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ALL ABOUT ELECTRONICS.
So, in this video, we will see how to use
this op-amp as summing amplifier and using
this op-amp configuration how we can add the
different input voltages.
So, now in the earlier video of inverting
and non-inverting op-amp configuration, we
have applied the single input either to the
non-inverting or the inverting op-amp terminals.
Now, in this video, we will apply the multiple
inputs either to the inverting or the non-inverting
op-amp terminal and we will see how this configuration
can be used as summing amplifier.
So, first we will see the inverting summing
amplifier and in this configuration, we will
apply the multiple inputs to the inverting
input terminal.
And we will find the expression of the output
voltage in terms of the different input voltages.
So, as you can see here, we have applied the
input voltages V1, V2 and V3 to this inverting
terminal via resistors R1, R2, and R3.
And here let's assume that current I1, I2,
and I3 are flowing through this resistor R1,
R2, and R3.
And let's say the current If is flowing through
this resistor Rf.
And here, we are assuming the op-amp as the
ideal op-amp.
So, no current is going inside this op-amp.
And let's consider this node as node X.
Now, in the earlier video of inverting op-amp
configuration, we have seen the concept of
virtual ground.
And in that concept, we had seen that whenever
we are applying the negative feedback to any
op-amp, then there is virtual short exist
between the inverting and the non-inverting
terminals.
So, if one terminal is at ground potential,
then another terminal will also be at ground
potential.
Or we can say that it will act as a virtual
ground.
So, this node will have zero potential.
So, now let's apply KCL at this node X.
So, applying KCL we can write I1 +I2 +I3,
that is equal to this current If.
Now, as this terminal is at ground potential,
so the current I1 will be equal to V1 minus
zero divides by R1.
Likewise, this current I2 will be equal to
V2 minus zero divides by R2.
And likewise, current I3 will be equal to
V3 minus zero divides by R3.
Now, this current If will be equal to zero
minus Vout divide by this feedback resistor
Rf.
That is zero minus Vout, divide by this feedback
resistor Rf.
So, if we simplify this expression then we
can write it as
V1 divide by R1, plus V2 divide by R2, plus
V3 divide by R3 that is equal to minus Vout
divide by Rf.
And if we further simplify it then we can
write it as,
Vout that is equal to minus, Rf divide by
R1 times V1, plus Rf divide by R2 times V2,
plus Rf divide by R3 times V3.
Now, we know that if we apply the single input
to this non-inverting op-amp configuration,
then the output voltage Vout can be given
as -minus Rf divide by R1 times the input
voltage.
Now, here instead of single input voltage
we have applied the multiple input voltages.
So, the output voltage will be equal to the
algebraic sum of the individual responses.
So, this is the expression of the output voltage
in case of the inverting summing amplifier.
Now, in this inverting summing amplifier configuration,
let's assume that the value of R1, R2, and
R3 is same.
So, in that case, the output voltage Vout
will be equal to -Rf divide by R times, V1
plus V2 plus V3.
So, now here if the value of Rf is equal to
R, then, in that case, our output voltage
vout will be equal to -(V1 plus V2 plus V3).
So, in this way, we can use this op-amp as
an adder and we can add the multiple input
voltages.
So, now suppose in this configuration, if
R1, R2, and R3 are different then, in that
case, the ratio of this feedback resistor
over this resistors R1, R2 and R3 will also
be different.
So, in that case along with the addition,
we can also perform the scaling operation.
So, in that case, our output voltage Vout
will be equal to -(A*V1 +B*V2 +C*V3).
Where A, B, and C represents the ratio of
this Rf divide by R1, Rf divide by R2 and
Rf divide by R3.
So, along with addition, we can also perform
the scaling operation.
So, along with this scaling and addition operation,
we can also perform the averaging operation
using this configuration.
So, now let's see how we can perform the averaging
of the different input voltages using this
configuration.
So, let's once again assume that the R1, R2,
and R3 are having the same value.
And let's say the ration of this Rf divide
by R is equal to 1 divide by n.
Where n represents the number of input voltages
that are being applied to this inverting terminal.
So, in that case, our output voltage Vout
will be equal to, minus Rf divide by R times,
V1 plus V2 plus V3.
And we put the value of this Rf by R as 1divide
by n, that is 1 divided by 3 in this case,
then our output voltage will be equal to minus
V1 plus V2 plus V3, divide by 3.
So, in this way, our output voltage will be
the average of the three different input voltages.
So, we can also perform the averaging operation
using this configuration.
So, now let's see some more applications in
which this inverting summing amplifier can
be used.
So, as I already told you, this summing amplifier
can be used for the operation of addition,
averaging as well as the scaling.
Apart from that, it can also be used to provide
the DC offset to the input signals.
So, suppose if your input voltages are AC
signals or let's say it is coming from some
sensor, and if we want to apply some DC offset,
then using this configuration we can also
apply some DC offset to the incoming signals.
Likewise, this inverting summing amplifier
can also be used for the digital to analog
conversion.
And apart from that, it can also be used for
the mixing the different audio signals.
So, these are the different applications in
which this summing amplifier can be used.
Now, theoretically, we can add the n number
of input voltages to this op-amp configuration,
but practically if you see the number of input
voltages depends on the power dissipation
as well as the total current that can be supplied
by the op-amp.
Because here if you see, the total current
If is the summation of the individual currents
I1, I2, I3 upto In.
So, this value should not exceed the maximum
value that is supported by the op-amp.
Now, one more interesting fact about this
inverting op-amp configuration is that all
the individual voltage sources are isolated
with respect to each other.
And that is because of the concept of virtual
ground.
Because if you consider the individual voltage
sources at a time, then at that time, the
remaining voltage sources will act as a short
circuit.
So, if we consider this voltage source V1,
that is acting alone, then at that time, this
voltage source V2 and V3 will act as a short
circuit.
Or we can say that they are at ground potential.
Now, because of the virtual ground, this node
is also at the ground potential.
So, we can say that effectively this R2 and
R3 does not exist in the circuit.
So, the effective impedance that is seen by
the voltage source V1 is the series resistance
R1 of that voltage source.
So, we can say that there is no interference
between the different voltage sources.
And that is the biggest advantage of this
inverting summing amplifier.
Now, the output of this inverting summing
amplifier is the negative voltage.
But suppose if we want the positive voltage
then what we can do, we can connect one more
inverting op-amp which is having unity gain.
So, as you can see here, suppose if we connect
one more inverting op-amp, which is having
unity gain at this point then the output voltage
will be positive.
So, by using one more inerting op-amp, we
can get the positive output voltage.
So, this is all about the inverting summing
amplifier.
Now, let's see what happens when we apply
the multiple input voltages at the non-inverting
input terminal.
So, now in this non-inverting summing amplifier,
we have applied the two input voltages at
this non-inverting end.
And we will find the output voltage in terms
of the input voltages V1 and V2.
Now, we know that in case of the non-inverting
configuration, the output voltage is equal
to one plus Rf divide by Ra times the voltage
at this node.
Let's say that is equal to Vplus.
So, as we have applied the multiple inputs
at this end, so first of all, we need to find
this voltage Vplus.
And we can do so by applying the principle
of superposition.
So, what we will do, we will consider the
one voltage source at a time and we will find
the voltage at this node.
And later on, we will combine the individual
responses to get the final response.
So, first of all, let's assume that this voltage
source V1 is acting alone and we have removed
this voltage source V2.
That means V2 is equal to zero.
So, in that case, let's say the voltage at
this point is equal to V1 plus.
So, V1plus will be equal to R2 divide by (R1
plus R2) times this voltage V1.
Likewise, when we consider this voltage source
V2 is acting alone, and V1 is equal to zero.
In that case, the voltage at this point is
let's say V2plus.
So, V2plus will be equal to R1 divide by (R1
plus R2) times this voltage V2.
So, in this way, when this voltage source
V1 and V2 are acting alone, then, in that
case, the V1plus will be equal to this value
and V2 plus will be equal to this value.
So, the overall voltage Vplu will be equal
to the summation of the individual voltages.
that is (V1plus) plus (V2 plus)
That is equal to R2 divide by (R1 plus R2)
time V1, plus R1 divide by (R1 plus R2 ) times
V2.
So, this will be voltage at this because of
the voltage V1 and V2.
Now, we know that the output voltage is equal
to one plus Rf divide by Ra times the voltage
at this end.
That is equal to Vplus.
So, our final output voltage will be equal
to one plus Rf divide by Ra times, R2 divide
by (R1 plus R2 ) times V1, plus R1 divide
by (R1 plus R2 ) times V2.
So, this will be the final voltage that we
get at the output of this non-inverting summing
amplifier.
So, now let's simplify this equation a bit
and let's assume that this R1 is equal to
R2 is equal to R.
So, when R1=R2=R, then, in that case, the
output voltage vout will be equal to one plus
Rf divide by Ra times V1 plus V2, divide by
2.
Now, if we consider Rf = Ra, in that case,
the output voltage Vout will be equal to V1
+V2.
That is the addition of the voltage V1 and
V2.
So, in this way, using this non-inverting
summing amplifier also, we can perform the
addition.
Now, let's take the case when we have three
different input voltages that are connected
to this non-inverting terminal.
So, as you can see here, we have three different
input voltages V1, V2, and V3, which are connected
to this non-inverting terminal via resistor
R1, R2, and R3.
So, here also, we will consider the one voltage
source at a time and we will find the voltage
at this non-inverting terminal.
And later on, we will add the individual responses
to get the voltage at this non-inverting terminal.
So, first of all, let's assume that this voltage
source V1 is acting alone and V2 and V3 are
equal to zero.
So, in that case, the circuit will look like
this.
So, the voltage at this point let's say is
equal to V1 plus.
So, V1 plus will be equal to (R2 in parallel
R3) divide by (R1 plus + (R2 in parallel R3
) ) times this voltage V1.
So, this is the voltage that you get at this
point when this voltage source V1 is acting
alone.
likewise, we can have voltage V2 plus that
is equal to,
(R1 parallel R3) divide by this R2 plus (R1
parallel R3) times this voltage V2.
this is the expression when the voltage source
V2 is acting alone.
And likewise, we can have V3 plus that is
equal to (R2 parallel R1) divide by R3 plus
(R2 parallel R1) times this voltage V3.
So, the total voltage that appears at this
point will be the summation of this individual
voltages.
So, as you can see here, in this non-inverting
configuration, as the number of input voltages
is increasing, the complexity is also increasing.
And to here reduce the complexity, let's assume
that R1, R2, and R3 are equal.
So, when all the three values are same, then
the value of V1plus will be equal to V1/3.
Likewise, the value of V2plus will be equal
to V2/3.
And likewise, V3 plus will be equal to V3/3.
So, in short, when R1, R2, and R3 are equal
in that case, Vplus will be equal to (V1 +V2
+V3 )/3
And the output voltage Vout will be equal
to
(1+ (Rf/Ra)) times Vplus.
Or we can say that (1+(Rf/Ra)) (V1 +V2+V3)/3
So, now suppose if the value of 1+ (Rf/Ra)
that is equal to 3, in that case, Vout will
be equal to V1+V2+V3.
So, as you can see here, in this configuration
as the number of input voltage increases,
the complexity is also increasing.
And here, the individual input voltage that
appears at this point does not only depend
on the value of this V1 and R1, but it also
depends upon the different series resistors
of the individual voltage sources.
Like, here, in this case, it also depends
upon the value of this R2 and R3.
So, in this configuration, we can say that
the individual voltage sources are not isolated
with respect to each other.
And they have their own influence on the input
voltage that appears at this non-inverting
terminal.
So, that is the reason, in most of the practical
applications, this inverting summing amplifier
is more preferred over this non-inverting
summing amplifier.
So, that is all about the inverting and the
non-inverting summing amplifier.
So, I hope in this video, you understood about
this inverting and the non-inverting summing
amplifiers.
So, if you have any question or suggestion,
do let me know in the comment section below.
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