Why does it take over 4 seconds to eliminate audio popping in this circuit?

Question

I made a simple audio mixer circuit with its output going to headphones. Right now I'm only testing with a single audio signal at the input, which is music from a smartphone. I hear a pop sound after I apply power (2 x AA) to the circuit, unless power has already been applied for at least approximately five seconds. Even if I leave power applied for a minute, and then I disconnect and reconnect power, there is a pop sound. What about this circuit would take five seconds to stabilize? VCC stabilized nearly instantly when I measured on the scope.

I'm using a dual op amp IC, the LM4808 (http://www.ti.com/lit/ds/symlink/lm4808.pdf).

There are four 470 uF decoupling caps in parallel at the output (for a desired cutoff), and a 1k resistor to Ground that I included as a discharge resistor, but it doesn't seem to have any effect.

Answer 1

Here's what the data sheet says about Cb (the 10 uF capacitor in your circuit): -

Bypass Capacitor Value

Besides minimizing the input capacitor size, careful consideration should be paid to the value of the bypass capacitor, CB. Since CB determines how fast the LM4808 settles to quiescent operation, its value is critical when minimizing turn-on pops. The slower the LM4808's outputs ramp to their quiescent DC voltage (nominally 1/2 VDD), the smaller the turn-on pop. Choosing CB equal to 1.0μF or larger, will minimize turn-on pops. As discussed above, choosing Ci no larger than necessary for the desired bandwidth helps minimize clicks and pops.

So I would definitely experiment with this value and initially try 1 uF. Also, it makes little sense to use both amplifiers (not op-amps BTW) of this package - attach your headphone decoupling capacitors to the first stage. Cascading two stages might make the popping problem worse.

You should also note that having nearly 2000 uF coupling your headphones (32 ohms or thereabouts) has a high pas cut-off of only 2.5 Hz and this is too low for audio. Make it more like 220 uF and you might find the pop is reduced.

Answer 2

The RC time constant of your output stage is \$\tau=4*470\mu F * 1k = 1.88s\$, and it takes 5 time constants for the voltage to be 99% of the steady state voltage. Basically, the pop happens because it takes a couple of time constants for the output capacitor to charge up to the DC voltage of the output stage, such that when you reapply power there is no (noticeable) sharp transition.

Answer 3

Shouldn't the discharge resistor prevent that problem?

No. The discharge resistor is part of the problem.

there is no popping when I wait at least five seconds after applying power, and then connect headphones to the amplifier output.

That confirms what is happening. The output capacitors are charging up through the (unknown reference designator) resistor.

When power is applied, there is zero charge on the output caps. The opamp output snaps up to Vcc/2 volts, and now the caps start to charge up to this value through the resistor. If you put a scope across the resistor, you will see a voltage spike with an exponential decay. Connecting the phones shortens the decay time, but now some of the charging current is going through the earphone coil, causing the diaphragm to jump, producing an audible pop.

Decreasing the cap array value will decrease the amplitude of the pop, but not eliminate it. But if you want flat response down to x Hz, the output stage corner frequency should be x/2 Hz for a 1 dB voltage drop and tolerable phase shift, and x/10 Hz for "clean" audio.

High power audio amplifiers have a speaker disconnect relay driven by a time delay circuit to prevent potentially speaker-damaging pops on power up. While those amps usually have a bipolar power system that eliminates the output coupling capacitor, there is no guarantee that the two supplies will come up perfectly symmetrically, causing an output pop for a different reason.