Please explane me why OPAMP is used in the adder circuit? why i can't add voltage without opamp? Anyway current doesn't go into opamp so it goes trought RF resistor

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This OpAmp topology adds voltages, not currents. – Nick Alexeev Apr 11 '14 at 17:20
@NickAlexeev: Actually, and fundamentally, this circuit is adding currents. The sum of the currents coming in from the left equals the current going out (through the feedback resistor) on the right. The resistors are used to convert voltages to currents. – Dave Tweed Apr 11 '14 at 17:23
@DaveTweed I know. Internally, it works with currents. Externally, V1 through V4 are voltages. – Nick Alexeev Apr 11 '14 at 17:26
For both discussion of op amp circuits -- including the "infinite-gain amplifier" theory (though he goes through that fairly fast) and real-world considerations involved in a very wide range of sample circuits, Jung's IC Op Amp Cookbook was a darned good reference. – keshlam Apr 11 '14 at 19:09

The concept of the "op-amp virtual earth" is very important to understanding why the op-amp is used as a mixing/adding circuit: -

I've stolen the OP's picture and marked in red where the virtual earth is. Anything directly connected to the -input is called "virtual earth". OK that doesn't explain why so, here goes... If the voltage at the +input is 0V (aka earth or ground) then the voltage at the -input HAS to also be 0V.

This may seem a strange thing to say until you think about the general model of an op-amp - it has infinite gain (or at least very high gain) and, if the voltage difference between the two inputs were measurable, then the OP-AMP output would be end-stopped against one of the power rails.

But, this doesn't happen because of the process of negative feedback - the OP-AMP produces a voltage on its output that is just right (goldilocks value) for making the -input voltage (via Rf) exactly the same as +input. This is called negative feedback.

So how does this help the mixer - it means that all the inputs (V1 to V4) are connected to resistors that appear to go straight to 0V - this means the current thru each input resistor is NOT dependent on the other inputs and their currents - they all appear to go to ground or earthy or 0V.

This means it is a true mixer.

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very good explanation! thank you – Rocker Apr 11 '14 at 18:06

The reason the opamp is used is that it holds the summing node (the one connected to the inverting input) at "virtual ground". This allows the input currents (flowing through R1 ... R4) to be completely independent of each other, creating a true arithmetic sum.

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why i can't add voltage without opamp?

You can!

Remove the op-amp and the feedback resistor. The voltage at the node connecting all four resistors is, by voltage division and superposition,

$$V_{sum} = V_1 \frac{R_2||R_3||R_4}{R_1 + R_2||R_3||R_4} + V_2 \frac{R_1||R_3||R_4}{R_2 + R_1||R_3||R_4} + V_3 \frac{R_1||R_2||R_4}{R_3 + R_1||R_2||R_4} + V_4 \frac{R_1||R_2||R_3}{R_4 + R_1||R_2||R_3}$$

But note that this result assumes an effective open-circuit for the load of our resistor only voltage summing circuit. Assuming we cannot ignore the load resistance, the result becomes

$$V_{sum} = V_1 \frac{R_2||R_3||R_4||R_L}{R_1 + R_2||R_3||R_4||R_L} + V_2 \frac{R_1||R_3||R_4||R_L}{R_2 + R_1||R_3||R_4||R_L} + V_3 \frac{R_1||R_2||R_4||R_L}{R_3 + R_1||R_2||R_4||R_L} + V_4 \frac{R_1||R_2||R_3||R_L}{R_4 + R_1||R_2||R_3||R_L}$$

Now, with the ideal op-amp and feedback resistor, we have, independent of the load resistance,

$$V_{OUT}= -R_F(\frac{V_1}{R_1} +\frac{V_2}{R_2} +\frac{V_3}{R_3} +\frac{V_4}{R_4} )$$

So, let me turn your question around: why would you want to add voltage without opamp?

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Strictly speaking, the circuit you provided is not an adder due to two reasons:

1) The output in your circuit is of the opposite polarity to the inputs. Thus, an inverter must be added.
2) For the circuit to be an adder, values of R1, R2, R3 and R4 (and ideally Rf) must be the same.

Please see the amended version of the adder below with formulas explaining operation:

simulate this circuit – Schematic created using CircuitLab

$${i_1} = {{{V_1}} \over R}, {i_2} = {{{V_2}} \over R}, {i_3} = {{{V_3}} \over R}$$

$$I = {i_1} + {i_2} + {i_3}$$

$$0 - {V_{out,i}} = I \times R$$ $${V_{out,i}} = - I \times R = - ({i_1} + {i_2} + {i_3}) \times R = - ({V_1} + {V_2} + {V_3})$$

So the output of the first op-amp is an inverted sum of the inputs - not exactly what we need.

Adding one more amplifier with gain of -1 and applying the formula for the gain, we get an adder:

$${A_v} = {{{V_{out}}} \over {{V_{out,i}}}} = - {r \over r} = - 1,{\rm{ }}{V_{out}} = - {V_{out,i}}$$ $${V_{out}} = {V_1} + {V_2} + {V_3}$$

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The question is really about why an op amp is needed, though -- not what the transfer function is. – Null Feb 5 at 16:09
@Null, I agree. The title of the question is "Operational amplifier adder". I thought, that showing a circuit and calling it an adder, while it is not an adder, strictly speaking, required this remark. – Sergei Gorbikov Feb 6 at 8:07