Mentioned in
the previous article, an operational amplifier is usually connected to
components in a feedback circuit. In this article, we will discuss one of two
amp configurations that are most widely used.
As you know, operational
amplifiers can be used in a vast array of circuit configurations and one of
the most simple configurations to use is the
inverting amplifier. The amplifier only requires the operational amplifier IC
and a few other small components. Inverting amplifiers are also used as summing
amplifiers, which sums the voltage present on multiple inputs and combines them
into a single output voltage. Some examples are summing several signals with
the same gain in an audio mixer or converting binary numbers to a voltage in a
digital-to-audio-converter (DAC).
The below figure, Fig 1.1
illustrates the amplifier's inverting configuration. There are an operation
amplifier and two resistors R1 and R2 inside of the inverting configuration. The second
resistor, R2 is
connected from terminal 3 (output terminal), back to terminal 1, which is the
inverting (-) input terminal. As R2 is connected in this manner, we can apply negative
feedback, which is the process of "feeding back" a small portion of
the output signal back into the input terminal. However, in order to make the
feedback negative, the output signal must be fed into the inverting terminal of
the operation amplifier. Also, if R2 was connected between the output terminal 3 and the
input terminal 2 (noninverting input), there would be positive feedback. Now
the output signal is fed back into the noninverting (+) input, creating a
positive feedback into the operational amplifier. If you look at the diagram
closely, R2 also
closed the loop surrounding the operational amp. In addition to R2 in this configuration, terminal 2
has been grounded and connected to the resistor R1 in between terminal 1 and the
input signal source of voltage v1. Between terminal 3 and the ground is the point from which we will take
the output since there is an impedance level that is ideally zero. Due to this,
voltage v0 does
not depend on the impedance value of the current that is supplied to the load
impedance that is connected between the ground and terminal 3.
Looking at Figure 1.1, we can
now analyze the circuit to find what is
called the closed-loop gain G, which can be written as
G≡v0v1G≡v0v1
While analyzing this circuit, the operational amplifier will
be considered to be ideal for measurement purposes. Figure 1.2 (a) illustrates
the equivalent closed-loop circuit, from which we can deduce that the
gain A is extremely large, that of which is ideally infinite. By
assuming that the circuit is producing a finite voltage at the output terminal
3, then we can say that between the operational amplifier's input terminals
that the voltage is negligible and ideally zero. The output voltage v0 can be defined as
v2−v1=v0A=0v2−v1=v0A=0
At the inverting input
terminal, the voltage, v1 is given by v1 = v2. This is so because as the gain A is approaching
towards infinity, v1 is
approaching v2 and
ideally equivalent. Having found v1, Ohm's law can now be applied to find the current i1 through R1.
i1=v1−v1R1=v1−0R1=v1R1i1=v1−v1R1=v1−0R1=v1R1
We need to know where this
current will flow. Since the operation amplifier has an ideal infinite input
impedance, which means there can be zero current drawn, it cannot flow here;
instead, i1 will
flow through R2 into
terminal 3. From this, Ohm's law can be applied atR2 to find the value of v0.
v0=v1−i1R2v0=v1−i1R2
=0−v1R1R2=0−v1R1R2
Then,
v0v1=−R2R1v0v1=−R2R1
This is the closed-loop gain.
This gain is just the ratio of the two resistor values R2 and R1. There is a minus sign on the right
side because the closed-loop inverting amplifier is providing a signal
inversion. For example, if the ratio R2/R1 = 10 and a sine wave signal of 2 V is applied at the
input terminal (v1), there
will be a sine wave signal of 20 V (peak-to-peak) and phase-shifted exactly 180
degrees with respect to v1. Due to the fact that there is a minus sign incorporated in the ratio
of the closed-loop gain, the configuration is called the inverting
configuration.
A significant point worth
talking about is that this closed-loop gain depends solely on external passive
components (R1 and R2). This will allow the closed-loop gain
to be made as accurate as needed by selected different passive components, such
as resistors capacitors, or inductors, of appropriate value. Also,
because of this dependency, the closed-loop gain is ideally independent of the
operational amplifier gain. To summarize: the amplifier started out having a
large gain A, and thus through applying a negative
feedback, a closed-loop gain R2/R1 has been obtained that is much smaller than the gain
but it is now stable and also predictable. I.e. gain is being traded for
accuracy.
To clarify what was just
said, an expression can be derived for the closed-loop gain as long as the
operational amplifier's open-loop gain A is a finite value. Figure 1.3
illustrates this clarification. By denoting the output voltage as v0, the voltage between the two input
terminals of the operational amplifier is now vo/A. Also, because the noninverting input terminal is grounded,
this voltage at the inverting input terminal must now be -v0/A. We can write an expression
for the current i1 that
passes through R1 as
i1=v1−(−v0/A)R1=v1+v0/AR1i1=v1−(−v0/A)R1=v1+v0/AR1
The operational amplifier's
infinite input impedance drives the current i1 to flow completely through R2. Now the output voltage, v0 can be found by
v0=−v0A−i1R2v0=−v0A−i1R2
=−v0A−(v1+v0/AR1)⋅R2=−v0A−(v1+v0/AR1)⋅R2
By summing terms, the
closed-loop gain G can be found by
G≡v0v1=−R2/R11+(1+R2/R1)/AG≡v0v1=−R2/R11+(1+R2/R1)/A
(1.1)
By noting that as A approaches
∞∞
, G approaches
the ideal value of -R2/R1. Looking at Fig 1.3, it can be seen that as A approaches
∞∞
, that the
voltage at the inverting input terminal will approach zero. If you recall, this
was the assumption used earlier to analyze the
operational amplifier when it was assumed to be ideal. Finally, we will note
that Eq. 1.1 shows that in order to minimize the dependency of the closed-loop
gain G on the value of the open-loop gain A, the
closed-loop gain needs to be much less than the value of the open-loop gain.
To conclude this article, the
inverting configuration has been discussed and explained. I hope that you have
gained a better understanding of the purpose of this amplifier as well as how
it is designed. Whether it is an audio mixer or a digital-to-audio converter,
you'll find an inverting amplifier within it. Two things are to be remembered
when talking about inverting amplifiers: no current flows through either input
terminal and that the differential input voltage is zero (v1=v2=0). From these two rules, we derived an equation to
calculate the closed-loop gain of the inverting amplifier. If you have any
questions or comments, please leave them below!