Page 273 - Mechatronics with Experiments

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JWST499-Cetinkunt
JWST499-c05
ELECTRONIC COMPONENTS FOR MECHATRONIC SYSTEMS 259
and the output voltage is affected by the output impedance,
R L
V out = K amp V in (5.79)
R + R out
L
Notice that the ideally, when R → ∞ and R → 0, V = v and V = K v , where the
in out in s out amp s
input and output loading effects are zero. In reality, the finite input and output impedance
introduces so-called loading errors to the signal amplification. This is minimized when
the input impedance is very large relative to the source impedance, and output impedance
is very small relative to the load impedance. The net gain of the amplifier in a real
amplifier is
( )( )
R L R in
v = K amp s (5.80)
v
o
R + R out R + R s
in
L
where the portion of the amplifier gain
R L R in
R + R R + R (5.81)
L out in s
is due to the input and output loading effects. In order to minimize the effect of input
loading errors,

R ≫ R s (5.82)
in
Similarly, in order to minimize the output loading errors,
R out ≪ R L (5.83)

Furthermore, more accurate analysis of the amplifier would take the frequency depen-
dance of the input and output impedance into account (Z (jw) instead of R , Z out (jw) instead
in
in
of R out ).
Example Let us consider a sensor, a signal amplifier, and a measurement device, all
connected in series as shown in Figure 5.8. The sensor provides a voltage proportional to the
measured physical variable. Let us call that voltage v (t) = 10 V at steady-state condition.
s
For simplicity let us assume that the gain of the amplifier is unity, K amp = 1.0. Determine
the voltage measured at the measuring device (i.e., digital voltmeter or oscilloscope) for
the following two conditions,

1. R = 100 Ω, R = 100 Ω, R = 100 Ω, R out = 100 Ω
L
in
s
2. R = 100 Ω, R = 100 Ω, R = 1 000 000 Ω, R out = 1 Ω
s
in
L
For the first case, the measured voltage is
( )( )
R L R in
v
v out = K amp s (5.84)
R + R o R + R s
L
in
( 100 )( 100 )
= 1 v s (5.85)
100 + 100 100 + 100
= 0.25 ⋅ v (5.86)
s
= 2.5 V (5.87)
which shows an error of 75% of the correct value of the voltage.