Cautions - Kenwood RA-920 Instruction Manual

Resistance attenuator
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CAUTIONS

1 . T h e signal v o l t a g e a p p l i e d t o I N P U T s h o u l d be l o w e r
t h a n 17 V r m s .
2 . T o a v o i d leakage o f signal a n d m i n i m i z e t h e e f f e c t o f
phase d e v i a t i o n , be sure t o use s h i e l d e d cables o n t h e
i n p u t a n d o u t p u t o f t h e u n i t . These cables s h o u l d be as
s h o r t as p o s s i b l e . T h i s is p a r t i c u l a r l y i m p o r t a n t
pulse signals are used as a signal s o u r c e .
3 . W h e n t h e u n i t is used w i t h t h e case g r o u n d f l o a t e d f r o m
t h e c i r c u i t g r o u n d , t h e p o t e n t i a l d i f f e r e n c e ( D C + A C
p e a k )
b e t w e e n
these t w o
±
6 0 0 V .
4 . D i a l
s e t t i n g
Select a p p r o p r i a t e dials f o r t h e desired a t t e n u a t i o n at
a n y
f r e q u e n c y .
N o t e t h a t t h e m a r k i n g s o n each
i n d i c a t e a p p r o x i m a t e settings o f a t t e n u a t i o n
e r r o r . A t y p i c a l e x a m p l e o f a t t e n u a t i o n e r r o r a n d f r e -
q u e n c y c h a r a c t e r i s t i c e r r o r is s h o w n i n T a b l e 1 , w h e r e
1 0 0 k H z signal is a t t e n u a t e d t o 1 0 0 d B , e
at 1 k H z o f each d i a l a n d £
c h a r a c t e r i s t i c .
T h e o v e r a l l e r r o r d e p e n d s o n t h e n u m b e r o f dials t o be
used as e x p l a i n e d b e l o w .
I) W h e n t h e d i a l s are set t o 3 0 d B x 2 = 6 0 d B , 10 d B x 3
= 3 0 d B , 1 d B x 9 = 9 d B a n d 0.1 d B x 10 = 1 d B ,
t h e n t h e o v e r a l l e r r o r (e)
f r e q u e n c y c h a r a c t e r i s t i c is o b t a i n e d f r o m t h e f o l l o w -
w h e n
p o i n t s s h o u l d
n o t
e x c e e d
d i a l
i n c l u d i n g
is t h e e r r o r
3
is t h e e r r o r o f f r e q u e n c y
f
caused b y each d i a l a n d
i n g e q u a t i o n :
£ =
±V(0.15
2
+ 0 . 1
+ ( 0 . 0 1
+ 0 . 0 5
)
= ± 0 . 2 8 d B
2
2
II) W h e n t h e dials are set t o 3 0 d B x 2 = 6 0 d B , 1 0 d B x
4 = 4 0 d B , t h e n :
£ = ± V ( 0 . 1 5
+ 0 . 1
) + ( 0 . 1
2
2
A s w i l l be u n d e r s t o o d f r o m t h e a b o v e e q u a t i o n s , t h e
less t h e n u m b e r o f d i a l s , t h e h i g h e r t h e a c c u r a c y .
T a b l e 1 . T y p i c a l E x a m p l e o f A t t e n u a t i o n E r r o r a n d
F r e q u e n c y C h a r a c t e r i s t i c E r r o r
E r r o r
3 0 d B step 10 d B step 1 d B step
± 0 . 1 5 d B
± 0 . 1 d B
± 0 . 1 d B
± 0 . 1 d B
£ f
) + ( 0 . 1
+ 0 . 1
) + ( 0 . 1
+ 0 . 0 5
2
2
2
2
+ 0 . 1
) = 0 . 2 3 d B
2
2
0.1 d B step
± 0 . 1 d B
± 0 . 0 1 d B
± 0 . 0 5 d B
± 0 . 0 5 d B
;
2
7

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