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I want to use a balanced line driver of the type DRV134 or THAT1646. With a bypass switch I want to bypass the IC and send the signal to another circuit instead. In order to null the output of the driver IC in case the switch is in the bypass position, I thought of shunting the input to ground via 10k. Is this the correct way of handling this situation? Is it the correct resistor value? The driver source is an op-amp with no resistor or cap between them.

schematic

simulate this circuit – Schematic created using CircuitLab

(the other pole of this switch switches the input of the other circuitry, respectively)

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You don't need a resistor if you don't want to use one. Inside the DRV134 the input feeds an inverting op-amp configuration and that is stable with a signal (or 0V) being applied to that input.

The THAT1646 gives no information that I could find that explains what might be happening inside the chip. I would suggest that if you want more help you show a fuller circuit diagram and indicate precisely how you would interface to both chips - they don't seem to be that compatible having read their data sheets.

The data sheet says it is pin compatible but what realistically does this mean - maybe in very common configurations they can be interchanged.

Alternatively just try it - I don't think you'll damage anything. For anyone that doubts about the chip internals, here's a picture: -

enter image description here

Note that pin 4 connects via a 30k resistor to a virtual earth formed by A1 along with two more similar circuits making the input impedance 10k.

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  • \$\begingroup\$ but in case the switch is closed (towards drv134), wouldn't the signal disappear through ground without the resistor? that's what I don't understand. note: i'm planning on using the drv134, not the that anyway :) \$\endgroup\$ – jilski Oct 22 '15 at 14:55
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    \$\begingroup\$ If the switch is closed, the voltage appearing on the output of the TL081 will also appear on the input of the DRV134 (with or without the 10k resistor) \$\endgroup\$ – Andy aka Oct 22 '15 at 14:56
  • \$\begingroup\$ I see, but I don't understand why the potential doesn't become zero between the input and ground if you don't add a resistor. Doesn't the current escape to ground without any barrier then? Thanks for clarifying! \$\endgroup\$ – jilski Oct 22 '15 at 15:33
  • \$\begingroup\$ If you simply remove the 10K resistor, and put nothing in its place, there is no route for the current to "escape" to ground (except via the normal path through the IC). \$\endgroup\$ – Peter Bennett Oct 22 '15 at 23:38
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    \$\begingroup\$ @jilski: if you remove R1, DO NOT replace it with a wire - just remove the resistor, and do not make any connection between the input (pin 4) and ground. If you add a wire between pin 4 and ground, you will short the signal to ground, and nothing will work. \$\endgroup\$ – Peter Bennett Oct 23 '15 at 18:50

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