Analog Application Note
AAN-2
Driving Capacitive Loads
The load impedance that a capacitor presents to an amplifier decreases as the frequency increases. The frequency that
matters here is not the applied signal frequency, but rather the frequency response of the amplifier used. High speed
amplifiers are more sensitive to capacitive loading because the load impedance is lower (harder to drive) than for a
lower speed amplifier. What this means is that layouts and loads you can get away with in a 1MHz bandwidth amplifier
will often cause problems for a higher speed amplifier.
Why is driving a capacitor a problem?
Amplifiers have a non-zero output impedance. The output impedance, combined with the load capacitor and other
components, puts an additional pole in the feedback loop.
AAN-2 Driving Capacitive Loads
Introduction
REV 0.0.1
A pole is introduced corresponding to the capacitor time constant. This time constant is determined by the resistance
seen by the capacitor—the parallel combination of RO, RL and (Rf+Rg). Moving the pole to higher frequency requires
lowering one or more of the resistor values.
The new pole is in addition to the normal loop response of the amplifier. At best, it can seriously degrade phase margin,
at worst it will cause oscillation. The situation can be improved by decreasing the value of RL, but this can cause other
problems with signal fidelity and power.
Exar Corporation
48720 Kato Road, Fremont CA 94538, USA
www.exar.com
Tel. +1 510 668-7000 - Fax. +1 510 668-7001
Analog Application Note
Example: CLC2600 with Capacitive Loading
The plot below shows the effect of capacitive load on the
CLC2600 op amp. These measurements use the same schematic as above (RO is not an external component, it is internal
to the amplifier), Rf=Rg=510 Ohms, RL=100. 4
2
1
CL=0
CL=1pF
CL=2pF
CL=5pF
CL=10pF
CL=20pF
0
-1
-2
-3
-4
-5
RS is selected in the CLC2600 data below by choosing
the smallest value of resistor that keeps peaking below
1dB. The resistor values don’t need to be exactly as
shown—a little higher resistance will result in less peaking and a little less bandwidth.
1
0
Rf = Rg = 510 Ohms
RL = 100 Ohms
-6
-7
0.1
1
10
100
Normalized Gain (dB)
Normalized Gain (dB)
3
How well does it work?
1000
Frequency (MHz)
CL = 500pF
Rs = 9Ω
-2
-3
CL = 100pF
Rs = 20Ω
-4
CL = 50pF
Rs = 30Ω
-5
-6
worse.
VOUT = 0.2Vpp
-7
0.1
What can you do about it?
The best improvement can be had by reducing or removing the
capacitive load, but this usually isn’t a possibility. The easiest
thing to do is to add a series resistor (RS) between the op amp
output and the load capacitor. At first this sounds like it might
make matters worse, but this additional external resistance is
outside the feedback loop instead of inside it:
CL = 10pF
Rs = 40Ω
1
10
100
1000
Frequency (MHz)
Using the correct value of series resistor allows driving a wide
range of load capacitor values. The series resistor value depends mostly on the amplifier bandwidth; check the amplifier
product datasheet for value recommendations.
For Further Assistance:
Exar Corporation Headquarters and Sales Offices
48720 Kato Road
Tel.: +1 (510) 668-7000
Fremont, CA 94538 - USA
Fax: +1 (510) 668-7001
www.exar.com
NOTICE
EXAR Corporation reserves the right to make changes to the products contained in this publication in order to improve design, performance or reliability. EXAR Corporation assumes no responsibility for the use of any
circuits described herein, conveys no license under any patent or other right, and makes no representation that the circuits are free of patent infringement. Charts and schedules contained here in are only for illustration
purposes and may vary depending upon a user’s specific application. While the information in this publication has been carefully checked; no responsibility, however, is assumed for inaccuracies.
EXAR Corporation does not recommend the use of any of its products in life support applications where the failure or malfunction of the product can reasonably be expected to cause failure of the life support system or
to significantly affect its safety or effectiveness. Products are not authorized for use in such applications unless EXAR Corporation receives, in writing, assurances to its satisfaction that: (a) the risk of injury or damage
has been minimized; (b) the user assumes all such risks; (c) potential liability of EXAR Corporation is adequately protected under the circumstances.
Reproduction, in part or whole, without the prior written consent of EXAR Corporation is prohibited.
©2007-2013 Exar Corporation 2/2
Rev 0.0.1
REV 0.0.1
Increasing load capacitance causes increased peaking. Going
much above 20pF will cause the amplifier to oscillate. If the
load resistance is increased or removed, the peaking gets
CL = 1000pF
Rs = 5Ω
-1
AAN-2 Driving Capacitive Loads
5
Rs reduces the phase shift added by Cload, providing isolation
between the amplifier and the load capacitance. Selecting the
right value of Rs will control the peaking caused by the load
capacitor with some reduction in bandwidth.