#32 Operational Amplifiers #32 — Inverting op-amp gain

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

1. 1. Question and circuit

   

   Inverting op-amp question

   V_i → R₁ → minus input

   Plus input → ground

   R_f feeds V_o back to the minus node

   $V_{\mathrm{i}}=5 V$

   $R_{1}=1 k\Omega, R_{\mathrm{f}}=2 k\Omega$

   Find V_o and A_v

   Narration transcript

   Here is the exact question for this lesson. We have an inverting op amp. The input signal is V i. It passes through R one into the minus input node. The plus input is grounded. A feedback resistor R f brings the output back to that same minus node. First, we want the general relation between V output and V input. Then, in the example, we use V i equal to five volts, R one equal to one kilo ohm, and R f equal to two kilo ohms. We will find V output and the voltage gain, step by step.

2. 2. Conditional virtual ground

   

   Conditional virtual ground

   $v_{+}=0 V$

   Negative feedback is required

   Operation must be linear and unsaturated

   Ideal model: v₋≈v₊

   $v_{-}\approx 0 V$

   It is not a physical ground connection

   Narration transcript

   The central idea is virtual ground. Let us say it carefully, because this is where many students get lost. The plus input is physically connected to ground, so v plus is zero volts. With negative feedback and an ideal op amp in its linear region, the op amp drives its output until v minus becomes almost equal to v plus. Therefore the minus input node is approximately zero volts. It is not a wire to ground, but its voltage is almost zero. That is why we call it virtual ground.

3. 3. Current continuity

   

   Current continuity at the node

   Ideal input current: i₋=0

   R₁ current must continue through R_f

   $i=(V_{\mathrm{i}}-0)/R_{1}$

   $i=(0-V_{\mathrm{o}})/R_{\mathrm{f}}$

   $V_{\mathrm{i}}/R_{1}=-V_{\mathrm{o}}/R_{\mathrm{f}}$

   The minus sign preserves polarity

   Narration transcript

   Now look at the current. In the ideal model, no current enters the op amp input. So the current that comes through R one has only one place to go: through R f toward the output. That is why the two resistor currents are equal. We write the left current as V i minus zero, divided by R one. We write the feedback current as zero minus V output, divided by R f. Since the currents are the same, those two fractions are equal.

4. 4. Gain derivation

   

   Derive the inverting gain

   $V_{\mathrm{i}}/R_{1}=-V_{\mathrm{o}}/R_{\mathrm{f}}$

   $(R_{\mathrm{f}}/R_{1})V_{\mathrm{i}}=-V_{\mathrm{o}}$

   $V_{\mathrm{o}}=-(R_{\mathrm{f}}/R_{1})V_{\mathrm{i}}$

   $A_{\mathrm{v}}=V_{\mathrm{o}}/V_{\mathrm{i}}$

   $A_{\mathrm{v}}=-R_{\mathrm{f}}/R_{1}$

   Minus sign: 180° phase reversal

   Narration transcript

   Now solve that equation without skipping the sign. V i over R one equals negative V output over R f. Multiply both sides by R f. R f over R one times V i equals negative V output. Move the minus sign to the other side, and we get V output equals negative R f over R one times V input. The voltage gain is therefore A v equals V output over V input, which is negative R f over R one. The minus sign is the inversion.

5. 5. Numerical example

   

   Numerical example

   $R_{\mathrm{f}}/R_{1}=2 k\Omega/1 k\Omega=2$

   $A_{\mathrm{v}}=-2$

   $V_{\mathrm{o}}=(-2)(5 V)=-10 V$

   i=5 V/1 kΩ=5 mA

   (5 mA)(2 kΩ)=10 V

   Current and gain checks agree

   Narration transcript

   Now plug in the numbers from the example. R f over R one is two kilo ohms divided by one kilo ohm. The kilo units cancel, so the ratio is two. Therefore V output equals negative two times V input. If V input is five volts, then V output is negative ten volts. Equivalently, the input current is five volts divided by one kilo ohm, which is five milliamps. The same five milliamps through two kilo ohms gives ten volts across R f, and the output side must be negative.

6. 6. Supply check

   

   Supply-rail check

   $V_{S+}=+12 V, V_{S-}=-12 V$

   Ideal demand: V_o=−10 V

   −12 V<−10 V<+12 V

   The ideal rail test passes

   Exceeding a rail causes saturation and clipping

   Check practical output swing in the data sheet

   Narration transcript

   Before we accept the answer, check the supply rails. The op amp is powered from plus twelve volts and minus twelve volts. Our calculated output amplitude is ten volts. Ten volts is inside the twelve volt limit, so the ideal result is possible here. If the calculation had asked for, say, fifteen volts, the real output would clip near the rail. In this example, negative ten volts is safe.

7. 7. Waveform

   

   Read the waveform

   f_o=f_i=1 kHz

   $|A_{\mathrm{v}}|=2$

   $V_{\mathrm{o,pk}}=2V_{\mathrm{i,pk}}$

   A_v<0 → phase reversal

   Positive input gives negative output

   5 V peak → ideal 10 V peak

   Narration transcript

   What does that mean for the sine wave? The frequency does not change: it is still one kilohertz. The amplitude doubles because the magnitude of the gain is two. But the minus sign flips the waveform. So when the input is positive, the output is negative; when the input is negative, the output is positive. In short, V output is an inverted sine wave with twice the amplitude of V input.

8. 8. Method summary

   

   Method summary

   1. Verify conditional virtual ground

   2. Continue current using i_in=0

   3. Solve KCL with its sign

   $A_{\mathrm{v}}=-R_{\mathrm{f}}/R_{1}$

   Here A_v=−2 and V_o=−10 V

   Final check: rails and practical output swing

   Narration transcript

   The inverting op amp is one of the most important op amp patterns. Read it with three ideas. First, the plus input is grounded. Second, negative feedback makes the minus node a virtual ground. Third, because op amp input current is zero, the current through R one must continue through R f. From those three ideas, the formula follows naturally: V output equals negative R f over R one times V input. For this example, the gain is negative two and the output is negative ten volts for a five volt input.

Source video: Circuit Theory #32 | Inverting Op-Amp: Gain Formula and 5 V Example (4:46)