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

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.

• You have at least 3 places where the residual charges may be adequate to cause popping. – analogsystemsrf Feb 27 '18 at 2:59
• Where? Why would it take that long? – donut Feb 27 '18 at 3:02
• In most cases a single 470 uF cap at the output is enough. Unless you have 32 bit audio with bass down to 2 HZ, in which case this circuit would not be good enough. – Sparky256 Feb 27 '18 at 3:37
• Please keep your "answers" to yourself or post them where they belong. As you can see, because of these non-answers, OP is already trying to initiate a discussion in the comment thread. They are not "helpful" to this website. – pipe Feb 27 '18 at 3:38
• Read: Pop & Click in Audio Amplifiers and I think you will immediately see the discussion on page 2 striking directly into the heart of at least one problem you have. – jonk Feb 27 '18 at 4:22

## 2 Answers

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.

• The headphones are an impedance in parallel with the 1k resistor, and they have a nominal impedance of 32 ohms, so wouldn't that make the time constant only 0.058s? – donut Feb 27 '18 at 3:38
• Hmm, yeah I guess so. – C_Elegans Feb 27 '18 at 3:56
• ... and it is the sudden current through the headphones that is causing the pop. – Transistor Feb 27 '18 at 7:14
• Shouldn't the discharge resistor prevent that problem? – donut Feb 27 '18 at 7:53
• Also, there is no popping when I wait at least five seconds after applying power, and then connect headphones to the amplifier output. – donut Feb 27 '18 at 8:05

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.

• A slower ramp up allows more time for the output capacitor to charge up, thus reducing the 'pop'. But when the output capacitor is far larger than necessary... – Bruce Abbott Feb 27 '18 at 9:27
• Ok I'll try 1 uF, but I'm not sure why it should theoretically work based on the datasheet because it says "1 uF or larger" and this capacitor "determines how fast the LM4808 settles to quiescent operation", so it would be a slower ramp up to the quiescent DC voltage if I'm using 10 uF. I don't understand the consequence of using a larger cap here. I'm cascading two op amps in order to de-invert the output from the first stage (it's just a requirement for my application). Why would you put decoupling caps at the output of the first stage? In addition to the second stage? – donut Feb 27 '18 at 18:03
• I'm suggesting you only use 1 stage and forget about the 2nd stage. Also use smaller capacitiance for the headphone such as 220 uF. You don't need any more than that. – Andy aka Feb 27 '18 at 18:10
• I have to use the second stage in order to de-invert the output from the first inverting summing amplifier stage. Is it a problem to not have an input capacitance to the second stage? Should there be a 1 uF input capacitor like in the first stage? ...I need a large capacitance at the output because it needs to work with very low impedance headphones with around 10 ohms nominal impedance, for a 10 Hz cut off. – donut Feb 27 '18 at 18:37
• @donut why not use a non-inverting summing amplifier? – BeB00 Jul 20 at 21:12