What is an adder circuit?

This time, we'll start with a circuit explanation! In an adder circuit, multiple input signals are connected to the inverting input terminal of an operational amplifier via resistors, and the non-inverting input terminal is connected to a reference voltage (generally GND). The output voltage Vout is returned to the inverting input terminal via resistor R2, and this configuration creates negative feedback. With negative feedback, the output is limited so that the voltage at the inverting input terminal approaches the voltage at the non-inverting input terminal (reference voltage).

As a result, the currents flowing from each input through the resistors merge before reaching the inverting input terminal, flow through the feedback resistor to the output side, and an output voltage corresponding to their sum is obtained. Because this circuit adds up multiple input signals and outputs the result, it is called an "adder circuit".

Adding circuit using OP97

When the input voltage 1 of the adder is Vin1, the input voltage 2 is Vin2, and the output voltage is Vout, the relationship between Vin1, Vin2 and Vout is as follows:

The gain of an adder circuit, like inverting and non-inverting amplifiers, depends on the values of R1 and R2, and, as with the inverting amplifier circuit, the key point is that the polarity is reversed!

Let's build an adder circuit on a breadboard!

Let's get started building the circuit! It basically has the same configuration as an inverting amplifier circuit, but there's a big difference: it accepts not only sine wave signals but also DC signals, so we need to be careful.

Let R1-1 be the resistor connected to Vin1, and R1-2 be the resistor connected to Vin2. Set the resistance values as follows: [R1-1=1kΩ, R1-2=4.7kΩ, R2=4.7kΩ, load resistance 10kΩ].

Breadboard connection example of an adder circuit using OP97

In reality. Adding circuit

Thisis the pinout forthe ADALM2000.

This time,

• 1+/1-(Oscilloscope channel 1/​ ​GND of oscilloscope channel 1)

・ 2+/2-(Oscilloscope channel 2/​ ​GND of oscilloscope channel 2)

・GND

・ V+/V-(Positive power supply / Negative power supply)

• W1 (Signal Input 1)

• W2 (Signal Input 2)

We will be using the pins shown. Please note that this time, unlike before, we will be inputting two types of signals, so the W2 pin will also be required!

Connect channel 1 of the oscilloscope to the input side and channel 2 to the output side, and then observe the waveforms of each.

Let's start Scopy and begin the setup! You can start the setup by selecting the tab on the left side of the screen.

As with the previous time, we will configure two settings: "Signal Generator" and "Power Supply". The configuration conditions are as follows:

Power supply voltage: ± 5V

• Input signal (W1)

              Frequency: 1kHz

              Amplitude: 2Vp-p

• Input signal (W2)

DC0V and DC1V

Let's examine the input and output voltage waveforms as before, paying attention to what waveforms are observed when two different signals are input, and what changes occur when the input signal (W2) is changed from DC0V to DC1V!

The input signal (W2) can be set as follows: ① Select CH2 in "Signal Generator" ②, and then ③ set the DC input from the Constant tab.

Input signal (W2) settings

Let's start by looking at the DC 0V input!

Actual measurement! Adding circuit!

Let's take a look at the waveform when a 1kHz​ ​sine wave and DC 0V are input.

Input/output voltage waveforms of DC0V input

That waveform looks familiar, doesn't it? That's right, it's exactly the same waveform we saw with an inverting amplifier circuit with a gain of-4.7!

Under the conditions R1-1=1kΩ, R1-2=4.7kΩ, and R2=4.7kΩ,

Therefore, the input is DC0V, meaning Vin2 = 0.

As a result, we were able to obtain an output voltage that was a-4.7 multiplier to the Vin1 (2Vp-p) signal. This means that when Vin2 is DC0V, the structure is the same as a simple inverting amplifier circuit for Vin1.

​LTspice waveform of a DC0V adder circuit

I was able to obtain a similar waveform using LTspice as well.

(Note: Since the OP97 component does not exist in LTspice, the evaluation was performed using the substitute OP07.)

So, how does the waveform change when we input DC1V to Vin2?

Let's change CH2 of the Signal Generator to DC1V as shown below and look at the waveform!

DC1V setting

Input/Output Voltage Waveforms of DC 1V Input

This time, we obtained a waveform with only the lower peak voltage clipped. Compared to the waveform with a DC0V input, it appears that the upper peak voltage has decreased and the lower peak voltage has increased. Because the gain is-4.7 times, the ideal swing of the output voltage with a DC0V input is ± 4.7V, but when a DC1V input is applied, the upper peak voltage becomes approximately 3.7V, a decrease of 1V.

It clearly appears to be caused by a DC input.

Let's consider this using the formula we just discussed. In this case, we are inputting DC1V, so Vin2 = 1.

This is the result we obtained. In other words, a -1V​ ​DC component was added to the Vin1​ ​sine wave (2Vp-p) signal amplified by-4.7 times, resulting in a waveform that was shifted 1V to the negative side overall.

Therefore, the upper peak voltage likely decreased from 4.7V to 3.7V, while the lower peak voltage probably increased from-4.7V to-5.7V.

When I checked the waveform in LTspice, it showed a waveform similar to the one I obtained, with a peak voltage of 3.7V at the top and increased clipping at the bottom.

Let's try changing the resistance value!

When I input DC1V, the lower peak voltage was clipped, so I couldn't confirm whether the waveform had actually shifted 1V to the negative side overall. Therefore, this time I will reduce the resistance value to lower the gain and obtain a fully swinging waveform to check the overall picture!

The resistors we will be changing this time are R1-2 and R2, which are currently 4.7kΩ, and we will be changing each of them to 1.5kΩ.

To begin with, we will check whether, when a DC​ ​0V input is received, the output voltage waveform is a sine wave (2Vp-p) of Vin1 with a gain of 1.5 times and inverted polarity.

Input/output voltage waveforms after changing to 1.5kΩ

We were able to obtain an output voltage waveform where the input signal of Vin1 was correctly inverted and amplified by 1.5 times!

Now, let's input DC1V and observe the waveform changes! If our previous analysis is correct,-1V will be added to the waveform that swings between ± 1.5V, shifting the overall waveform 1V in the negative direction. As a result, we should obtain an output voltage waveform with a peak voltage of 0.5V at the top and-2.5V at the bottom.

Input/output voltage waveforms for 1.5kΩ, DC1V input

LTspice waveform with 1.5kΩ and DC1V input.

As expected, we obtained waveforms with an upper peak voltage of 0.5V and a lower peak voltage of-2.5V!

Bonus: Let's try putting in DC2V.

As a bonus, let's try inputting 2V DC while keeping the resistance at 1.5kΩ. In principle, nothing changes, so the DC component of-2V will be added to the sine wave amplified by-4.7 times, shifting the waveform 2V in the negative direction overall, resulting in a waveform with an upper peak voltage of-0.5V and a lower peak voltage of-3.5V. Let's see the results!

Input/output voltage waveforms for 1.5kΩ, DC2V input.

LTspice waveform with 1.5kΩ and DC2V input.

Your prediction was spot on! This time, the lower end stayed within-3.5V and-4V, so there was no clipping, but if you input DC3V, you'll probably get a waveform where the lower end clips.

This concludes my review of the OP97, but I encourage everyone to try the DC3V input as well!

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