I would like to ask let's say I'm biasing the input of an op amp by using a simple voltage divider DC with an input Ac voltage which is placed with a capacitor. The output seems to be amplified but only the Ac signal is amplified. Why is that so? Shouldn't DC biased voltage be amplified together?
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\$\begingroup\$ Remember that the op-amp is amplifying the difference between the inputs... \$\endgroup\$– Ian BlandCommented Oct 16, 2016 at 23:35
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3\$\begingroup\$ Post a schematic or your question will probably be closed. You can do so by editing your question (click on the edit link) and then after that, use the button at the top (looks like schematic symbols) to add a schematic. \$\endgroup\$– user57037Commented Oct 16, 2016 at 23:39
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\$\begingroup\$ But the other input is connected to the ground ? I used a pspice and the stimulation show that at the +input consist of both superimposed Ac and DC and the other one is grounded with a feedback resistance \$\endgroup\$– M.LCommented Oct 16, 2016 at 23:41
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\$\begingroup\$ The inverting input is not grounded. Due to the 200K feedback resistor, and the 10uF cpacitor in the path to ground, the inverting input will be held to the same DC level as the output. \$\endgroup\$– Peter BennettCommented Oct 16, 2016 at 23:57
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\$\begingroup\$ Your schematic is kind of confusing, but ignoring the stuff connected to the + terminal the DC gain from the + terminal to the output is unity, so I wouldn't expect the DC part of whatever is at the + terminal to be amplified. \$\endgroup\$– John DCommented Oct 16, 2016 at 23:57
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1 Answer
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Just to tidy things up - (the answer has been given by Peter,Mkeith and Spehro et al.)
R1 and R2 form a voltage divider - as they are equal this means that half the supply voltage is at their mid point. R3 connects this voltage to the non inverting input (less a negligible drop for input bias).
C1 allows only AC input to the non inverting input which is offset by the 1/2 V.
The op amp is configured as a non inverting amp. The gain is determined by R4,C2 and R5. For mid range gain the impedance of the capacitor is negligible so the AC gain is 1 + R5/R4 = 21.
Looking at the DC gain the capacitors are effectively open circuits. The configuration is now a unity voltage follower. The only input is the 1/2V from the voltage divider (R1,R2). The output will be 1/2 V. The value of R5 doesn't come into the calculation and could be replaced with a wire link.
What the circuit will do is to give a DC output of 1/2V and add to that an AC signal 21 times the input AC input signal.
