LM386 output Question - Page 1
Forum section: electronics / beginners

author: skinny
author_rank: Contributor
timestamp: November 21, 2014, 01:44:15 am
content: Hi Everyone,I've just started watching the EEV Blog videos and thought I might visit the forums. For my first post, I have a question.On the LM386 datasheet in the Typical Applications, is a setup to run the an amp at a variety of gain conditions. On the output of the more common configurations, there is a 0.05uF capacitor in series with an 10 ohm resistor going to ground. (I've included one of the configurations as an attachment.)As best as I understand it, the capacitor is to get rid of any unwanted high frequency noise, but what is the resistor doing there? It's such a low value, that I wouldn't think it would make a difference. Any insight?Thanks for any help that anyone can provide and have a great day.

author: Circlotron
author_rank: Super Contributor
timestamp: November 21, 2014, 01:53:10 am
content: As the signal output frequency gets higher and higher the impedance of the loudspeaker also gets higher. The circuit will likely break into oscillation at some high frequency. The impedance of the series R+C gets lower with frequency, approaching 10 ohms. This keeps a load on the output at high frequencies, helping the amplifier to behave itself.

author: skinny
author_rank: Contributor
timestamp: November 21, 2014, 02:35:20 am
content: Thanks for the response. You have brought up a couple more questions for me.1) Why would the impedance of the loudspeaker continue to get higher and higher with an increase in frequency? What characteristics of the speaker cause this effect? Would it still matter if you were using the LM386 as a preamp to feed into another stage?2) If only the capacitor was used and the impedance began to get extremely low, wouldn't the aforementioned high impedance of the loudspeaker keep the amp from going into oscillation?Thanks again for the help.

author: Syntax_Error
author_rank: Regular Contributor
timestamp: November 21, 2014, 03:43:12 am
content: 1. The inductance of the loudspeaker coil causes its inductive reactance component of imdepance to dominate at high frequencies. If not driving a loudspeaker, and driving a resistive load instead, this would not be an issue.2. I think without the resistor there, the impedance of the capacitor branch will fall, approaching zero as frequency climbs. This would overload the LM386.3. These are how I, as an amateur, see the situation. I do not answer from a position of authority or deep experience.*Edited to correct spelling.

author: LukeW
author_rank: Frequent Contributor
timestamp: November 21, 2014, 04:37:31 am
content: http://en.wikipedia.org/wiki/Boucherot_cellhttp://en.wikipedia.org/wiki/Zobel_network(A Boucherot cellisa Zobel network, it's just a name that's sometimes used for a Zobel network in the specific context of impedance correction on a loudspeaker.)Similar circuits are often encountered in lots of amplifier/speaker systems for the same general reason, it's not something that's specific to just an LM386.I hope that's helpful.

author: nuno
author_rank: Frequent Contributor
timestamp: November 21, 2014, 01:47:01 pm
content: So the amp-speaker puts out lower volume at higher frequencies?

author: skinny
author_rank: Contributor
timestamp: November 21, 2014, 11:58:01 pm
content: I just read up on the Boucherot cell through the link. That is one cool concept. Thanks a lot! Exactly what I needed.

author: skinny
author_rank: Contributor
timestamp: November 22, 2014, 11:22:00 pm
content: Once again, thanks for the help. When I build something, I realize that many times I don't exactly know how everything works. I try to explain it to myself. I've just started putting it in video form. The piece you guys helped me with was the last thing I needed.http://youtu.be/egDSHUxFGV0

author: janengelbrecht
author_rank: Regular Contributor
timestamp: November 22, 2014, 11:43:52 pm
content: Lm386 is designed to drive 16 Ohm Loudspeakers. So lets assume we have that.The Boucherot cell then has a R with value = 16 Ohm (or 22 could be used!).The C value is calculated as: L / R^2 ....The trick is then to get the L value for a given Loudspeaker