2
\$\begingroup\$

I have to use a LM384 amp circuit (shown below) for a school project and need to explain how it works.

I've read the datasheet (http://www.ti.com/lit/ds/symlink/lm384.pdf) and thus have a guess about some components but others remain a mystery.

  • Vin potentiometer is only a voltage divider to adjust input voltage
  • .1uF capacitor (above) is a decoupling capacitor to avoid noise in the power supply
  • 5uF capacitor has the same purpose
  • RC branch + 500uF capacitor : That is the question. I don't really get the purpose of this.

Is the 500uF capacitor combined with 8 Ohm load just making a high pass filter with a cutting frequency of 1/(2*piRC) = 40 Hz ? Then what would be the point of this parallel RC branch?

Any help appreciated, thanks a lot!

Circuit

\$\endgroup\$
  • \$\begingroup\$ Have you considered simulating this in Spice, and adjusting the components in question to see how they affect the frequency response? \$\endgroup\$ – Nick Johnson May 15 '15 at 22:49
4
\$\begingroup\$

Let's look at the two things that you are asking about.

1) think about how the amplifier works. Many amplifiers use a bipolar power supply so that they can produce both positive and negative waveforms at the output.

Your amplifier uses only a single positive power supply rail. So now think about what DC voltage is present at the output pin of the amplifier chip. Then think about why a large capacitor is needed in that spot.

2) That series RC network from the amplifier output to ground is called a "Zobel Network". Look it in Google.

\$\endgroup\$
  • \$\begingroup\$ So if I'm right : 1) The amp would produce a 11V DC offset on the output. Therefore, they use a capacitor to bring it back around 0V. 2) Zobel network (or Boucherot cell in audio applications) aim to correct the inductive load impedance which may start oscillating at high frequencies. This helps flattening the frequency response. Thank you! \$\endgroup\$ – charlesh May 16 '15 at 11:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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