#34 Operational Amplifiers #34 — Summing amplifier and unity follower

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. Unity follower

   

   Unity voltage follower

   v_o wired directly to v₋

   Ideal negative feedback: v₋≈v₊

   $v_{+}=v_{\mathrm{i}} \Rightarrow v_{\mathrm{o}}\approx v_{\mathrm{i}}$

   $A_{\mathrm{v}}=v_{\mathrm{o}}/v_{\mathrm{i}}=1$

   High Z_in, low Z_out

   Purpose: isolate stages without loading

   Narration transcript

   Before we dive into the summing amplifier, let's quickly look at a special case of the non-inverting amplifier. The unity follower. In this configuration, the output is connected directly back to the inverting input, with no resistors in the feedback path. This means the voltage gain is exactly one. V out equals V in. It acts as a buffer, providing high input impedance and low output impedance, which is perfect for isolating stages in a circuit.

2. 2. Summing circuit

   

   Inverting summing amplifier

   V₁→R₁, V₂→R₂, V₃→R₃

   All inputs meet at the minus node

   $v_{+}=0 \Rightarrow v_{-}\approx 0$

   Ideal input current: i₋=0

   R_f returns output to the summing node

   Each input receives its own weight

   Narration transcript

   Now let's look at the summing amplifier, one of the most useful op-amp circuits. Here we have three input voltages, V1, V2, and V3, each connected through its own input resistor, R1, R2, and R3, to the inverting input of the op-amp. The feedback resistor Rf connects from the output back to this same node.

3. 3. KCL derivation

   

   General equation from KCL

   $i_{1}=V_{1}/R_{1}, i_{2}=V_{2}/R_{2}, i_{3}=V_{3}/R_{3}$

   $i_{1}+i_{2}+i_{3}=i_{\mathrm{f}}$

   $V_{1}/R_{1}+V_{2}/R_{2}+V_{3}/R_{3}=-V_{\mathrm{o}}/R_{\mathrm{f}}$

   $V_{\mathrm{o}}=-R_{\mathrm{f}}(V_{1}/R_{1}+V_{2}/R_{2}+V_{3}/R_{3})$

   Weight of input k: R_f/R_k

   Condition: negative feedback and no saturation

   Narration transcript

   Let's derive the output voltage formula using Kirchhoff's Current Law at the inverting input node. Since this is a virtual ground, the voltage here is approximately zero. The current through each input resistor is simply the input voltage divided by its resistance. By KCL, the sum of input currents equals the feedback current. Rearranging, we get: V out equals negative the sum of Rf over R1 times V1, plus Rf over R2 times V2, plus Rf over R3 times V3. Notice the negative sign, this is an inverting summing amplifier.

4. 4. Example 14.5(a)

   

   Example 14.5(a)

   $V_{1}=1 V, V_{2}=2 V, V_{3}=3 V$

   $R_{1}=500 k\Omega; R_{2}=R_{3}=R_{\mathrm{f}}=1 M\Omega$

   $R_{\mathrm{f}}/R_{1}=2, R_{\mathrm{f}}/R_{2}=R_{\mathrm{f}}/R_{3}=1$

   $Σ_{\mathrm{w}}=2(1)+1(2)+1(3)=7$

   $V_{\mathrm{o}}=-Σ_{\mathrm{w}}$

   $V_{\mathrm{o}}=-7 V$

   Narration transcript

   Let's solve Example 14.5 part a. We have V1 equals plus 1 volt, V2 equals plus 2 volts, V3 equals plus 3 volts. The resistors are R1 equals 500 kilohms, R2 and R3 each equal 1 megohm, and Rf equals 1 megohm. Substituting into our formula: V out equals negative of 1000k over 500k times 1, plus 1000k over 1000k times 2, plus 1000k over 1000k times 3. That gives us negative of 2 plus 2 plus 3, which equals negative 7 volts.

5. 5. Example 14.5(b)

   

   Example 14.5(b)

   $V_{1}=-2 V, V_{2}=3 V, V_{3}=1 V$

   $R_{1}=200 k\Omega, R_{2}=500 k\Omega, R_{3}=R_{\mathrm{f}}=1 M\Omega$

   Weights: 5, 2, 1

   $Σ_{\mathrm{w}}=5(-2)+2(3)+1=-3$

   $V_{\mathrm{o}}=-(-3 V)$

   $V_{\mathrm{o}}=+3 V$

   Narration transcript

   Now part b, with different values. V1 equals negative 2 volts, V2 equals plus 3 volts, V3 equals plus 1 volt. R1 is 200 kilohms, R2 is 500 kilohms, R3 is 1 megohm, and Rf remains 1 megohm. V out equals negative of 5 times negative 2, plus 2 times 3, plus 1 times 1. That's negative of negative 10 plus 6 plus 1, which equals negative of negative 3, giving us positive 3 volts. Notice how the output can be positive when the weighted sum of inputs is negative.

6. 6. Weighted waveforms

   

   Analog weighting and addition

   V₁ weight: R_f/R₁

   V₂ weight: R_f/R₂

   V₃ weight: R_f/R₃

   Weight first, then add

   Invert the sign at the output

   Op amp = simple analog calculator

   Narration transcript

   Here's a visual comparison of the input and output waveforms. The three input signals are each weighted by their respective gain factors and then summed with an inverted sign at the output. This is essentially what an analog computer does, it performs mathematical addition using op-amp circuits.

7. 7. Equal resistors

   

   Equal-resistor special case

   $R_{1}=R_{2}=R_{3}=R_{\mathrm{f}}$

   $R_{\mathrm{f}}/R_{1}=R_{\mathrm{f}}/R_{2}=R_{\mathrm{f}}/R_{3}=1$

   $V_{\mathrm{o}}=-(V_{1}+V_{2}+V_{3})$

   Unity-magnitude gain for each input

   Audio mixing and signal processing

   Pure inverting adder

   Narration transcript

   An important special case: when all input resistors equal the feedback resistor, R1 equals R2 equals R3 equals Rf, the output simplifies to V out equals negative the sum of V1 plus V2 plus V3. This is a pure inverting adder with unity gain on each input. This configuration is widely used in audio mixing and signal processing.

8. 8. Method summary

   

   Method summary

   Follower: A_v=1 and buffering

   Summer: weight each input by R_f/R_k

   $V_{\mathrm{o}}=-Σ_{\mathrm{k}}(R_{\mathrm{f}}/R_{\mathrm{k}})V_{\mathrm{k}}$

   Equal resistors: V_o=−Σ_k V_k

   The virtual-node rule is conditional

   Final check: supply rails and saturation

   Narration transcript

   To summarize: The summing amplifier adds multiple weighted inputs with sign inversion. The key formula is V out equals negative of Rf over R1 times V1 plus Rf over R2 times V2 plus Rf over R3 times V3. Each input has its own gain factor determined by the ratio of Rf to its input resistor. And when all resistors are equal, you get a simple inverting adder.