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The following circuit is of an opamp ac amplifier..

The text says:

If only ac signals are being amplified, it is often a good idea to "roll off' the gain to unity at dc, especially if the amplifier has large voltage gain, in order to reduce the effects of finite "input off- set voltage.

How exactly the capacitor help in putting the gain of amplifier to unity at dc inputs(input bias currents)..i mean..look the capacitor is attached in parallel acting as low pass filter i.e shorting all AC to ground..which means attenuated dc output voltage will pass to the inverting input of the opamp causing dc amplification(opamp action)..which is against what is said in the text...

please help....

enter image description here

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3 Answers 3

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The capacitor blocks DC therefore the only parts left (at DC) to consider are the op amp and R2. R1 is in series with an "infinite" impedance caused by C1 and is therefore not part of the DC analysis.

With only R2 present (at DC) the op-amp gain is unity.

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  • \$\begingroup\$ for unity gain...it should be a voltage follower..as R2 has some value..it means it would attenuate the dc output voltage ..which results the generation of some common mode signal(difference of signal at inputs)...which further will result in amplication....then how come gain is unity.... \$\endgroup\$
    – partykid
    Sep 6, 2015 at 11:42
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    \$\begingroup\$ "R2 would attenuate the DC output voltage". But the attenuation is R2/(input impedance) = R2/infinity = ... 0. As there is no attenuation of the feedback voltage, this is a voltage follower. \$\endgroup\$
    – user_1818839
    Sep 6, 2015 at 12:52
  • \$\begingroup\$ @Brain......great man.....you did it... \$\endgroup\$
    – partykid
    Sep 6, 2015 at 13:11
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Here is the gain formula (ideal opamp assumed):

G=1+R2/(R1+1/jwC)

As you can see, for w=0 the denominator approaches infinity and the second part of the sum disappears. Hence G=1. Note that the value of R2 plays no role since we assume (for an ideal opamP) that there is no current into the inv. input terminal (no voltage drop across R2).

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  • \$\begingroup\$ As R1 is not a part of DC analysis as Andy said...and there is no role of R2 as you said...and if there is no role of R2 there is no feedback ...which alternatively means that the circuit is unstable due to very large open loop gain of the opamp... \$\endgroup\$
    – partykid
    Sep 6, 2015 at 11:59
  • \$\begingroup\$ Two errors: (1) We have full (100%) feeedback because R2 couples back the output voltage to the inv. input without any loss (no drop across R2 because the current is zero). It is the VALUE of R2 that does not matter. But a dc path must exist! (2) If there would be no feedback (without any feedback path, R2 approaches infinity) the circuit could not been used as a fixed amplifier (offset voltage cause output saturation) - however, the circuit would NOT be unstable. Without any feedback no circuit is "unstable" but it cannot be used as desired. \$\endgroup\$
    – LvW
    Sep 6, 2015 at 14:31
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At DC, a capacitor is an open circuit. As frequency approaches infinity, a capacitor becomes a short circuit.

If you consider the above circuit at DC, the open-circuited capacitor essentially removes R1 entirely, leaving you with a non-inverting unity-gain buffer.

As the frequency increases, the capacitor tends to look more and more like a short circuit. If you replace the capacitor with a short what you're left with is a non-inverting amplifier with:

Gain = 1 + R2/R1

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