Opamp adder circuit: Where does the "-" come from?

In summary: You have to be very careful about the signs; just one error can ruin the whole equation.In summary, the conversation discusses the use of KCL to analyze an adder circuit and the importance of being consistent with the polarities of currents when making equations. It also mentions the virtual ground property of opamps and the potential for sign errors when adding voltages in loops.
  • #1
altruan23
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5
Homework Statement
I was analyzing the adder circuit and I dont understand from where does the - come in the Vout equation.
Relevant Equations
KCL
So i used KCL and both currents are flowing into the node, and then leaving together to go to the resistor R3.
So my eq can be seen in the picture. I was looking in a book and they had a minus infront of the parantheses.
Is the current flowing from R3 into the node??
1647963385469.png
 
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  • #2
altruan23 said:
Homework Statement:: I was analyzing the adder circuit and I don't understand from where does the - come in the Vout equation.
Relevant Equations:: KCL

So i used KCL and both currents are flowing into the node, and then leaving together to go to the resistor R3.
So my eq can be seen in the picture. I was looking in a book and they had a minus infront of the parantheses.
Is the current flowing from R3 into the node??
View attachment 298767
I'm not able to read your image (in the future, please type your work into the forum using LaTeX -- see the LaTeX Guide link below the Edit window).

But intuitively, when the input voltage goes positive, the output of the opamp needs to go negative to keep it's "-" input at ground (using the "virtual ground" property of an opamp with its "+" input at ground). Does that make any sense to you?
 
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  • #3
When you make an equation from KCL at a node, you must be careful to be consistent with the polarities of the currents. Try again with the following rule: All currents flowing into a circuit node sum to zero.
 
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  • #4
isnt it that the sum of the currents flowing into the node, is the same as the sum of the currents flowing out of the node. and schouldnt the sum of both currents flow to the R3?
 
  • #5
so that means i1+i2= i3 and that is what my KCL is saying.
i set i3 as leaving the node, because there's a virtual ground so it needs to go to R3.
 
  • #6
altruan23 said:
isnt it that the sum of the currents flowing into the node, is the same as the sum of the currents flowing out of the node. and schouldnt the sum of both currents flow to the R3?
Yes, you can use "into" or "out of". But your equation has some polarities as into and others out of. You must choose one version and remain consistent.

So, that "- node" is at zero volts, which we could name ## V_{neg} \equiv 0 ## (the ideal op-amp with negative feedback assumption). So, the current flowing into the that node from ##V_1 ##, ## I_1 \equiv \frac{(V_1 - V_{neg})}{R1} = \frac{(V_1)}{R1} ##. Same for all of the other currents into the that node ##I_n \equiv \frac{(V_n - V_{neg})}{Rn} = \frac{(V_n)}{Rn} ##.

If you choose the "into" rule than any current polarity that is flowing "out of" the node (as you've defined by your arrows) has to have a - sign added to correct the polarity.

Yes, the sum of ## I_1 + I_2 = I_3 ## given the polarities you've defined. This is equivalent to ## I_1 + I_2 - I_3 = 0 ## which is KCL at the ##V_{neg}## node. ## I_1 ## and ## I_2 ## flow into the node, ## I_3 ## flows out of the node, so it gets a - sign in KCL. Also, ## I_3 \equiv \frac{(V_{neg} - V_o )}{R3} = \frac{(-V_o)}{R3} ##, again with your polarity choice.
 
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  • #7
altruan23 said:
so that means i1+i2= i3
I still prefer to write it as I1 + I2 + I3 = 0 with all of the current arrows pointing into the node. That helps me to avoid making brain-fade type of mistakes.
 
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  • #8
berkeman said:
I still prefer to write it as I1 + I2 + I3 = 0 with all of the current arrows pointing into the node. That helps me to avoid making brain-fade type of mistakes.
Me too. The problem with sign errors is that it's conceptually easy to understand, but really easy to slip up and actually do it wrong. It is really helpful to just have a rule that you always try to follow. With experience, we learn that these mistakes are easier to avoid than to find and correct.
 
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  • #9
Also, in more complicated networks it is often impossible to define all of the currents flowing into nodes. Sometimes current flows out of one node and then into another. So, you just have to be really careful at the beginning to properly account for the polarities when you make the equations. There's an identical problem with KVL, adding voltages in loops.
 
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Related to Opamp adder circuit: Where does the "-" come from?

1. Where does the "-" come from in an Opamp adder circuit?

The "-" in an Opamp adder circuit represents the inverting input of the operational amplifier. This input is used to subtract the input signals from the non-inverting input.

2. How does an Opamp adder circuit work?

An Opamp adder circuit works by using the inverting input of the operational amplifier to subtract the input signals from the non-inverting input. The output of the circuit is the sum of all the input signals, with the "-" input acting as a negative multiplier.

3. Can an Opamp adder circuit handle more than two input signals?

Yes, an Opamp adder circuit can handle multiple input signals. The number of inputs that can be added together depends on the number of input terminals on the operational amplifier.

4. What are the advantages of using an Opamp adder circuit?

An Opamp adder circuit has several advantages, including high accuracy, low cost, and the ability to handle multiple input signals. It also has a high input impedance, meaning it does not draw much current from the input sources.

5. Are there any limitations to using an Opamp adder circuit?

One limitation of an Opamp adder circuit is that it requires a dual power supply, which may not be available in all applications. It also has limited output voltage swing, meaning it may not be suitable for high-voltage applications.

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