The op-amp has four terminals to which connections can be
made. Inputs attach to nodes
a and
b , and the output is node
c . As the circuit model on the right
shows, the op-amp serves as an amplifier for the difference ofthe input node voltages.
Op-amps not only have the circuit model shown in
[link] ,
but their element values are very special.
The
input resistance ,
,
is typically
large , on the order of
1 MΩ.
The
output resistance ,
,
is
small , usually less than 100 Ω.
The
voltage gain ,
,
is
large , exceeding
.
The large gain catches the eye; it suggests that an op-amp could
turn a 1 mV input signal into a 100 V one. If you were to buildsuch a circuit--attaching a voltage source to node
a , attaching node
b to the reference, and looking at the output--you would be
disappointed. In dealing with electronic components, you cannotforget the unrepresented but needed power supply.
It is impossible for electronic components to yield voltagesthat exceed those provided by the power supply or for them to
yield currents that exceed the power supply's rating.
Typical power supply voltages required for op-amp circuits are
. Attaching the 1 mv signal not only would fail
to produce a 100 V signal, the resulting waveform would beseverely distorted. While a desirable outcome if you are a rock&roll aficionado, high-quality stereos should not distort
signals. Another consideration in designing circuits withop-amps is that these element values are typical: Careful
control of the gain can only be obtained by choosing a circuitso that its element values dictate the resulting gain, which
must be smaller than that provided by the op-amp.
Op-amp
The top circuit depicts an op-amp in a feedback amplifier
configuration. On the bottom is the equivalent circuit, andintegrates the op-amp circuit model into the circuit.
Inverting amplifier
The feedback configuration shown in
[link] is the most common op-amp circuit for obtaining what is knownas an
inverting amplifier .
provides the exact input-output relationship. In choosing element
values with respect to op-amp characteristics, we can simplify theexpression dramatically.
Make the load resistance,
,
much larger than
.
This situation drops the term
from the second factor of
[link] .
Make the resistor,
,
smaller than
,
which means that the
term in the third factor is negligible.
With these two design criteria, the expression(
[link] )
becomes
Because the gain is large and the resistance
is small, the first term becomes
,
leaving us with
If we select the values of
and
so that
,
this factor will no longer depend on the op-amp's inherentgain, and it will equal
.
Under these conditions, we obtain the classic input-outputrelationship for the op-amp-based inverting amplifier.
Consequently, the gain provided by our circuit is entirely
determined by our choice of the feedback resistor
and the input resistor
.
It is always negative, and can be less than one or greaterthan one in magnitude. It cannot exceed the op-amp's inherent
gain and should not produce such large outputs that distortionresults (remember the power supply!). Interestingly, note that
this relationship does not depend on the load resistance. Thiseffect occurs because we use load resistances large compared
to the op-amp's output resistance. Thus observation meansthat, if careful, we can place op-amp circuits in cascade,
without incurring the effect of
succeeding circuits changing the behavior (transfer function)of previous ones; see
this problem .
Questions & Answers
if three forces F1.f2 .f3 act at a point on a Cartesian plane in the daigram .....so if the question says write down the x and y components ..... I really don't understand
a fixed gas of a mass is held at standard pressure temperature of 15 degrees Celsius .Calculate the temperature of the gas in Celsius if the pressure is changed to 2×10 to the power 4
