I'm trying to design an amplifier, which amplifies line level audio to 5 Vpp, so I can sample it with a microcontroller. My plan is to do this in two steps: amplify the signal to 5 Vpp and then add a 2.5 V offset. I'm stuck at the first step.

I put together a simple inverting amplifier like so:

Op Amp Circuit

I'm using the LM324N op-amp. I'm powering it using a PC power supply (-12 V for negative rail and 12 V for positive rail). Ground, pictured in the circuit is 0 V.

I'm testing the setup with a sine wave (from my phone) as input. The amplifier works normally at higher frequencies, but the gain starts dropping at around 300 Hz. At 20 Hz, the gain is basically 1. For example:

1000 Hz:

Normal at 1000 Hz

40 Hz:

Lower gain at 40 Hz

What am I doing wrong? Is this normal behaviour for this op-amp? As I understand, op-amps start behaving abnormally at high frequencies, not at low. Is something wrong with my input?

  • \$\begingroup\$ Perhaps you should investigate what sort of low pass filter your input configuration created...and alter some values. \$\endgroup\$
    – Ecnerwal
    Feb 12, 2016 at 13:57
  • \$\begingroup\$ 1. nitpick: that's an inverting amplifier, not a non-inverting amplifier. 2. why do you think you need another step to apply 2.5V offset? Just make a potential divider for the non-inverting input to hold it at 2.5V (I assume your supply rails will be at 0V and 5V?) The inverting input will then off course float up to 2.5V when there is no signal, but the capacitor will prevent any DC leakage to the input. \$\endgroup\$ Feb 12, 2016 at 13:59
  • \$\begingroup\$ @steveverrill 1. Oops fixed it. 2. I did actually just do that. I only used PC power supply for testing. Now I'm running Op Amp from 12V(final project is going to run off a car battery), I'm getting 5V from a linear regulator and using resistive divider to get 2.5V to offset the signal. I'm actually pretty happy how this setup turned out :P \$\endgroup\$
    – snow
    Feb 12, 2016 at 15:05

3 Answers 3


If you ignore the input capacitor and look at the input impedance into your op-amp circuit from the left of the 200k resistor, the input impedance is the 200k resistor. This is because the op-amp is configured as a virtual earth amplifier.

In other words there is 200k loading your capacitor and the 3db high pass cut off point is when Xc = 200k ohms (in magnitude). This equates to: -

Frequency = \$\dfrac{1}{2\pi RC}\$ = 79.6 Hz

In other words below this frequency the signal level coming out of your amp falls at 6 dB per octave hence, at about 40 Hz the output signal will be about 6 dB down and at 20Hz the output signal will be 12 dB down.

At 20 Hz, the gain is basically 1.

Midband gain is about 5 (1 Mohm / 200kohm) and this, in decibel terms is about 14 dB hence, at 20 Hz there is a net gain of about 2 dB (1.26:1 in real numbers). Make C bigger in value.

  • 1
    \$\begingroup\$ I replaced the capacitor with 1uF and it works like charm. Thanks a bunch. \$\endgroup\$
    – snow
    Feb 12, 2016 at 11:54

The coupling capacitor C1 of 10 nF is too small. You made a high-pass filter with -3 dB frequency of 1/(2pi*RC) = 80 Hz where R is the 200 k resistor and C is the 10 nF capacitor.

If you make C1 100 nF or more, the problem will be solved.


The impedance of a capacitor is frequency dependent: \$X_C=\frac{1}{2\pi f C}\$,
i.e. \$X_C\$ is high for low frequencies and low for high frequencies.
For 1kHz \$X_{C1}\$ is 16k\$\Omega\$ (→ higher level),
for 41Hz \$X_{C1}\$ is 388k\$\Omega\$ (→ lower level).

Replacing C1 by 10µF or bigger should help. The frequency dependency of \$X_C\$ will be still there but it won't matter any more because \$X_C\$ will be 1000-times smaller and in any case (for audio frequencies) much smaller than R1.


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