SI9961A Просмотр технического описания (PDF) - Vishay Semiconductors
Номер в каталоге
Компоненты Описание
Список матч
SI9961A Datasheet PDF : 10 Pages
| |||
Si9961A
Vishay Siliconix
Frequency Compensation:
The VCM transconductance (in siemens) of this simplified
case may be expressed in the s (Laplace) plane as:)
gv =
Where
1
Lv
s
+
Rv
Lv
Rv = VCM resistance in ohms
LV = VCM inductance in henrys
s is the Laplace operator
In this case, the transconductance pole is at - Rv/Lv. It is
desirable to cancel this pole in the interest of stability. To do
this, a compensation amplifier is cascaded with the VCM and
its driver. The transfer function of this amplifier is:
( ) Hc = A ×
s+
1
RL × CL
s
Where RL = Compensation amplifier feedback
resistor in ohms
CL = Compensation amplifier feedback
capacitor in farads
A = Compensation amplifier and driver
voltage gain at high frequency
If RL x CL is set equal to Lv/Rv, then the combined open loop
transconductance in siemens becomes:
gto
=
A
s × Lv
In this case, the transconductance has a single pole at the
origin. If this open loop transfer is closed with a transimped-
ance amplifier having a gain of B ohms, the resultant closed
loop transconductance stage has the transfer function (in
siemens) of:
A
gtc
=
Lv
s+
A× B
Lv
Where B = Current feedback transimpedance amplifier gain
in ohms.
The entire transconductance now contains only a single pole
at - A*B/Lv. A and B are chosen to be considerably higher
than the servo bandwidth, to avoid undue phase margin
reduction.
As a typical example, in the referenced schematic, assume
that Rsa and Rsb = 0.5 Ω, R5 = R3 = 10 kΩ, VCM inductance
(Lv) = 1.5 mH, VCM resistance (Rv) = 15 Ω. Hence:
Rv = 15 Ω
Lv = 1.5 mH
B=2Ω
A = 16 x RL/10000
CL = Lv/(Rv x RL) = 100 x 10- 6/RL farads
www.vishay.com
6
Gain Optimization:
There are three things to consider when optimizing the gain
(A) above. The first is servo bandwidth. The main criterion
here is to avoid having the transconductance amplifier cause
an undue loss of phase margin in the overall servo (mechan-
ical + electrical + firmware) loop. The second is to avoid con-
firguing a bandwidth that is more than required in view of
noise and stability considerations. The third is to keep the
voltage output waveform overshoot to a level that will not
cause cross-conduction of the output FETs.
The first two problems can be considered together. Let us
assume a disk drive with a spindle RPM of 4400 and with 50
servo sectors per track. The sample rate is therefore:
fs
= 50 ×
440
60
This is a sample frequency of 3667 Hz
As a rule of thumb, the open loop unity gain crossover fre-
quency of the entire servo (mechanical + electrical + firm-
ware) loop should be less than 1/10 of the sample frequency.
In this example, the servo open loop unity gain crossover fre-
quency would be less than 367 Hz. If we allow only a 10°
degradation in phase margin due to the transconductance
amplifier, then a phase lag of 10° at 367 Hz is acceptable.
This results in a 3 dB point in the transconductance at :
f3db
=
367
tan (10)
or a 3 dB point in the transconductance at 2081 Hz.
The pole in the closed loop transconductance (- A * B / Lv)
should then be 2081 * 2 * π = 13075. This means that A = 9.8.
From the above equation for A, RL = 6.2 kΩ. This sets the
minimum gain limit governed by the servo bandwidth require-
ments. The gain should not be much greater than this, since
increased noise will degrade the servo response.
The third problem, keeping the transconductance amplifier
voltage output wave form overshoot to a level that will not
cause the wrong output FETs to conduct, can be evaluated
by deriving the voltage transfer function of the closed loop
transconductance amplifier from input voltage to output volt-
age (Vin to output A and B on the reference schematic).
This is :
Hto =
A×
s+ p
s+ x
Where
p = 1/RL x CL) or Rv/Lv Comp amplifier
zero/VCM pole
x = A x B/Lv closed loop pole
Document Number: 70014
S-40845-Rev. H, 03-May-04
Share Link: 