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 .
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
from theory: distance [miles] = speed [mph] × time [hours]
info #1
speed_Dennis × 1.5 = speed_Wayne × 2
=> speed_Wayne = 0.75 × speed_Dennis (i)
info #2
speed_Dennis = speed_Wayne + 7 [mph] (ii)
use (i) in (ii) => [...]
speed_Dennis = 28 mph
speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5.
Substituting the first equation into the second:
W * 2 = (W + 7) * 1.5
W * 2 = W * 1.5 + 7 * 1.5
0.5 * W = 7 * 1.5
W = 7 * 3 or 21
W is 21
D = W + 7
D = 21 + 7
D = 28
Salma
Devon is 32 32 years older than his son, Milan. The sum of both their ages is 54 54. Using the variables d d and m m to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
however, may I ask you some questions about Algarba?
Amoon
hi
Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67.
Check:
Sales = 3542
Commission 12%=425.04
Pay = 500 + 425.04 = 925.04.
925.04 > 925.00
Munster
difference between rational and irrational numbers
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
