# Non investing summer op amp

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Such circuit gives the addition of the applied signals at the output. Hence it is called Summer or adder circuit. Depending upon the sign of the output, the Summing Amplifier circuits are classified as inverting summing amplifier and non inverting summing amplifier. In this circuit, all the input signals to be added are applied to the inverting input terminal, of the op-amp.

The circuit with two input signals is shown in the Fig. As point B is grounded, due to virtual ground concept the node A is also at virtual ground potential. Now from input side,. Infact in such a way, n input voltages can be added. Thus the magnitude of the output voltage is the sum of the input voltages and hence circuit is called as summer or adder circuit. Due to the negative sign of the sum at the output it is called inverting summing amplifier.

It shows that there is phase inversion. The circuit discussed above is inverting summing op amp, which can be noticed from the negative sign in the equation 6. The circuit diagram of a summing amplifier is as shown in the figure above. Instead of using a single input resistor, all the input sources have their own input drive resistors. A circuit like this amplifies each input signal. The gain for each input is given by the ratio of the feedback resistor R f to the input resistance in the respective branch.

Let R 1 be the input impedance and V 1 be the input voltage of the first channel. It is already been said that a summing amplifier is basically an Inverting Amplifier with more than one voltage at the inverting input terminal. The output voltage for each channel can be calculated individually and the final output voltage will be the sum of all the individual outputs. To calculate the output voltage of a particular channel, we have to ground all the remaining channels and use the basic inverting amplifier output voltage formula for each channel.

If all the channels are grounded except the first channel, then output for first channel is given by:. Similarly, if all the channels are grounded except the second channel, then output for second channel is given by:. The output signal is the algebraic sum of individual outputs or in other words it is the sum of all the inputs multiplied by their respective gains.

But if all the input resistances are chosen to be of equal magnitude, then the Summing Amplifier is said to be having an equal-weighted configuration, where the gain for each input channel is same. Sometimes, it is necessary to just add the input voltages without amplifying them. In such situations, the value of input resistance R 1 , R 2 , R 3 etc. As a result, the gain of the amplifier will be unity.

Hence, the output voltage will be an addition of the input voltages. Theoretically, we can apply as many input signals to the input of the summing amplifier as required. However, it must be noted that all of the input currents are added and then fed back through the resistor R f , so we should be aware of the power rating of the resistors.

Here, the input voltages are applied to the non-inverting input terminal of the Op Amp and a part of the output is fed back to the inverting input terminal, through voltage-divider-bias feedback. The circuit of a Non-Inverting Summing Amplifier is shown in the following image. For the sake of convenience, the following circuit consists of only three inputs, but more inputs can be added.

First and foremost, even though this is also a Summing Amplifier, the calculations are not as straight forward as the Inverting Summing Amplifier because there is no advantage of virtual ground summing node in the Non-Inverting Summing Amplifier.

To understand the working of a Non-Inverting Summing Amplifier, we have to divide the circuit into two parts:. If V IN is the combination of all the input signals, then this is applied at the non-inverting terminal of the Op Amp. From the above circuit, we can calculate the output voltage of the Non-Inverting Amplifier with V IN as input and R f and R i as the feedback divider resistors as follows:.

Let us calculate the V IN1 portion of the V IN and by simple mathematics, we can easily derive the other two values i. Coming to V IN1 , when V 2 and V 3 are grounded, their corresponding resistors cannot be ignored as form a voltage divider network. Then the input resistors are selected as large as possible to suit the type of the op-amp used.

Three audio signals drive a summing amplifier as shown in the following circuit. What is the output voltage? The sounds from different musical instruments can be converted to a specific voltage level, using transducers, and connected as input to a summing amplifier. These different signal sources will be combined together by the summing amplifier and the combined signal is sent to an audio amplifier.

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Let's check, V out was connected to two resistors, and the bottom is connected to ground, and this was connected there. So what our goal is right now, we want to find V out as a function of V in. That's what we're shooting for. So let's see if we can do that.

Let's give our resistors some names. Let's call this R1, and R2, our favorite names always, and now everything is labeled. Now and we can label this point here, and this point we can call V minus, V minus. So that's our two unknowns. Our unknowns are V not, V out, and V minus, so let's see if we can find them. So what I'm going to do is just start writing some expressions for things that I know are true.

Alright, that's what this Op-amp is telling us is true. Now what else do I know? Let's look at this resistor chain here. This resistor chain actually looks a lot like a voltage divider, and it's actually a very good voltage divider.

Remember we said this current here, what is this current here? It's zero. I can use the voltage divider expression that I know. In that case, I know that V minus, this is the voltage divider equation, equals V out times what?

Times the bottom resistor remember this? R2 over R1 plus R2, so the voltage divider expression says that when you have a stack of resistors like this, with the voltage on the top and ground on the bottom, this is the expression for the voltage at the midpoint. Kay, so what I'm going to do next is I'm going to take this expression and stuff it right in there.

Let's do that. See if we got enough room, okay now let's go over here. Let's keep going, let's keep working on this. Alright, so now I'm going to gather all the V not terms over on the left hand side. Let's try that. V plus is V in. Okay let's keep going I can factor out the V not. Alright so we're getting close, and our original goal, we want to find V out in terms of V in.

So I'm going to take this whole expression here and divide it over to the other side, so then I have just V not on this side, and V in on the other side. Make some more room. Alright so that's our answer. That's the answer. That's V out equals some function of V in. Now I want to make a really important observation here.

This is going to be a real cool simplification. Okay, so this is the point where Op-amp theory gets really cool. Watch what happens here. We know that A is a giant number. A is something like 10 to the fifth, or 10 to the sixth, and it's whatever we have here, if our resistors are sort of normal-sized resistors we know that a giant number times a normal number is still going to be a very big number compared to one.

So this one is almost insignificant in this expression down here, so what I'm going to do, bear with me, I'm going to cross it out. I'm going to say no, I don't need that anymore. So if this, if this number here, if A is a million, 10 to the sixth, and this expression here is something like one half then this total thing is one half 10 to the sixth or a half a million, and that's huge compared to one.

So I can pretty safely ignore the one, it's very, very small. Now when I do that, well look what happens next, now I have A top and bottom in the expression, and I can cancel that too. So the A goes away, now this is pretty astonishing. We have this amplifier circuit and all of a sudden I have an expression here where A doesn't appear, the gain does not appear, and what does this turn into? This is called V not equals V in, times what?

Times R1 plus R2, divided by R2. So our amplifier, our feedback circuit came down to V out is V in multiplied by the ratio of the resistors that we added to the circuit. A non-inverting amplifier generates an amplified output signal that is in phase with the applied input signal. A non-inverting amplifier works like a voltage follower circuit because this circuit uses a negative feedback connection.

So it gives a part of the output signal as feedback to the inverting input terminal instead of giving a complete output signal. The complement of this op-amp is inverting op-amp which generates the output signal that is degrees out of phase. This circuit is ideal for impedance buffering applications due to high input and low output impedance. The non-inverting op-amp circuit diagram is shown below.

In this circuit configuration, the output voltage signal is given to the inverting terminal - of the operational amplifier like feedback through a resistor where another resistor is given to the ground. Here, a voltage divider with two types of resistors will provide a small fraction of the output toward the inverting pin of the operational amplifier circuit.

These two resistors will provide necessary feedback to the operational amplifier. Here, the R1 resistor is called a feedback resistor Rf. Because of this, the Vout depends on the feedback network. The Current rule states that there is no flow of current toward the inputs of an op-amp whereas the voltage rule states that the op-amp voltage tries to ensure that the voltage disparity between the two op-amp inputs is zero.

From the above non-inverting op-amp circuit, once the voltage rule is applied to that circuit, the voltage at the inverting input will be the same as the non-inverting input. So the applied voltage will be Vin. So the voltage gain can be calculated as,. Therefore the non-inverting op-amp will generate an amplified signal that is in phase through the input. In a non-inverting operational amplifier circuit, the input impedance Zin can be calculated by using the following formula.