Datasheet

R5 =
R1 + R2
R2
©
§
¹
·
B
gm
x
FB
COMP
SS
6
7
5
C8
R1
R2
C10
C9
R5
To Vout
2
fsw
fn =
1
10 x S x Ro x Co
fp =
+
0.5
2 x S x L x fsw x Co
1
2S x Re x Co
fz =
B
fzc fpc1
FREQUENCY
GAIN (dB)
fpc
(0Hz)
-20dB/dec
-
20
dB
/
dec
0dB/dec
LM26001, LM26001Q
SNVS430G –MAY 2006–REVISED MAY 2006 www.ti.com
The control-output transfer function consists of one pole (fp), one zero (fz), and a double pole at fn (half the
switching frequency).
Referring to Figure 22, the following should be done to create a -20dB /decade roll-off of the loop gain:
1. Place a pole at 0Hz (fpc)
2. Place a zero at fp (fzc)
3. Place a second pole at fz (fpc1)
The resulting feedback (compensation) bode plot is shown below in Figure 23. Adding the control-output
response to the feedback response will then result in a nearly continuous -20db/decade slope.
Figure 23. Feedback Transfer Function
The control-output corner frequencies can be determined approximately by the following equations:
(17)
(18)
(19)
Where Co is the output capacitance, Ro is the load resistance, Re is the output capacitor ESR, and fsw is the
switching frequency. The effects of slope compensation and current sense gain are included in this equation.
However, the equation is an approximation intended to simplify loop compensation calculations. To derive the
exact transfer function, use 0.2V/V sense amp gain and 36mVp-p slope compensation.
Since fp is determined by the output network, it shifts with loading. Determine the range of frequencies
(fpmin/max) across the expected load range. Then determine the compensation values as described below and
shown in Figure 24.
Figure 24. Compensation Network
1. The compensation network automatically introduces a low frequency pole (fpc), which is close to 0Hz.
2. Once the fp range is determined, R5 should be calculated using:
(20)
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