E1C06 09/14/2010
11:55:5 Page 227
that R 1 þ R 2 ¼ R T . The resistances R m and R 1 form a parallel loop, yielding an equivalent resistance,
R L , given by
R L ¼
R 1 R m
R 1 þ R m
ð6:31Þ
The equivalent resistance for the entire circuit, as seen from the voltage source, is R eq :
R eq ¼ R 2 þ R L ¼ R 2 þ
R 1 R m
R 1 þ R m
ð6:32Þ
The current flow from the voltage source is then
I ¼
E i
R eq
¼
E i
R 2 þ R 1 R m = R 1 þ R m
ð
Þ
ð6:33Þ
and the output voltage is given by
E o ¼ E i À IR 2
ð6:34Þ
This result can be expressed as
E o
E i
¼
1
1 þ R 2 =R 1
ð
Þ R 1 =R m þ 1
ð
Þ
ð6:35Þ
The limit of this expression as R m tends to infinity is
E o
E i
¼
R 1
R 1 þ R 2
ð6:36Þ
which is just Equation 6.8. Also, as R 2 approaches zero, the output voltage approaches the supply
voltage, as expected. Expressing the value found from Equation 6.8 as the true value E o =E i
ð
Þ
0
, the
loading error e I may be given by
e I ¼ E i
E o
E i
À
E o
E i
0 !
ð6:37Þ
Here the loading error goes to zero as R m ! 1. Hence, a high meter resistance reduces loading
error in voltage measurements.
B
A
I
R 2
R 1
E i
R m
E o
Figure 6.15 Instruments in parallel to signal path form an
equivalent voltage-dividing circuit.
6.5 Loading Errors and Impedance Matching 227