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Is a current return path a good (or reasonable) design practice with an AC-coupled transimpedance amplifier?

Horowitz & Hill states that "you must provide a return to ground for the (very small) input current" and an article at analog.com on common design errors highlights this as a common design error.

It is not clear to me if a current return path is also required with an ac-coupled transimpedance amplifier. When I look at Horowitz & Hill, the ac-coupled transimpedance amplifier section of the circuit shown in Figure 8.87 does not have a current return path after the blocking capacitor. It could have been omitted for clarity because the figure shows the ac-coupled as an optional configuration.

FWIW, I am not designing a circuit. I am trying to understand some unusual behavior in some equipment and my working theory is that the blocking capacitor is building up a charge over time (as indicated in the analog.com article).

I have included a circuit diagram (revised to show the recommended resistor from the lower plate of C1 to ground).

[Revised TIA Circuit[2]

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    \$\begingroup\$ Can you show the schematic of the TIA? Maybe the feedback resistor provides a DC path for the bias current, but without seeing your actual circuit it's hard to know. \$\endgroup\$
    – John D
    Commented Jul 25, 2017 at 3:13
  • \$\begingroup\$ I have updated the question to include the circuit schematic \$\endgroup\$
    – Jim D
    Commented Jul 25, 2017 at 12:41
  • \$\begingroup\$ I believe that the signal (AC) current return path they speak of is the Earth symbols in your diagram. The DC level is maintained at ground by the negative feedback circuit. Certainly charge might build up with leaky components or other strange things but should eventually decay to zero. If you have a capacitive transducer (piezo?) it should also have some kind of DC reference/bypass either through the circuit or a HIGH value bleeder resistor. \$\endgroup\$
    – KalleMP
    Commented Jul 25, 2017 at 21:03

2 Answers 2

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Every op amp has a differential amplifier at his input.

enter image description here

And every transistor needs a "base" current to flow to work as the amplifier.

enter image description here

And this is why any op amp needs the DC path for his bias current (the inputs cannot be left floating).

In your example, the non-inverting input is connected to GND.

And the return path for inverting input current is "closed inside" the opamp.

The current will flow from the opamp input via feedback resistor into the opamp output node and to the GND via op amp output transistor.

This diagram tries to show the situation

enter image description here

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  • \$\begingroup\$ I am not sure I understand your explanation. Figure 2 in the analog.com article shows a path from C1 to ground via R1 as the correct design. Your explanation appears to suggest that path is not necessary. Is the absence of a resistor from the non-inverting input to ground the key? \$\endgroup\$
    – Jim D
    Commented Jul 25, 2017 at 13:43
  • \$\begingroup\$ @JimD In your example, a non-inverting input is directly connected to GND. Hence the bias current will have a path to GND. As for the inverting input, the feedback resistor (R2) and the op amp output will provide the path for the bias current for inverting input. So, no extra resistor is needed. But you can add a resistor from C1 "lower" plate to ground, to "set" DC condition for C1 capacitor. With this additional resistor C1 capacitor (the input) is no longer floating. \$\endgroup\$
    – G36
    Commented Jul 25, 2017 at 13:55
  • \$\begingroup\$ So any charge buildup on C1 will drain to ground without propagating to the ADC (see amended circuit diagram). I also included the addition of a resistor (R5) you recommended. \$\endgroup\$
    – Jim D
    Commented Jul 25, 2017 at 15:21
  • \$\begingroup\$ What type of a op amp will you be using? \$\endgroup\$
    – G36
    Commented Jul 25, 2017 at 15:36
  • \$\begingroup\$ A THS3001 link \$\endgroup\$
    – Jim D
    Commented Jul 26, 2017 at 2:10
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Yes----you need a resistor to the Vi(+)

enter image description here

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  • \$\begingroup\$ Can you add some details? I don't see the resistor you mention in the diagram you provided. \$\endgroup\$
    – JRE
    Commented Jul 25, 2017 at 3:59
  • \$\begingroup\$ You have to add that resistor. From Vin+ to GND. \$\endgroup\$ Commented Jul 25, 2017 at 4:14
  • \$\begingroup\$ That's not a transimpedance amplifier, so this answer doesn't apply. \$\endgroup\$
    – John D
    Commented Jul 26, 2017 at 15:14
  • \$\begingroup\$ The schematic shows a circuit with no-DC-path, hence the value in answering the question. \$\endgroup\$ Commented Jul 26, 2017 at 16:22
  • \$\begingroup\$ The OP's schematic shows a circuit with a clear DC path through the feedback resistor. That's all that's needed for the bias current. \$\endgroup\$
    – John D
    Commented Jul 26, 2017 at 20:08

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