#33 Operational Amplifiers #33 — Non-inverting op-amp gain

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

1. 1. Question and circuit

   

   Non-inverting op-amp question

   V_i → plus input

   R_f: V_o to minus input

   R₁: minus input to ground

   $\displaystyle V_{i}=6 V$

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

   Find V_o and A_v

   Narration transcript

   Here is the exact question for this lesson. We have a non-inverting op amp. The input signal V i goes directly into the plus input. The minus input is not grounded directly; instead, it sees a feedback divider made from R f and R one. First, we want the general relation between V output and V input. Then, in the example, V i is six volts, R one is one kilo ohm, R f is two kilo ohms, and the supply rails are plus and minus twenty four volts. We will find the output voltage and the voltage gain, step by step.

2. 2. Conditional virtual copy

   

   Conditional virtual copy

   $\displaystyle v_{+}=V_{i}$

   Negative feedback is required

   Operation must be linear and unsaturated

   Ideal model:

   $\displaystyle v_{-}\approx v_{+}$

   $\displaystyle v_{-}\approx V_{i}$

   This node is not virtual ground

   Narration transcript

   The central idea is almost the same as before, but with one important change. In the inverting amplifier, the plus input was grounded, so the minus node became a virtual ground. Here the plus input is not ground. It is V i. With negative feedback and an ideal op amp in its linear region, the output moves until v minus becomes almost equal to v plus. Since v plus equals V i, the minus node becomes almost equal to V i. So this time it is not virtual ground. It is more like a virtual copy of the input voltage.

3. 3. Feedback divider

   

   Feedback divider

   Ideal input current:

   $\displaystyle i_{-}=0$

   R_f and R₁ are in series from output to ground

   $\displaystyle v_{-}=\frac{R_{1}}{R_{1}+R_{f}}\cdot V_{o}$

   $\displaystyle v_{-}\approx V_{i}$

   $\displaystyle V_{i}=\frac{R_{1}}{R_{1}+R_{f}}\cdot V_{o}$

   The divider is unloaded

   Narration transcript

   Now look at the two resistors. The op amp input current is zero, so the minus input does not steal current from the divider. That means R f and R one behave like an ordinary voltage divider between V output and ground. The voltage at the middle node is the fraction R one over R one plus R f, multiplied by V output. But that same middle node is v minus, and feedback makes it approximately equal to V i. This is the whole trick.

4. 4. Gain derivation

   

   Derive the non-inverting gain

   $\displaystyle V_{i}=\frac{R_{1}}{R_{1}+R_{f}}\cdot V_{o}$

   $\displaystyle V_{o}=\frac{R_{1}+R_{f}}{R_{1} }\cdot V_{i}$

   $\displaystyle V_{o}=\left(1+\frac{R_{f}}{R_{1}}\right)V_{i}$

   $\displaystyle A_{v}=\frac{V_{o}}{V_{i}}$

   $\displaystyle A_{v}=1+\frac{R_{f}}{R_{1}}$

   Positive sign: no phase reversal

   Narration transcript

   Now solve the formula slowly. We just said that V i equals R one over R one plus R f, multiplied by V output. To isolate V output, divide by that fraction. Dividing by R one over R one plus R f is the same as multiplying by R one plus R f over R one. So V output equals R one plus R f, over R one, times V input. Then we split the fraction into one plus R f over R one. Therefore the voltage gain is A v equals one plus R f over R one. Notice the sign: it is positive. This circuit does not invert the signal.

5. 5. Numerical example

   

   Numerical example

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

   $\displaystyle A_{v}=1+2=3$

   $\displaystyle V_{o}=\left(3\right)\left(6 V\right)=18 V$

   i=18 V/(3 kΩ)=6 mA

   (6 mA)(1 kΩ)=6 V

   Divider and gain checks agree

   Narration transcript

   Now plug in the example numbers. R f over R one is two kilo ohms divided by one kilo ohm. The kilo ohm units cancel, so the ratio is two. The gain is one plus two, which is three. If V input is six volts, V output is three times six volts. Three times six is eighteen, so V output is eighteen volts. As a quick check, the total divider resistance is three kilo ohms. Eighteen volts over three kilo ohms gives six milliamps. That current through R one, one kilo ohm, gives six volts at the minus node, matching the input.

6. 6. Supply check

   

   Supply-rail check

   $\displaystyle V_{S+}=+24 V, V_{S-}=-24 V$

   Ideal demand:

   $\displaystyle V_{o}=+18 V$

   $\displaystyle -24 V<18 V<+24 V$

   The ideal rail test passes

   With ±12 V the output would clip

   Check practical output swing in the data sheet

   Narration transcript

   Before we accept the answer, check the supply rails. The example op amp is powered from plus twenty four volts and minus twenty four volts. Our calculated output is plus eighteen volts. Eighteen volts is inside the twenty four volt limit, so the ideal answer can fit. If the same circuit only had plus and minus twelve volt rails, eighteen volts would be too large and the output would clip. Here, with plus and minus twenty four volts, the answer is safe.

7. 7. Waveform

   

   Read the waveform

   $\displaystyle f_{o}=f_{i}$

   $\displaystyle A_{v}=+3$

   $\displaystyle V_{\mathrm{o,pk}}=3V_{\mathrm{i,pk}}$

   A_v>0 → same phase

   Zero crossings occur together

   6 V peak → ideal 18 V peak

   Narration transcript

   What does that mean for a sine wave? The frequency does not change. The output has the same phase as the input, because the gain is positive. The amplitude becomes three times larger because the gain is three. So when the input is positive, the output is also positive. When the input crosses zero, the output crosses zero at the same time. This is why the circuit is called non-inverting.

8. 8. Method summary

   

   Method summary

   1. Identify the input at the plus terminal

   2. Verify the conditions for v₋≈V_i

   3. Write the output divider

   $\displaystyle A_{v}=1+\frac{R_{f}}{R_{1}}$

   Here A_v=3 and V_o=18 V

   Final check: rails and practical output swing

   Narration transcript

   The non-inverting op amp is read with three ideas. First, the input goes to the plus terminal. Second, feedback makes the minus terminal almost equal to that input, not necessarily zero. Third, the two feedback resistors form a voltage divider from output to ground. From those ideas, the formula follows naturally: V output equals one plus R f over R one, times V input. For this example, the gain is three and a six volt input gives an eighteen volt output.

Source video: Circuit Theory #33 | Non-Inverting Op-Amp: Positive Gain and 6 V Example (4:55)