Save 3 dB of output power using feedback to set the output impedance
You can use this with high-side or low-side output current sensing.
Vic Jordan, Carson City, NV; Edited by Paul Rako and Fran Granville -- EDN, October 6, 2011
It’s common practice to series-terminate an op amp to match the impedance of the load. This practice causes a 3-dB loss of output power in the termination resistance, however (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.
With this technique, you add 6 dB of
the reciprocal feedback to set the output
impedance, obtaining a return loss of
more than 30 dB. You add a series-current-sensing resistor, another op amp,
and a limiting resistor (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.
To match the amplifier’s 1Ω load,
the series-feedback circuitry must divert one-half of the input current from the
negative input node of the op amp. The
original 1A input current that flows
through RF 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%.In telecommunications lines, for
longitudinal balance, the impedance
of both conductors to ground
should be the same. Longitudinal balance
protects against crosstalk and
60-Hz-induced noise. It is also important
at the higher frequencies that DSL
(digital-subscriber-line) service uses.
Telecom companies commonly use
transformers to provide longitudinal
balance of 80 to 120 dB. The transformers
also isolate transients, such as
those due to lightning. You can apply
this technique using transformer coupling
and low-side current sensing (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: IIN=(VIN−
V−)/1Ω=VIN/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=VIN/1Ω+VOUT/2.7Ω+0.37VOUT/ RLOAD. You can express the circuit function in vector and matrix terms: (I)=(ADMITTANCE)×(V). You can also expand for the appropriate states of current:
Be careful writing your own equations;
two dependent equations can
easily lead to an incorrect answer. An
HP-48 calculator solves them using the
“least-squares” method, but it does not
check for a determinant condition of
zero to warn you of nonindependent
equations. You can use the HP-48 to
sum two real matrices to form a complex
one. This approach comes in handy
when you include reactive elements in
the circuit models. If you prefer using a
computer rather than a paper napkin,
you can also use Spice to analyze this
circuit.Three equations are used to analyze the circuit of Figure 3. You can express the input current as a function of the input resistance: IIN=(VIN−V−)/RIN=VIN/ RIN. As in the previous example, you sum the currents at the amplifier's negative pin to zero, 0=VIN/RIN+VOUT/28 kΩ+(V4−V−)/900Ω, and then sum the currents at the V4 node: 0=(V4 − V−)/900Ω+(V4−VOUT)/RLOAD+V4/20Ω. You express the currents as a vector:
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 V4 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.
Talkback
-
The circuitry Fig.2 as shown could not work due to missing resistor divider at non inverting input of the second amplifier. The goal is to measure voltage drop across resistor 0.03 Ohm, so you need a differential amplifier. It is obvious. Two resistors usually the same values as 100 Ohm and 370 Ohm must be added.
2. With my all respect to author and EDN I would like to defend college professors and I am pretty sure that 10 minutes Spice simulation in advance could help the author to detect the problem and verify equations.
3. The author did suggest a good alternative idea, but as we can see, the publication does not meet expectation.
Vladimir Doubovis - 2011-12-11 09:25:45 PST -
I apologize to Vladimir for misleading him. The author of the Design Idea normalized the resistor values to 1 ohm to show the principles involved. I considered changing them to 50 or 600 ohms, to make the circuit ready for a real application, but in the end, it is risky to tell the art department to make changes in the schematic and then try to match the changes in the edits text. So, no, it is not intended to have 1 ohm input impedance, just multiply all the values by 50 or 600 or 75 or whatever common impedance you need to use. You can tell that this is a valid design since it is not a Spice experiment from a college professor, but a real circuit used by a working engineer. Once again, sorry to mislead you with the normalized values. With real values, it is easy to find and op-amp that will work. I should have made this clear in the text and I did not.
Paul Rako - 2011-4-11 00:28:50 PDT -
1.The article looks a very misleading and goals do not achieved.
2.To terminate input needs to have a source with matching to Rin impedance. But it could be possible only if you will use an ideal Op Amp in Fig.1. In other cases the potential of pin 2 of Op Amp will be not equal to zero and termination does not work.
3. Needs also consider huge power loses in Rf resistor, almost 2.7W for Fig.1 and Fig.2 and in Op Amps. It is very interesting what kind of Op Amp was used in experiment.
4.For schematics shown on Fig.2 Op Amp output due very low values of resistors Rin and Rf should drive about 2A current (about 1A through resistor Rf and 1A through load).
5.It is not required to use HP-48 calculator and to write down matrix equations that probably not correct to understand that circuitry on Fig.2 does not work for shown components values. In simple words, the second Op Amp has a total gain about +1 and as a result of 0.3 resistor from its output to inverting input of the first Op Amp the output voltage at 1 Ohm load will be only about 1/4 from input voltage.
Vladimir Doubovis - 2011-1-11 12:58:42 PDT -
Excellent work, for alternative ideas google: "Positive Feedback Terminates Cables".
Jerry Steele - 2011-6-10 05:11:54 PDT
