In a previous article, How to Derive the Summing Amplifier Transfer Function, I deduced the formula for the non-inverting summing amplifier with two signals in its input. But what if we have 3, 4 or an n number of signals? Can we add them all with one amplifier?
Theoretically, yes. Practically, it is a different story. There is a practical limit on how many signals can be summed up with one amplifier. When the number of input signals grows, each signal component in the sum decreases in value. By the end of this article you will understand why.
Figure 1
We already saw that, for a summing amplifier with two input signals (Figure 1), the transfer function is
 |
(1) |
If we need to add 3 signals, the circuit schematic looks like the one in Figure 2. What is the transfer function of this summing amplifier with 3 inputs?
Figure 2
Using the Superposition Theorem, we will first leave just V1 in this circuit. V2 and V3 are made zero, by connecting R2 and R3 to ground (Figure 3).
For an ideal Op Amp, we can consider that the input current in the non-inverting input is zero. With this assumption in mind, resistors R1, R2 and R3 make a voltage attenuator, with R2 and R3 in parallel. Therefore Vp is
 |
(2) |
where with R2 || R3 I noted the parallel value of R2 and R3.
With just the input source V1, the Op Amp output is noted with Vout1 and can be written as
 |
(3) |
or, after replacing Vp with expression (2),
 |
(4) |
Similarly, we can write Vout2 and Vout3 when the only input signals are V2 and V3 respectively.
 |
(5) |
 |
(6) |
Adding equations (4), (5), and (6) as the Superposition Theorem says, the transfer function of a non-inverting summing amplifier with 3 input signals becomes:
 |
(7) |
Now, I have to replace the parallel symbol || with the actual mathematical expression. For simplicity (at least this is how it looks to me), I will use the power of negative one rather than fractions. Therefore Vout is
 |
(8) |
What about a summing amplifier with 4 inputs or with 5? Better, let’s derive the summing amplifier transfer function for n inputs (Figure 4). That way, one can use this formula in a simulation program or a math program like Mathcad to determine the output level for a certain pattern of signals in the amplifier input.
I will use equation (8) to derive the transfer function for n inputs. Equation (8) can be written in a more convenient way, more compact, so that the n inputs will become evident.
Figure 4
We can see that, in the first term, V1 multiplies a fraction that can be written as in (9).
 |
(9) |
where k counts the number of input resistors.
With this notation, the transfer function of the summing amplifier with 3 inputs becomes
 |
(10) |
or, simplified,
 |
(11) |
and where j counts the number of signals.
Equation (11) can be easily expended to n input signals. All we have to do now is to replace the number 3 with n.
Therefore, the transfer function of the summing amplifier with n input signals becomes:
 |
(12) |
Q. E. D.
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It is easy to see that, as the number of the input signals increases, the parentheses’ value decreases (they are at the power of negative one), so the fraction decreases. As a consequence, as the number of the input signals increases, each sum component decreases. The input signals’ weight in the sum can become very small and approach the noise floor in the system. As we will see in a future article, the inverting summing amplifier is better fit for a large number of input signals.
Still, deducing the summing amplifier transfer function with n inputs is fun, isn’t it?
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Categories: Analog Design, Operational Amplifier Formulas, Summing amplifier, Superposition Theorem
oh man. and fun it is. took me awhile but i finally got it. thanks for this great article.
Very interesting the proof of this transfer function. A summing amplifier with multiple input signals can be useful in many applications. You can create an Excel spreadsheet and plug in voltage levels and play with the resistors values until you get it right.
Thanks for a job well done.
Matt
If you have n input signals in the summing amplifier, what happens with the resistive imbalance between the 2 inputs of the op amp? How do you match the input resistance for each input? And is it necessary?
Yes, you need to match the total value of the resistors in the non-inverting input with those in the inverting input. The input bias currents will generate a voltage drop on these resistors. If the total resistor values are not equal, the voltage drop will generate an offset error, which will be multiplied by the total gain of the amplifier.
For the summing amplifier with n input signals you need to make sure that
R1 || R2 || … || Rn = Rf1 || Rf2.
Easier said than done, but doable. At minimum, try to bring them as close as possible, or use an Op Amp with low input currents.