MIC2155_0911 Datasheet, PDF (32/35 Pages) Micrel Semiconductor – Two-Phase, Single-Output, PWM Synchronous Buck Control IC

Micrel, Inc.
Unlike the voltage output amplifier used for Channel 1
compensation, a transconductance amplifier is used for
the Channel 2 compensation since only a pole/zero
combination is required for compensation. The
transconductance amplifier transfer function is:
Gea(s) = gm × ^{1+ s × Rz1× Cz1}
s × Cz1
Where:
Rz1 and Cz1 are the external components connected to
the COMP2 pin
gm is the transconductance of the internal amplifier.
The pole and zero frequencies are:
f_POLE
=
gm
2 × π × Cz1
f_ZERO
=
1
2 × π × Rz1× Cz1
The gain of the modulator is:
g_MOD
=
1
V_M
Where:
V_M is the peak-to-peak amplitude of the internal
sawtooth.
The gain of the feedback circuit is output current divide
by V_EA
H = R_L2
The filter transfer function is the output current over the
applied voltage:
G_FILTER(s)
=
V_IN − V_OUT
s ×L2
The open loop transfer function is:
G_OL2(s) = G_EA(s) × G_MOD × H × G_FILTER(s) =
gm × (1+ s × Rz1× Cz1)× R_L2 × (V_IN − V_OUT )
(s × Cz1)× Vm × (s ×L2)
MIC2155/2156
The loop is inherently stable because the phase shift is
only 90 degrees. The error amplifier pole and zero is
selected to achieve a desired crossover frequency. In
this example, the desired crossover frequency is 50kHz.
The transfer function of the filter, modulator and
feedback is plotted in Figure 28.
50
V_IN = 12V
V_OUT = 1.8V
L = 1.5µH
0
Gain
-50
Phase
-100
10
100 1k 10k 100k 1M
FREQUENCY
Figure 28. Current Sharing Loop Gain/Phase
The gain boost required at 50kHz is 28dB which is a
gain of 25. The gain for frequencies above the zero is:
G_MID = Rz1× gm
For a typical gm = 1.25mS, solving for Rz1:
Rz1 = ^GMID = ²⁵ = 20kΩ
gm 1.25mS
Set the zero frequency to be 1/5 of the crossover
frequency:
Cz1 =
1
2 × π × Rz1× fz1
Cz1 =
1
= 800pF
2 × π × 20k ×10k
The compensated open loop gain/phase plot is shown in
Figure 29.
November 2009
32
M9999-111209-B