The use of electrolytic capacitors as coupling capacitors in audio circuits

Discussion in 'Homework Help' started by Mango Symphony, Sep 22, 2011.

  1. Mango Symphony

    Thread Starter New Member

    Mar 1, 2011
    I have been studying circuit diagrams for simple op-amp based audio amplifiers. Many times I have seen an electrolytic capacitor used as a coupling capacitor in a circuit where neither end of it appears to have a higher potential than the other. All textbooks mention that electrolytics must be biased correctly i.e. that one end must carry the higher potential, to avoid damage to the capacitor.

    So why then, in a circuit where it is unknown what potentials will exist at either end, is it acceptable to use a polarised component in this way? Surely the audio signal, which is charging and discharging this capacitor, spends as much time positive as it does negative, and if this is the case, no electrolytic capacitor would be 'happy' subjected to these constant reversals of potential across it?

    Senior Member

    Jun 29, 2010
    If you will connect the base of transistor directly to input signal.
    that mean you are shorting the base directly to Vcc and Gnd.
    Which is not right condition for transistors in active region as they have supply in midle of Vcc and Gnd ( Swing).
    The capacitors act as an insulators for DC( i think)
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
    From the DC point of view we always connect electrolytic capacitor properly.
    And you must remember that when we have AC signal in steady state condition, then the average current which flow through capacitor is equal 0A.
    Or if you look form Xc point of view, you will see that AC signal create very small voltage drop across Xc.
    And also kept in mind that electrolytic capacitor can easily tolerate small amount of the reverse-bias voltage across capacitor.
    Only when reverse-bias voltage will be larger then 1V the capacitor will be destroyed.
    Last edited: Sep 22, 2011
    endolith likes this.
  4. steveb

    Senior Member

    Jul 3, 2008
    Can you show an example? We really need to see the particular example to decide if a mistake was made or if the approach represents a sound design choice. In such cases, it might be better to use the old trick of placing two electrolytic caps in series plus to plus, with the two negative terminals going to the circuit and the input source.
    Last edited: Sep 22, 2011
  5. MrChips


    Oct 2, 2009
    Electrolytics are generally acceptable in coupling low voltage AC signals because the average DC bias is low. Always try to place the +ve end of the cap to the signal that you think is more positive and you will be ok. If in doubt you can always use a voltmeter to measure the average DC voltage across the capacitor.
  6. Keith Park_1470879252

    New Member

    Aug 10, 2016
    Even though this post has been very old. I'd like to mention something.

    The reason why electrolytic capacitors are used for bypass(coupling capacitor) is due to positive offset vias.
    Assume that the input signal swings between -1.1v and +1.1v (2.2Vp-p, or 0.775Vrms : standard line level 0db)
    And, +9v and GND are supplied for an opamp. The opamp is not able to handle minus voltage input.
    Therefore, signal should be offset to +4.5 (9V / 2) for maximum headroom. Then, signal will swing between +3.4v and 5.6v (4.5v +- 1.1v)
    To acheve offset vias that I mentioned above, coupling capacitor should be used.
    Since we know that input signal is always lower voltage than coupled signal(offset), we don't have to use bipolar capacitor anymore.
    Typically, electrolytic caps have higher capacity compare to its size.

    There will be the exactly opposite operation at the output stage.
  7. MrAl

    Distinguished Member

    Jun 17, 2014

    Yes, interesting observation by the OP.

    Most of the time we have the electrolytic in a place in the circuit where it will have a positive DC voltage at its positive terminal and less positive at its negative terminal.
    For example the output of a low power amplifier, where the (+) terminal is connected to the output of the amp and the (-) terminal is connected to the speaker (+) terminal and the speaker (-) terminal is connected to the negative supply, and also the output of the amp is biased to 1/2 of Vcc for zero input signal. But what if we have a dual supply and we bias the output of the amp to 0v and connect the speaker (-) terminal to 0v also. That means when the amp is first turned on the cap has 0v across it, then charges positively for a positive input, then starts to discharge as the signal goes through zero, then goes negative, then the cycle repeats.
    For a 1kHz sine and 1000uf cap and 10 ohm resistor, the voltage across the cap goes up to about +32mv then as the exponential dies out the positive half cycles go lower and lower until it dies out completely and then we see a sine about +16mv down to -16mv for a 1v peak AC amplifier output. So the voltage across the cap does go plus and minus, and so does the current through the cap at about plus and minus 100ma. For a 10v signal it would be around plus and minus 160mv, and for a 100v signal it would be around plus and minus 1.6v which is getting up there but that would be a high power amp which probably would not use an electrolytic cap.
    If we used a 100uf cap the voltage across the cap would be about 10 times higher, so we see that there may be times when an electrolytic is not a good option.
    For higher power a bridged output is better which does not need any cap coupling. A bridged output also produces more power per supply volt than a single ended output, so there are advantages.