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I would like to know how I could practically measure the input impedance of the following circuit:

enter image description here


I update my post by adding the rest of the circuit.

enter image description here

This is what my circuit looks like. The aim of my test is to measure a differential input impedance 10 kohms ±10% in the bandwidth (300-2000 Hz).

I hope that is clear enough for you so that you could help me.

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  • \$\begingroup\$ Welcome! At DC? Multimeter. \$\endgroup\$
    – winny
    May 30, 2022 at 12:05
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    \$\begingroup\$ Is this the sole circuit ? is the output impedance large compared to the rest of the component values ? \$\endgroup\$
    – citizen
    May 30, 2022 at 12:05
  • \$\begingroup\$ @winny , at AC not DC. It is actually an input impedance of an audio channel including an anti-aliaising filter. I have to make sure that the input impedance (represented by the three resistances in the previous picture) of the audio channel is a 10Kohms. \$\endgroup\$
    – Y.OU
    May 30, 2022 at 12:13
  • \$\begingroup\$ Thanks. Where are the DSP_TEST_ANA_* points connected? \$\endgroup\$
    – devnull
    May 30, 2022 at 12:28
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    \$\begingroup\$ @Y.OU - Hi, On Stack Exchange the procedure is for someone who asks a question (the "OP") to edit the question to add any new information. The only time that an OP would write an answer below is if they have solved the problem themselves, no more help is needed, and the question can be closed. Therefore I moved your extra info from an "answer" to become an update in the question. Please see the tour and the help center for more rules and etiquette details. Thanks. \$\endgroup\$
    – SamGibson
    May 30, 2022 at 12:29

1 Answer 1

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The usual way to measure an input impedance is to drive the input with a signal, in series with two different known impedances, and measure the change in output level when changing drive impedance.

The most convenient impedances to use are zero, and one in the ballpark of your expected input impedance. If the output halves when going to the second impedance, then your external impedance is equal to your internal one. Otherwise you just have to write and solve the equation for voltage division into the input with arbitrary impedances.

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  • \$\begingroup\$ I don't understand why exactly two different known impedances?! when making some researches, i have seen methods using one variable resistor . What is the difference between this method and yours? \$\endgroup\$
    – Y.OU
    May 30, 2022 at 16:23
  • \$\begingroup\$ @Y.OU One variable resistor is a way of switching between two different known impedances with one component. I'll bet the method said 'turn the resistor down to zero, measure output, increase resistor until output halves, remove resistor and measure it'. Am I right? Alternatively if you know the gain accurately, you can do it in one further measurement (knowin the gain accurately substitutes for the first reference measurement) \$\endgroup\$
    – Neil_UK
    May 30, 2022 at 16:43
  • \$\begingroup\$ Yes you're right. Which gain you mean ? \$\endgroup\$
    – Y.OU
    May 30, 2022 at 16:53
  • \$\begingroup\$ @Y.OU the gain of the amplifier from the input to the output, so you don't need to measure the reference output. (knowin the gain accurately substitutes for the first reference measurement) \$\endgroup\$
    – Neil_UK
    May 30, 2022 at 18:32

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