Op-amp circuit has lower gain at low frequencies

Question

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

I put together a simple inverting amplifier like so:

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:

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.