manual (1993). A detailed examination of the "inloop compensation" method. Free.
"Currentfeedback amplifiers, Part 1 and Part 2", by Erik Barnes, Analog Dialogue 303 and 304 (1996), now consolidated in Ask The Applications Engineer (1997). Available on our Web site.