Bridged audio power amplifier for 8 Ohm load but 2x50 Ohm loudspeaker - Page 1
Forum section: electronics / beginners

author: Bigman
author_rank: Regular Contributor
timestamp: April 23, 2015, 02:09:55 pm
content: Hello,BackgroundI have a transimpedance-amplifier circuit which provides me following output-signals:DC: 0,7VAmplitude: max. 50mV or lessFrequencies: from 20Hz to 10kHzNow, I am looking for a solution how to amplifier those signals that I can make them hearable on an in-ear headset. I believe a bridged audio power amplifier such as TI's LM4902 is the way to go.My problemICs such as TI's LM4902 are providing their power output in a way such as "capable of delivering 265mW ... to an 8 Ohm load". Well, but I am not using an 8 Ohm load. Instead I will connect two 50 Ohm loudspeaker in parallel. Furthermore, 265mW seems far too much what I need. If I did my math right, I could already live with 0,246mW at each loudspeaker.QuestionMain question: Since my real load with 25 Ohm is totally different compared to the specs 8 Ohm, is this just a matter of loudness or does it also have other effects (e.g. on the bandwith).Secondary issue: How would you usually couple the 1st and 2nd stage in such a case. I assume with a simple capacitor.Last issue: later on, it would be great, if I could also control the volume at the loudspeaker. My assumption is, that this would usually be done by using for R3 an adjustable resistorAll this, you can also see from the attached picture.Thanks in advance for any kind tipps & hints.

author: flynwill
author_rank: Regular Contributor
timestamp: April 23, 2015, 03:10:32 pm
content: If you are driving earphones it's pretty unlikely you really need the bridged configuration. I'm guessing you're planning to run your device off of at least 3V and have a single-ended supply. Most headphones will be very loud with a 2-3V P-P signal feeding them.In any case to answer your questions, yes that TI chip would work just fine, and the bridged configuration does have the advantage of eliminating the need for output coupling capacitors, but with the downside that the "common" to your headphones has a live signal on it which could possibly cause issues if it is ever connected to something other than the earphones.There are any of a number of other op-amps out there that can drive headphone impedance loads just fine. You might look at the AD8656 as an example. It will require a few more parts, a pair of resistors and a capacitor to create a virtual ground at 1/2 the supply voltage. (If your supply power is noisy this might need to be a low-dropout voltage regulator), and you'll also need an output coupling capacitor and bleed resistor.You do need also need the coupling capacitor marked with ? in your schematic.Finally, you could make R3 a variable resistor as the volume control, this would not be the normal approach. Normal would be to use an audio-taper pot setup as a voltage divider at the input to the power amp stage -- CW side of the pot connected to the right side of the coupling capacitor, wiper connected to your R2, and the CCW side of the pot connected to the virtual ground (pin 2 on the ti chip). I don't have an easy way to send you a sketch, but hopefully you'll understand what I mean by that.

author: npelov
author_rank: Frequent Contributor
timestamp: April 23, 2015, 03:21:22 pm
content: Quote from: flynwill on April 23, 2015, 03:10:32 pmIt depends on the schematic. The coupling capacitor removes DC offset and if signal ground is tied to negative power supply then the power amp won't be able to produce the negative part of the waveform. A full schematic would tell if a coupling capacitor is needed.

author: Bigman
author_rank: Regular Contributor
timestamp: April 23, 2015, 03:36:39 pm
content: those are very kind a good replies. Many thanks especially to lynwill.at the end, I would love to be able to power this with a single, standard 18650 Li-battery, which can provide a voltage down to 3.6V.Based on the loudspeaker's spec, I assumed just 0,1Volts (RMS) would be already sufficient (the spec says 106dB for a 1kHz tone would be achieved with 0.26mW. Together with the impedance of 50Ohm, I believe according to sqrt(P*R)=sqrt(0.26 mW*50 Ohm)=0.1V is fine.Regarding bridged configuration: after I read several TI's spec, I got the impression "bridged configuration" is the right way to go ... those datasheets are only mentioning the advantages, but not the disadvantage. To be honest: I have to do more "google-research", to understand when to use which configuration (also to understand what it means "the common" has a live signal"). Again, I just chose bridged, because the TI-datasheets made me believe this is the best but I will try to catch up and do my homework.Regarding vol. control: based on your description (which I highly appreciate), I will try to update my sketch, so others may also be able to benifit from this sketch.Regarding coupling-capicitor: I also have to think through this, but also here: thanks for the hints (and give me a bit time to understand this).

author: npelov
author_rank: Frequent Contributor
timestamp: April 23, 2015, 03:51:56 pm
content: I'm sorry. I saw the typical application circuit of LM4902 in the datasheet. They decouple both inputs of the amplifier. So flynwill is right - you do need a decoupling cap.

author: Bigman
author_rank: Regular Contributor
timestamp: April 24, 2015, 10:16:39 am
content: I tried to do a little more homework and the following is also intended for people with similar problems and my find this later:Now, I wanted to better undertand what is meant by "single end" and "bridge mode". I believe the attached picture is a good starting point. Indeed, the benefit of having peak-to-peak voltage twice as high seems to be the major benefit of bridge mode ... but I still don't really understand the drawback as mentioned by flynwill. I assume with the single end solution the capacitor is somehow protecting the amplifier. And such kind of protection we don't have for the bridge mode.Regarding vol. control: first I had to learn the meaning of "audio taper pot" ... which was more a language translation issue. Well, for everybody who is not sourcing parts from an english speaking company: this seems to be just a potentiometer whose resistance changes logarithmic and not linear. Regarding the connection of such a pot with a bridge-mode, I also tried to make a sketch, but I am not certain if it's correct (so be carefull with it).CheersP.S.: for the attached picture single vs. bridge mode, I was inspired by following link:http://www.mikrocontroller.net/topic/357102#3992641

author: flynwill
author_rank: Regular Contributor
timestamp: April 24, 2015, 02:40:29 pm
content: You get all A's on your "homework" from me.Your drawings of bridged vs single ended and of the volume control are correct.It's not that the output capacitor is "protecting the amplifier" but rather it is blocking the DC on the output of the amplifier. Without it a (comparatively) large DC current would flow through your headset, and probably melt it rather quickly. The capacitor is a drawback in that it needs to be large enough to pass the lowest frequencies of interest. You don't say what your sound source is, but if we assume 10Hz as an acceptable LF cutoff, and 25 ohms for your headset the output cap will need to be at least 640 uF.A bridged output isn't the only way to get there. Traditional audio amplifiers are run from "split" power supplies that provide both a positive and a negative voltage (eg your typical home stereo amp will have +40 volt and -40 Volt supplies).The reason for using an audio taper (log) pot for volume control is that your ears do not respond linearly to increases or decreases in sound level. If you use a linear taper the result is that it will seem like you have almost no control at the lower end of the range the sound goes from too loud to nothing with a very small move.You haven't said what the source to your trans-impedance amp is, and why you wish to listen to it. I'm curious what you're building.

author: Bigman
author_rank: Regular Contributor
timestamp: April 24, 2015, 03:00:53 pm
content: ThanksI have found in the internet a pdf-file from a person who describes a lot about "through-the-air-communication". I will see if I can find the link and post it. Anyhow, I found the idea facinating what all you can do with a low-power-laser and a photodiode (from simple "alarm detection" up to sending data).Hence, I have bought a couple of photodiodes BPW34 and BP104S. As a light source, I am using VCSELs (OPV330). But I am totally new to electronics and this is just a hobby. Only to understand transistors and OpAmps took me almost a year. But I am trying to dig into it. I only have to take care that I am not trying to understand too much at once. E.g. on the one side, I am thinking about "what to make with the output-signal, e.g. putting it on a headset", but at the same time I have a hard time to get rid of potential DC-offset caused by ambient light.So, I am currently in the stage just to understand circuits and test it with my little devices. Where it all will end, I don't know. The sky is the limit.(P.S.: I know, for data transmission, I would have much higher frequencies. My current target is just to transmit something hearable.)

author: flynwill
author_rank: Regular Contributor
timestamp: April 24, 2015, 05:29:10 pm
content: Ah. Well that should be a fun beginner project. That laser will need some optics (a single convex lens will probably do it) to get a focused beam. You can improve your photo-detectors by adding a lens to them as well. I might suggest buying a cheap laser pointer to get a laser & optics together in a single package.To get reasonable linearity you will need a trans-impedance amplifier (Voltage in, current out) to drive the laser as well. Can be as simple as an op-amp taking it's feedback signal from a resister in series with the laser. You also want to bias the laser so that it is at 1/2 power with no signal.

author: Bigman
author_rank: Regular Contributor
timestamp: April 24, 2015, 05:32:50 pm
content: I just found the book, which started to inspire me:http://www.imagineeringezine.com/ttaoc-pdf/OTTAC-Handbook.PDF

author: Smokey
author_rank: Super Contributor
timestamp: April 25, 2015, 08:31:27 pm
content: The thing to keep in mind with audio amplifiers is that they are VOLTAGE SOURCES. They have a control loop that regulates the output voltage but hey don't actively regulate current. The loads (speakers) are defined by resistance (4ohm, 8ohm, etc) because the amplifier will try to deliver whatever current it needs to to keep the voltage at the set point. So since V=IR, if you have an 8ohm load and you have the amplifier output set to 10V, it will deliver 1.25A. That same setup with a 4ohm load will deliver 2.5A. Since most audio amplifiers don't have advanced SOA (safe operating area/over power) protection you have to be careful not to exceed the rated current/power of the amplifier by not using a load with too low resistance.The amplifier doesn't care what resistance load you use as long as it doesn't draw too much current. Highly reactive loads (capacitive/inductive) loads on the other hand can be an issue for some amplifiers.