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So I was reading the datasheet of TI's NE5532 and I noticed something that I don't quite understand:

Chapter 7.5 (page 5) says that the NE5532 has an input resistance (I guess input impedance?) of 30kOhm to 300kOhm. I don't really understand the internals of opamps but my initial thought was that this input resistance seems rather low for an opamp (most ones I've seen are 1MOhm - 1GOhm). From what I know opamps usually have a relatively high input impedance so that they do not affect the input signal by very much.

Chapter 9.1 (page 8) then shows an example circuit of a single ended to balanced converter (with bias) that looks like this: enter image description here

As far as I understand 9.1.2.2 (page 9) then mentions that this particular circuit used 36kOhm (+-2%) resistors (R1-R4).

I have to questions which relate to that:

  • Doesn't the low input resistance of the NE5532 totally screw up the "voltage dividers" that are used to create the inverter and the bias voltage with these particular resistor values? The opamp's input resistance isn't even 10 times as large as the input impedance.
  • What exactly does input resistance mean? I understand how resistor networks and Kirchhoff's law work, but if I look at the opamp's schematic in 8.2 I don't really understand anything: What is this (possibly virtual) input resistor connected to? How does it affect my circuit and how can circuits like above work properly with such low input resistance? If the opamp would have a an input resistance, which is a magnitude higher than the impedance of the input signal, I wouldn't worry, but like that I totally don't know how things work.
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The input resistance in the NE5532 is the differential input resistance and not the resistance from either input pin to a supply rail. That's why it barely affects the circuit because, the inputs are generally at the same voltage - negative feedback ensures that. If you want to know what effect the common mode impedances have on the target circuit you will have to make do with bias current specifications.

Modern chips will usually specify both differential and common-mode input impedances but the NE5532 isn't a modern chip in any sense of the word.

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  • \$\begingroup\$ Could you please tell me some modern alternative to NE5532? single supply is better, but I'm also courious about splited one. \$\endgroup\$ Jun 2 at 7:01
  • \$\begingroup\$ An opamp is chosen to suit the target circuit. There isn't an easy route for alternatives based on data sheet comparisons. \$\endgroup\$
    – Andy aka
    Jun 2 at 7:30
  • \$\begingroup\$ Ok you are strict, ;) but if you caoud take a look at this Q, I will be gratefull. electronics.stackexchange.com/questions/622002/… \$\endgroup\$ Jun 2 at 8:29
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I'm pretty sure you're looking at \$r_I\$. From he data sheet: "All characteristics are measured under open-loop conditions, with zero common-mode input voltage, unless otherwise specified."

\$r_I\$ represents the effective resistance between the positive and negative inputs (\$V_-\$ and \$V_+\$), for the amplifier in open-loop conditions.

When you wrap them with feedback, op-amps act to keep the difference between \$V_-\$ and \$V_+\$ as small as they can. This, in turn, raises the effective impedance.

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Yes the NE5532 and NE5534 input impedance is rather low. This is a consequence of the input stage design being tuned for relatively low noise (much lower than other opamps of its era).

This low resistance doesn't materially affect the voltage dividers around it, because if you choose the NE5532, you are concerned about Johnson noise, therefore you will be choosing relatively low values for them.

The input noise of this opamp is about the same as the Johnson noise of a 600 ohm resistor; choosing resistors more than about 10x that value throws away the biggest advantage of the opamp. (Though you may also choose it for its reasonable bandwidth or relatively high output drive capability at low distortion).

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