Electronic Components Datasheet Search
  English  ▼

X  

MIC2131 Datasheet(PDF) 17 Page - Micrel Semiconductor

Part # MIC2131
Description  High Voltage Synchronous Buck Control IC with Low EMI Option
PDF  20 Pages
Scroll/Zoom Zoom In 100%  Zoom Out
Manufacturer  MICREL [Micrel Semiconductor]
Direct Link  http://www.micrel.com
Logo MICREL - Micrel Semiconductor

MIC2131 Datasheet(HTML) 17 Page - Micrel Semiconductor

Back Button MIC2131 Datasheet HTML 12Page - Micrel Semiconductor MIC2131 Datasheet HTML 13Page - Micrel Semiconductor MIC2131 Datasheet HTML 14Page - Micrel Semiconductor MIC2131 Datasheet HTML 15Page - Micrel Semiconductor MIC2131 Datasheet HTML 16Page - Micrel Semiconductor MIC2131 Datasheet HTML 17Page - Micrel Semiconductor MIC2131 Datasheet HTML 18Page - Micrel Semiconductor MIC2131 Datasheet HTML 19Page - Micrel Semiconductor MIC2131 Datasheet HTML 20Page - Micrel Semiconductor  
Zoom Inzoom in Zoom Outzoom out
 17 / 20 page
background image
Micrel, Inc.
MIC2130/1
April 2008
17
M9999-042108-C
Example:
VIN = 24V; VOUT = 3.3V; IOUT = 10A; L = 7.3µH; COUT =
660µF; Resr = 40mΩ; Fsw = 150KHz
The gain and phase of the modulator and filter is:
GMod(s) x Gflt(s)
This is the gain
comp
OUT
V
V
(s) in Figure 12
A computer generated plot of GMod(s) x Gflt(s) is shown in
Figure 12.
Figure 12. Modulator Transfer Function
There is a -180° phase change near F0. At frequencies
greater then F0 the phase increases towards -90° due to
the zero of Fesr. The phase effects of poles and zeros
start a decade below and finish a decade above the
frequency of a pole or zero. Therefore, at the frequency
of a pole or zero the phase effect is only half of the final
value. At the complex pole 2.3kHz the phase is -90 and
would be -180 at 23kHz if not for the +90 phase lead of
the zero at around 6kHz due to the esr of the filter
capacitors. (Actually, the phase gain plots reach their
final values asymptotically).
By inspecting Figure 12 the DC and low frequency gain
of GMod = 20Log(0.85 x 24) = 26.2db; F0 = 2.3kHz; Fesr =
6kHz and Q is 13.6db.
The peak Gain equals the low freq gain plus the Q =
26.2 + 13.6 = 39.9db.
It is desired that T(s) (the open loop transfer function)
have a cross over frequency (Fco) of 1/10 the Switching
frequency at 15kHz. It is require that
πFco)
T(j2
∠
(the
phase of T(s) at Fco), to be greater than -180° by at
least the phase margin. By inspecting the Gain plot of
GMod(s) at 15kHz, GMod(s) has a gain of about 3.9db.
Therefore, to make T(j2πFco) = 1→ 0db;
T(s) = Gea(s) x GMod(s) x Hfb(s) = 1 at Fco
Hfb = Vref/Vout = 0.7/3.3 = 0.212 → -13.5db
|Gea|db = |T|db - |GMod|db -|Hfb|db = 0-3.9db – (-13.5db) =
9.6db ═>3.02 The error amp needs 9db of gain at Fco.
Therefore gm x ZComp = 3.02 at 15kHz.
The location of the error amp’s zero and poles are
selected in order to achieve the desired phase margin of
T(s). For the maximum phase boost at the cross over
Frequency (Fco), place the first Zero1 of the EA at
Fco/10 since the effect of its phase boost will be at the
maximum at Fco. Likewise, place the pole of the EA at
least 10 x Fco so the effects of its phase lag will be at a
minimum at Fco. Therefore, use R1 = 2k; C1 = 0.068µF;
C2 = 470pF.
gm Error Amplifier
Usually, it is undesirable to have high error amplifier gain
at high frequencies otherwise high frequency noise
spikes at large amplitude would be present at the output.
Hence, gain should be permitted to fall off at high
frequencies. At low frequency, it is desired to have high
open-loop gain to attenuate the power line ripple. Thus,
the error amplifier gain should be allowed to increase
rapidly at low frequencies.
The transfer function for the internal gm error amplifier
with R1, C1, and C2 at the comp pin is given by the
following equation:
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
⋅
⋅
+
⋅
+
⋅
⋅
⋅
+
⋅
=
2
1
2
1
1
1
)
2
1
(
1
1
1
)
(
C
C
s
C
C
R
C
C
s
C
s
R
g
s
G
m
ea
The above equation can be simplified by assuming
C2<C1,
()⎥⎦
⎤
⎢
⎣
⎡
⋅
⋅
+
⋅
⋅
⋅
⋅
+
⋅
=
s
C
R
C
s
C
s
R
g
s
G
m
ea
2
1
1
)
1
(
1
1
1
)
(
From the above transfer function, one can see that R1
and C1 introduce a zero and R1 and C2 a pole at the
following frequencies:
Fzero1= 1/2 π × R1 × C1 Fpole1 = 1/2 π × C2 × R1
Fpole@origin = 1/2 π × C1



Html Pages

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20


Datasheet Download

Go To PDF Page


Link URL



Does ALLDATASHEET help your business so far?  [ DONATE ] 

About Alldatasheet   |   Advertisement   |   Contact us   |   Privacy Policy   |   Link to Datasheet    |   Link Exchange   |   Manufacturer List
All Rights Reserved©Alldatasheet.com


Mirror Sites
English : Alldatasheet.com  |   English : Alldatasheet.net  |   Chinese : Alldatasheetcn.com  |   German : Alldatasheetde.com  |   Japanese : Alldatasheet.jp
Russian : Alldatasheetru.com  |   Korean : Alldatasheet.co.kr  |   Spanish : Alldatasheet.es  |   French : Alldatasheet.fr  |   Italian : Alldatasheetit.com
Portuguese : Alldatasheetpt.com  |   Polish : Alldatasheet.pl  |   Vietnamese : Alldatasheet.vn
Indian : Alldatasheet.in  |   Mexican : Alldatasheet.com.mx  |   British : Alldatasheet.co.uk  |   New Zealand : Alldatasheet.co.nz
Family Site : ic2ic.com  |   icmetro.com