# How to Drive Large Capacitive Loads with an Op-Amp Circuit

February 05, 2019 by Dr. Sergio Franco## Learn more about capacitive load compensation.

Learn more about capacitive load compensation.

There are applications in which an op-amp drives a heavily capacitive load: typical examples are sample-and-hold amplifiers, peak detectors, coaxial drivers, and drivers for certain types of A/D (analog-to-digital) converters.

Capacitive loads have a notorious tendency to destabilize negative-feedback circuits because of the pole formed by the capacitive load with the output resistance of the error amplifier. Being inside the feedback loop, this pole introduces *phase lag*, which erodes the phase margin of the system, leading to *peaking* in its AC response, and *ringing* in its transient response, or even to outright *oscillation*.

### Supporting Information

- Negative Feedback, Part 1: General Structure and Essential Concepts
- Negative Feedback, Part 5: Gain Margin and Phase Margin
- Negative Feedback, Part 9: Breaking the Loop

Before analyzing this source of instability in detail, let us quickly review the basics of circuit stability.

### Understanding Circuit Stability

We'll demonstrate the concept of circuit stability and instability with an example noninverting amplifier, shown in Figure 1(a) below.

**Figure 1. **(a) A noninverting amplifier and (b) graphical visualization of its loop gain |T|.

**Figure 1.**(a) A noninverting amplifier and (b) graphical visualization of its loop gain |T|.

In this example noninverting amplifier, any signal injected at the amplifier’s input port is first magnified by the open-loop gain *a* and is then attenuated by the feedback factor

*Equation 1*

*Equation 1*

Thus, the overall gain T experienced by the signal in going around the loop, aptly called the *loop gain*, is

*T = aβ*

*Equation 2*

*Equation 2*

Rewriting as *T = a*/*(*1/*β)*, taking the logarithms, and multiplying by 20 to convert to decibels, gives

|*T*|_{dB} = |*a*|_{dB} - |1/*β*|_{dB}

indicating that we can visualize the decibel plot of |*T*| as the *difference* between the decibel plots of |*a*| and |*1*/*β*|. This is depicted in Figure 1(b) for the case of a constant gain-bandwidth product op-amp, the most common op-amp type.

The frequency *f*_{x} at which the two curves intersect, aptly called the *crossover* *frequency*, plays an important role because it offers a quantitative indication of circuit stability via the *phase margin* *φ _{m}*, defined as

*φ _{m}* = 180° + ph[

*T*(

*jf*)]

_{x}**Equation 3**

**Equation 3**

where ph[T(*jf _{x}*)] represents the phase of

*T*at

*f*,

_{x}

ph[T(*jf _{x}*)] = ph[

*a*(

*jf*)] + ph[

_{x}*β*(

*jf*)| = ph[

_{x}*a*(

*jf*)] – ph[1/

_{x}*β*(

*jf*)|

_{x}

For the single-pole example shown, ph[1/*β*] = 0; ph[*a*] starts out at 0° at DC, drops to –45° at the pole frequency *f _{b}*, and tends asymptotically to –90° at higher frequencies. By inspection, we have in this case

*φ*≅ 180° – 90° = 90°. The AC and transient responses of this circuit will be similar to those of an ordinary

_{m}*R-C*network.

However, should *φ _{m}* get reduced for some reason, then the AC response will exhibit

*peaking*for

*φ*≤ 65.5°, and the transient response will exhibit

_{m}*ringing*for

*φ*≤ 76.3°. For

_{m}*φ*= 45°, the circuit exhibits a peaking of 2.4 dB, and ringing with an overshoot of 23%.

_{m}### Capacitively Loaded Op Amp Circuit

We now turn our attention to the capacitively loaded circuit of Figure 2.

**Figure 2.** A capacitively loaded *op-amp** circuit*

**Figure 2.**A capacitively loaded

This figure explicitly shows the output resistance *r _{o}* internal to the op-amp. A well-designed circuit will have R

_{1}+ R

_{2}>>

*r*, so we can ignore loading of the output node by the feedback network and say that

_{o}*r*and C

_{o}_{L}establish a pole frequency of

**Equation 4**

**Equation 4**

We can examine the effect of this pole from two different but equivalent viewpoints, for which we will look to Figure 3 for demonstration.

**Figure 3. **Combining the *r*_{o}*-C*_{L} network (a) with the amplifier, and (b) with the feedback network.

**Figure 3.**Combining the

_{o}

_{L}network (a) with the amplifier, and (b) with the feedback network.

- In Figure 3(a) we combine the
*r*-_{o}*C*network with the amplifier itself, so from the viewpoint of the feedback network_{L}*R*, this combination acts as a_{1}-R_{2}*composite amplifier*with open-loop gain

Clearly, the loop gain *T* has now *two* pole frequencies, *f _{b}* and

*f*, each contributing to ph[

_{p}*T*(

*jf*)] a phase shift approaching –90°, so the circuit will exhibit a phase margin approaching zero, with generally intolerable peaking and ringing.

_{x}

- In Figure 3(b), we combine the ro-CL network with the feedback network itself, so from the viewpoint of the amplifier’s dependent source, this combination appears to have the composite feedback factor

Note that as we plot |1/*β _{c}*|, the pole frequency

*f*becomes a de facto zero frequency. Considering that we now have ph[

_{p}*T*(

*jf*)] = ph[

_{x}*a*(

*jf*)] – ph[1/

_{x}*β*(

_{c}*jf*)], with ph[

_{x}*a*(

*jf*)] approaching –90° and ph[1/

_{x}*β*(

_{c}*jf*)] approaching +90°, we still have a system with a phase margin approaching zero.

_{x}### Frequency Compensation

For the circuit to function properly we need to *modify* its loop gain so as to restore an acceptable phase margin—a process called *frequency compensation*. The popular scheme of Figure 4 utilizes a small series resistance *R _{s}* to decouple the amplifier’s output pin from

*C*, and a small feedback capacitance

_{L}*C*to provide a high-frequency bypass from the output pin back to the inverting input pin, adjusted so as to neutralize the phase lag due to

_{f}*f*.

_{p}

**Figure 4.** A popular frequency-compensation scheme for the circuit of Figure 2.

**Figure 4.**A popular frequency-compensation scheme for the circuit of Figure 2.

The objective here is to *bend **downward* the rising portion of the |1/*β _{c}*| curve of Figure 2(b) so as to make it look like the |1/

*β*| curve of Figure 1(b), which we know to provide a phase margin of about 90°. To this end, we need to impose two conditions:

- The high-frequency asymptote must
*equal*the low-frequency asymptote, the latter being 1 +*R*/_{2}*R*. Considering that at sufficiently high frequencies both capacitors act as short circuits, it is apparent that the_{1}*R*-_{1}*R*network is disabled, so the high-frequency asymptote becomes 1 +_{2}*r*/_{o}*R*(recall that we are assuming a well-designed circuit with negligible loading by the_{s}*R*-_{1}*R*network). For the asymptotes to be equal, we thus need 1 +_{2}*r*/_{o}*R*= 1 +_{s}*R*/_{2}*R*, or_{1}

**Equation 5**

**Equation 5**

- With
*R*in place, the in-loop pole frequency changes to_{s}

Also, *C _{f}* forms a high-pass

*C*-

*R*network with the combination

*R*||

_{1}*R*(again we are assuming a well-designed op-amp circuit with

_{2}*R*||

_{1}*R*>>

_{2}*r*||

_{o}*R*), so its pole frequency is

_{s}

We wish the phase lead introduced by *C _{f}* to

*neutralize*the phase lag due to

*C*. This we achieve by imposing [1]

_{L}

Substituting the above expressions for *f _{1}* and

*f*and solving for

_{2}*C*gives

_{f}

**Equation 6**

**Equation 6**

Moreover, the –3 dB frequency of the closed-loop gain is [1]

**Equation 7**

**Equation 7**

### Verification Via PSpice

We wish to verify the above considerations using the example of Figure 5, having *R _{2}* = 2

*R*= 20 kΩ and

_{1}*C*= 100 nF. PSpice’s 741 macromodel has

_{L}*r*= 50 Ω, so we use Equations 5 and 6 (shown above) to calculate

_{o}*R*= 25 Ω and

_{s}*C*= 562.5 pF.

_{f}

**Figure 5.** PSpice circuit to plot the open-loop gain a and 1/β.

**Figure 5.**PSpice circuit to plot the open-loop gain a and 1/β.

Let us first check that compensation does indeed flatten out the 1/β curve. In order to visualize the situation from the viewpoint of the op-amp’s internal source, we need to bring the op-amp’s output resistance *r _{o}* outside, and subject it to a test voltage

*V*, as shown. Running the circuit first with

_{t}*R*= 0 and

_{s}*C*= 0, we get the

_{f}*rising*1/

*β*curve of Figure 6, which crosses the

*a*curve at 100.4 kHz, where we find ph[

*a*] ≅ –93.3° and ph[1/

*β*] ≅ +72.4°. Consequently, ph[

*T*] = ph[

*a*] – ph[1/

*β*] = –93.3° – (+73.4°) = –165.7°, so Equation 3 gives

*φ*= 14.3°, a recipe for intolerable peaking and ringing, as shown in Figure 8 (below).

_{m}

**Figure 6. **Plots obtained using the circuit of Figure 5. Here, a = V(OA)/V(T), and 1/β = V(T)/V(N).

**Figure 6.**Plots obtained using the circuit of Figure 5. Here, a = V(OA)/V(T), and 1/β = V(T)/V(N).

Next, we re-run PSpice with *R _{s}* and

*C*in place, as shown, and we get the flat 1/

_{f}*β*curve, which crosses the

*a*curve at 326.2 kHz, where ph[

*T*] = ph[

*a*] – ph[1/

*β*] = –100.7° – 0° = –100.7°, so now

*φ*= 180° – 100.7° = 79.3°, a much better margin. The compensated responses are shown in Figure 8. Equation 7 predicts a –3 dB frequency of 21.2 kHz, which is close to the measured value of 21.1 kHz. Note that the closed-loop AC response has two pole frequencies, 21.2 kHz, and 326.2 kHz.

_{m}

**Figure 7.** PSpice circuit to plot the closed-loop AC response. The closed-loop transient response is obtained by changing the input source to a 1.0-V step.

**Figure 7.**PSpice circuit to plot the closed-loop AC response. The closed-loop transient response is obtained by changing the input source to a 1.0-V step.

**Figure 8. **Closed-loop (a) AC responses and (b) transient responses obtained using the uncompensated and compensated versions of the simulation circuit.

**Figure 8.**Closed-loop (a) AC responses and (b) transient responses obtained using the uncompensated and compensated versions of the simulation circuit.

### Conclusion

This article has discussed the way in which a large capacitive load can reduce the stability of a negative-feedback amplifier. Compensation can be achieved by the addition of a resistor and capacitor, and the article presents a method of calculating appropriate values for these components.

It should be pointed out that the op-amp’s output impedance, denoted by *r _{o}*, is not always a well-known parameter; moreover, at high frequencies it may become reactive. Consequently, the above equations should be taken as a starting point, after which the designer may find it necessary to tweak the initial values to optimize the circuit for the particular application at hand.

#### References

[1] Design with Operational Amplifiers and Analog Integrated Circuits, 4th Edition by Dr. Sergio Franco

1 CommentVery useful circuit configuration, I use it all the time. I have a different way of looking at stability. As a Filter Wizard, I see filters in many circuits; this is basically a 2nd order filter circuit. A simple analysis will give the transfer function, with has a pole frequency and a pole Q. “Simply” solve for the Q and frequency that you think is appropriate for the application. Particularly appropriate when you are using this as a regulator circuit, it its output impedance versus frequency is important (as in audio).