I'm new to electronics and trying to get a grasp of the basics at the moment.

For this purpose, I'm reading the book Make: Electronics by Charles Pratt (in a German translation.)

In one experiment, he's building an astable multivibrator and uses its signal (boosted by another transistor) to light an LED, which is later replaced by a speaker (see picture below.)

I don't understand the use of the capacitor here. The book says it's there to block DC (and let AC pass, my addition,) but the multivibrator produces DC on/off signal and no AC signals anyway. In other words, I don't get what it is actually filtering away, since its other main feature (current storage function) is certainly not of great use here.

From: Pratt, Charles: Make: Electronics, 2nd edition (german), p. 126

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    \$\begingroup\$ A "DC on/off signal" is an alternating signal, and the AC component is precisely its variation as it turns on and off. \$\endgroup\$
    – nanofarad
    Aug 4, 2022 at 14:54
  • \$\begingroup\$ @nanofarad I thought AC was (exclusively) current changing its direction (due to changing polarity)... OK then it makes sense. So, if I would omit the capacitor, the speaker would generate a sequence of high and low noises, because the AC signals are not filtered away anymore, right? \$\endgroup\$ Aug 4, 2022 at 15:14
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    \$\begingroup\$ AC can be superimposed on a DC offset, which is basically what this is. The cap removes the DC component. \$\endgroup\$ Aug 4, 2022 at 15:17
  • \$\begingroup\$ Here, you have a sum of a DC signal (the offset) and AC signal (the variation). \$\endgroup\$
    – nanofarad
    Aug 4, 2022 at 15:17
  • \$\begingroup\$ Does this answer your question? Why place capacitors in front of the line to a (headphone) audio speaker? \$\endgroup\$
    – JYelton
    Aug 8, 2022 at 16:10

1 Answer 1


There is a lot that is "wrong" with the above image. Just note that while it may work, it is not ideal.

What's the use of the capacitor before the speaker here?

You'll see caps used commonly with speakers. That is because speakers are true AC devices; putting +10V on the coil moves it one way, and putting -10V on it moves it in the opposite direction.

The problem this cap solves, is that the DC power source being delivered to the speaker is only ever 0V or +V, not -V and +V. There is no negative voltage, so the speaker would only ever work half as well as it could.

Adding the capacitor "removes" this 0V bias, or in other words, allows what gets to the speaker to "shift" to an average value, which over time will end up being -V to +V (whatever the particular number of volts happens to be.)

If 0V to 10V were used to drive the speaker (no capacitor), it would sound bad and not be very loud. With the capacitor added it would see about -5V to +5V, but sound nicer and be much louder. So capacitors are commonly used in basic speaker circuits to give them AC from a pulsed DC source.

  • \$\begingroup\$ So what happens is, if I understand correctly: as long as there is a signal/current coming from the multivibrator, the cap is charging and the voltage at the speaker is positive. Once the signal ends, the cap is uncharging via the resistor and the voltage at the speaker is becoming negative / the flow of current is changing direction \$\endgroup\$ Aug 4, 2022 at 15:26
  • \$\begingroup\$ @Oidipous_REXX pretty much. Note that you want the capacitor to be so big that it never finishes charging and never finishes discharging - you want it to stay in the middle. And this average charge level where it stays around - is average voltage of the sound wave, and cancels it out. So that way the speaker always sees an average voltage of 0 i.e. it sees equal amounts of positive and negative voltages \$\endgroup\$
    – user253751
    Aug 4, 2022 at 16:10
  • \$\begingroup\$ For more background on Capacitors, try Kuphaldt's Lessons in Electric Circuits. \$\endgroup\$
    – rdtsc
    Aug 4, 2022 at 18:55
  • \$\begingroup\$ Could the DC bias cause the loudspeaker coil to overheat? \$\endgroup\$ Aug 4, 2022 at 18:56
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    \$\begingroup\$ @rdtsc After reading some pages, I think this book is really great, because it explains things well to newbies - many thanks for the hint! In support of Pratt's book and his (intentionally) simplified circuit drawings: his purpose is to teach by demonstration and experiments, so some of his buildups are necessarily not perfec/well done. Given some explanations could really be more extensive, I think it is still a good book for beginners. \$\endgroup\$ Aug 5, 2022 at 6:57

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