I have a Panasonic RX-ES29 boombox with a type C plug (ungrounded). I sometimes hear a relatively loud snap/pop sound in the right speaker of my headphones while I completely plug them into the headphones jack of the boombox when it's on. There's a little buzz when the tip of the jack is inserted but the loud pop comes when the jack is completely inserted. My headphones are Audio Technica ATH-AVC500.

I tried to capture that sound using a male to male cable, but after playing the captured one, it's not as loud as the one I hear by physically connecting headphones to it. You can see the pictures I took from Audacity here and there.

enter image description here

enter image description here

Audacity doesn't detect it as an audio clipping, but it looks like it is by the pictures that I linked above.

One of the strange things about it is that it does not always occur, I mean it's not always there; like sometimes when something discharges near your finger. I tried to unplug and plug several times just after hearing the loud sound but it's gone, exactly like when a charged object is discharged and it needs some time to charge up again.

I'm very curious to know an explanation for this weird behavior.

  • 13
    \$\begingroup\$ Panasonic seems to have absolved themselves of complaint for their design shortcoming. Manual says: Reduce volume before connection. Plug type: 3.5 mm stereo \$\endgroup\$
    – glen_geek
    Aug 10, 2017 at 19:45
  • 1
    \$\begingroup\$ The reason why the captured sound is not as loud is that you reach the limit of the recording device. \$\endgroup\$
    – yo'
    Aug 11, 2017 at 9:09
  • 1
    \$\begingroup\$ @yo' Actually, I captured it as a microphone because my computer doesn't have a Line In. \$\endgroup\$
    – Sepp A
    Aug 11, 2017 at 20:08

4 Answers 4


That's a "DC pop". A sound engineer dies somewhere every time you make one!


simulate this circuit – Schematic created using CircuitLab

Figure 1. Amplifier output and disconnected speaker.

Amplifiers fed from a single-rail supply hold their output at 1/2 supply when quiet. If the speaker was connected directly to the amplifier output there would be a DC current flowing through it continuously, heating it up and biasing the cone off centre. It would have limited travel in that direction and, so, would clip and distort much earlier than it should.

By adding a de-coupling capacitor the DC is blocked but the AC can pass through.

If the amplifier is switched on without the speaker connected as shown in Figure 1 the left side of C1 will be pulled to V+/2. The right side will follow suit. When the speaker is plugged in a current will flow to ground until the right side reaches zero volts.

A similar result will occur when plugging an audio source into a device with a DC blocking capacitor on the input.

Note also that if the devices are disconnected while powered up the capacitor may discharge back to power-up conditions due to capacitor leakage or PCB leakage.


simulate this circuit

Figure 2. Adding a discharge resistor.

As noted in the comments, we can add a discharge resistor to the circuit so that after some delay the right side of C1 has fallen to 0 V. Since this resistor will provide additional loading to the circuit we don't want to make it any lower than necessary.

One approach to finding a solution would be to decide how long we can wait for discharge.

enter image description here

Figure 3. Capacitor discharge curve. The capacitor discharges by 63% per RC time period.

A handy rule of thumb is that the "time constant" of an RC circuit is given by multiplying R and C. \$ \tau = RC \$. After \$ \tau \$ the voltage will have decayed by 63%. After \$ 3\tau \$ it will have decayed by 95% and \$ 5\tau \$, 99%.

Let's say we want to incorporate a resistor that will discharge by 99% in 1s: \$ 5 \tau = 1\;s \$ so \$ \tau = 0.2 \;s\$. In our example C1 is 470 µF so \$ R = \frac {\tau}{C} = \frac {0.2}{470\mu} = 420 \; \Omega \$. This is more than ten times the speaker impedance so it won't load the circuit too much.

  • 31
    \$\begingroup\$ +1 for "A sound engineer dies somewhere every time you make one!" \$\endgroup\$
    – Trevor_G
    Aug 10, 2017 at 19:36
  • 2
    \$\begingroup\$ @SeppA Add a high-value resistor (like 1000 ohm) on the output jack from signal to ground. Might help keep the sound engineer population alive. \$\endgroup\$
    – glen_geek
    Aug 10, 2017 at 19:55
  • 6
    \$\begingroup\$ I was thinking "add a resistor" but I'm sure there is an unpopulated resistor footprint on the output, because a bean counter looked over the schematic and asked "we sure won't pay for that!" \$\endgroup\$
    – bobflux
    Aug 10, 2017 at 20:23
  • 2
    \$\begingroup\$ @CapnJack Resistor value isn't critical...small enough to prevent the coupling capacitor from self-discharging...but much larger than the headphone impedance, so it doesn't suck a lot of power from the final amplifier. \$\endgroup\$
    – glen_geek
    Aug 11, 2017 at 15:02
  • 2
    \$\begingroup\$ @CapnJack That output coupling capacitor will be electrolytic, perhaps 470uf...an imperfection is self-discharge. A rough model would add a parallel large resistor to account for its self-discharge, perhaps 100k. The pop gets attenuated by the voltage-divider ratio. A 100 ohm resistor across the jack would attenuate a pop 999 times (a 1k resistor only 99 times). \$\endgroup\$
    – glen_geek
    Aug 11, 2017 at 15:45

"Exactly like when a charged object is discharged and it needs some time to charge up again."

That is exactly what you are hearing. The output is AC-coupled through a capacitor to the headphones to block any DC component of the signal. With the headphones first unplugged the capacitor had one side open circuit so it stays discharged.

When you plug in the headphones, the capacitor needs to charge to whatever the DC voltage is before it... and POP.. you hear that current as it energizes the speakers in the headphones.

Unplug and plug again, the capacitor retains that charge a long time.

Note the same effect can be heard through the speakers when first turning on an amp if special measures are not taken to limit it.

This can be significantly corrected by adding a largish resistor to ground across the outputs of the amp on the right of the cap.

ADD By the way, the reason you only hear it on one speaker is likely to do with the way jack plugs work. The right speaker is likely attached to the tip of the plug, as such, as you plug it in it first makes contact with the left channel, charging that cap, then makes contact with its own channel charging that, too. By the time the left speaker contact makes its connection, its output capacitor is already charged.

  • \$\begingroup\$ +1 for adding a largish resistor to ground across the outputs of the amp on the right of the cap. \$\endgroup\$
    – TimB
    Aug 10, 2017 at 22:10
  • 1
    \$\begingroup\$ Largish is a good technical term, simply meaning between > and >>. So, with an 8 ohm speaker, 80 - 800 ohm resistor. Given it is a discharge circuit, whatever is lying around would work ok. Audiophiles would calculate it... \$\endgroup\$
    – MikeP
    Aug 11, 2017 at 20:59
  • 1
    \$\begingroup\$ @MikeP - in my experience, audiophiles would be unlikely to have an amp that runs off a single supply rail, and therefore have a very small DC component, so probably wouldn't bother. I mean, adding that resistor in there turns the whole thing into a high-pass filter anyway, which means some of their signal is being lost! \$\endgroup\$
    – Jules
    Aug 11, 2017 at 22:40
  • \$\begingroup\$ @Jules - "Audiophiles" (or perhaps "audiophools" as the case may be) would probably object to having a cap in there at all. Thus, no high-pass and no "loss". But that also means that their speakers can receive DC offsets that don't do squat except heat up the voice coils. "Power compression" is when a speaker heats up, which increases resistance, and therefore draws less from a voltage-source amplifier for the same setting. \$\endgroup\$
    – AaronD
    Aug 12, 2017 at 1:48
  • 1
    \$\begingroup\$ @Trevor: The standard as I know it is Tip=Left, Ring=Right, and Sleeve=Ground/Return. So it would be the left channel per the standard that charges both caps. However, it's possible that the OP's wiring is crossed somewhere. It's hard to tell that without a test track or other known recording; the sound stage is simply mirrored and still makes sense. \$\endgroup\$
    – AaronD
    Aug 12, 2017 at 1:51

@Transistor is right, however there is a tiny subtlety here...

Jack connectors short everything to ground when you plug them in. When you slide it in, the various metal bits of the jack and the plug make contact in every possible combination before the jack finally settles into position.

There's a little buzz when the tip of the jack is inserted

I would guess that in this position, both channels are shorted together, and the opamps on both outputs go in and out of short circuit protection, or misbehave in other weird ways that cause this buzz.

When they go into protection, most likely the output goes to ground, so the cap ends up discharged... or maybe they clip to the rail, who knows.

And when the jack slides into place, both opamps come out of protection and snap their outputs to Vcc/2, therefore you hear something loud.

The weird thing is that your pop is the wrong polarity. It should be positive. Either your soundcard inverts polarity, or what I said above is true, and there is some weird clipping/instability going on as the outputs are shorted during insertion.

  • \$\begingroup\$ I captured it as a microphone, not a Line In and the cable that I used was very short and not a high-quality one. Maybe these have something to do with the weirdness you described. \$\endgroup\$
    – Sepp A
    Aug 12, 2017 at 11:24

A common problem I have found with earlier amplifiers is the large DC offset at the headphone jack on amplifier switch on. This was due to the capacitor coupled output stages on amps that only has a single rail for the power amplifier. AS a result the output is floating above ground and has to be blocked by a fairly large value capacitor. Amplifiers like the great Marantz 1060 is one suce amplifier that has destroyed a few headphones of mine in the past including more recently a pair of AKG K550's. The capacitor used to block the DC is fine for the loudspeakers that are anything as low as 4 ohm and rated in the tens of watts. But this DC into a modern fragile headphone voice coil can result in a burnt out driver. The only easy was s mentioned here, is to block the DC post the headphone attenuator with a capacitor of around 220-470uF. At this value there should be very little noticeable low frequency rolloff into a headphone >16 Ohms.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.