L292
Example :
a) Data
b) Calculation
c) Summarising
- Motors characteristics:
LM = 5 mH
RM = 5 W
LM / RM = 1msec
- Voltage and current characteristics:
Vs = 20 V
IM = 2 A
- Closed loop bandwidth : 3 kHz
VI = 9.1 V
- From relationship (4) :
Rs =
0.044
IM
VI
=
0.2 Ω
and from (1) :
Gmo =
2VS
RM VR
=
1 Ω −1
- RC = 1 msec [from expression (2) ].
- Assuming ξ = 1/ √2 ; from (7) follows :
ξ2
=
1
2
=
400 C
4RF CF • 0.2
- The cutoff frequency is :
fT
=
143 • 10−3
RF CF
= 3 kHz
- RC = 1.10-3 sec
-
1000 C
RF CF
=
1
- RF CF ≅ 47 µs
C = 47 nF
R = 22 KΩ
For RF = 510 Ω → CF = 92 nF
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