# Save 3 dB of output power using feedback to set the output impedance

**Figure 1**). Newer op amps operating on 3 and 5V have limited output swings, meaning that you should avoid using series-buildout resistors. Instead, you can use a series-feedback circuit to set the output impedance. John Wittman, then a senior staff engineer at GTE Lenkurt Electric Co, introduced this technique more than 40 years ago.

**Figure 2**). This example shows a high-side sensor and an unbalanced load. The forward amplifier is designed to have twice the needed gain when unloaded. In this example, the open-circuit gain is 2.7, and the input impedance is 1Ω. The input current is 1A, with an input signal of 1V.

_{F}then decreases to 0.5A, meaning that the output voltage is half of the open-circuit voltage. The output impedance is now 1Ω, and the series feedback is 6 dB, allowing you to match the output impedance to the load and still get almost the full voltage swing from the amplifier. You no longer waste half the output power in a series termination. This example uses a current-sense resistance value that is 3% of the output load, so power loss will be 3%. With careful design, you can lower the loss to less than 1%.

**Figure 3**). The design process is still the same, except that just two resistors can provide 6 dB of feedback.

You can formalize the analysis of the circuits using state

**equations**. For the circuit of

**Figure 2**, because the negative input of an op amp is at virtual ground, you can relate input voltage and current by inspection: I

_{IN}=(V

_{IN}− V−)/1Ω=V

_{IN}/1Ω. You can derive a second

**equation**employing the fact that the negative input of an op amp has high impedance, so the currents at that node must add to 0A.

Sum the currents at V−, except you reference the node currents back to the sense resistor, including the 0.3Ω resistor and the 0.03Ω sense resistor: 0=V

_{IN}/1Ω+V

_{OUT}/2.7Ω+0.37V

_{OUT}/ R

_{LOAD}. You can express the circuit function in vector and matrix terms: (I)=(ADMITTANCE)×(V). You can also expand for the appropriate states of current:

**equation**and solve for (V):

**Figure 2**, the forcing function is I

_{1}; the input current is 1A. Invert the admittance matrix and then multiply the current vector to find the voltage vector. You can use a Hewlett-Packard HP-48 calculator to do the hard work. It yields the result that V

_{IN}is 1V and calculates V

_{OUT}at −1.35V, one-half the unloaded gain of 2.7. You then repeat the analysis for a load resistance of 1000Ω:

_{1}of 1A, yields an open-load voltage vector, with V

_{IN}equal to 1V and V

_{OUT}equal to −2.7V, confirming that the design is correct.

Three

**equations**are used to analyze the circuit of

**Figure 3**. You can express the input current as a function of the input resistance: I

_{IN}=(V

_{IN}−V−)/R

_{IN}=V

_{IN}/ R

_{IN}. As in the previous example, you sum the currents at the amplifier's negative pin to zero, 0=V

_{IN}/R

_{IN}+V

_{OUT}/28 kΩ+(V

_{4}−V−)/900Ω, and then sum the currents at the V

_{4}node: 0=(V

_{4}− V−)/900Ω+(V

_{4}−V

_{OUT})/R

_{LOAD}+V

_{4}/20Ω. You express the currents as a vector:

**equation**determines (Y), the admittance matrix.

In this case, the input current should be 100 μA, and the load resistance should be 600Ω. Use an HP-48 calculator to invert the admittance matrix and multiply it by the current matrix. The resulting voltage matrix yields an input voltage of 1V, an output voltage of −1.4V, and a V

_{4}of −0.05V. Next, set the load to 10,000Ω. Assume that the magnetizing inductance of the transformer is infinite. You then repeat the exercise to check that the output voltage is |2.8V|.

You can match the maximum available signal power from the op amp to the load by changing the transformer turns ratio. Calculate the optimum op-amp signal output impedance to be the peak output swing voltage divided by the maximum peak capability of the op amp.

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