Question about designing Chip amps

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There exists a type of very popular wire boards where you enter each component in holes in a plastic board (with contacts underneath the holes) and connect the components with wire jumpers. These boards are less suited for amplifiers because the critical components are often not located close enough to the chip to avoid self-oscillation.

Do you have a symmetrical power supply?

I suggest you start making the amplifier (in mono version) described on page 2 of the LM1875 datasheet. It is named "Typical Applications".
Try to leave the leads of the components a bit long such that they can be re-used, EXCEPT for C3, C4 and R4. They need to be placed very close the the LM1875 with short leads, else you risk self-oscillation!
You must find a piece of aluminum profile or plate to fasten on the LM1875 as heatsink. Make sure it does not touch anything else because the LM1875 cooling pad is not insulated from one of the 5 pins.


I will shortly describe the function of each of the components:


R1/C1/R2 form the input circuit. C1 is called a coupling capacitor and serves to bring the AC signal (sound) to the LM1875 input, even if there is a DC difference between the input from the source and pin 1 of the LM1875. R1 makes sure C1 is charged to ground level on the input side, even without a source connected. R1 is not critical and may be 100K or so if that’s available. R2 is a bias resistor recommended by the IC manufacturer and serves to bring the IC input potential to a suitable level (here probably ground level). Together with C1, R2 determines the low frequency cut-off point for the amplifier, thus, how low frequencies the the amplifier will pass through.


C2/R3/R4 form the feedback circuit. C2 has got a high impedance compared to R3 at very low frequencies such that the amplifier will not try to amplify very low frequencies. In the audio range, the values of R3 and R4 determine the amplification. With 20K and 1K the amplification is 20K/1K=20 times. Somewhere in the datasheet it is described that the gain must be minimum 10 (20dB) for stability. Thus, you could leave R4 as 10K and the gain would be 10K/1K=10 times. If you use 4K7 as R4 the gain would be 4.7 times but then the manufacturer warn you the amplifier may oscillate. These components are always connected to what will act as the inverting input (here pin 2).


C3/C4/C7/C8 are power line decoupling capacitors. The small 100nF capacitors are very important for HF decoupling and should be placed as close to the LM1875 pins as possible to avoid self oscillation. The 100nF capacitors are often chosen as ceramic capacitors because they have good HF properties. C7 and C8 are not critical in value as long as they are minimum 100uF. 1000UF, 2200UF or even 4700uF are often used. They serve to keep the charge ready for the amplifier when the amplifier needs to reproduce a high volume signal at the output. They are buffer capacitors and of the electrolytic type.


R5 and C5 form what Neurochrome calls a Zobel network. It serves to compensate the amplifier for certain changes in the speaker impedance. On the Internet you can read more about Zobel networks. Why these component values? The manufacturer has studied the LM1875 needs carefully and concluded these values to be the best. We believe him.


That was it and you now have a first overview of the design.


One very useful rule being valid for an amplifier designed as a power Operational Amplifier (like the LM1875): The amplifier will put its output such that there is no voltage difference between the two input pins (pins 1 and 2). With your knowledge of regulation systems, you can try to make simple calculations of how the output will vary when the signal in the input (pin 1) is varied.


If you connect the components as described in this schematic, it is 99% certain you have made your first working DIY amplifier.


Let us know when you have made the circuit and we will advise you how to test it in a rather safe way.


That was very informative and I appreciate the time and effort you put in. Can I argue that R1 is the Input impedance? As you want a high impedance to ensure that the input voltage stays near the name?
 
A Complete Guide to Design and Build a Hi-Fi LM3886 Amplifier - Circuit Basics

Goes through a lot of the calculations for components.

Cheers,
Jeff

Yeah I have seen this before, he doesn't back up any of the equations but I am assuming this came from years of experience. I dont feel comfortable about using this method and just cookie cutter it. I am going to do the Typical Design application for the LM1875 as suggested for the Single power supply because I dont have a dual split one. From there I will do my own experiments and what not.
 
If you are just starting out you will want to learn some basic skills.

You will need some software to design your PCB layout. I spent two weeks trying different software packages that are available for free. Look for software that will output Gerber files so you can easily have boards made.

Doing simulations is a quick and cheap way of learning a lot in a very short time. I use LTspice and TINA-TI to get a better understanding of a circuit.

Creating a PCB from a datasheet schematic is a challenge. Many of the GND/0V connections are missing. Redraw the schematic with your layout in mind, it will make the next step a lot easier. Also, most of the datasheets have important information on layout and parts.

Don't start with the expectation of getting it right the first time. Take a simple design like the LM1875 and play with it.
 
Got some good news! SO I finally did it. I made the typical Application one with the single power supply and man it sounds so nice, however I cant get the output Voltage RMS due to not having a dummy load and a 1k Hz full blast might shatter my ear drum. It was the jumper wires along ahhaha


hmmm I might start to believe it might be my power supply voltage, as I was too scared to go any higher than 8V.. ops. As I was cranking up the voltage it was that distorting sound..as I went higher higher it was gone...thats my bad
 
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PRR

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...I measured the output RMS on my speaker using my volt meter, and I got around 6.2V RMS, however in simulation its not even close....

That sim seems to show 14V rms with a 30V supply.

Basic "does it add up??" thinking shows this is not likely (for Sine assumption). 14V rms is 40V peak to peak. Your single-rail supply must fit 40V (plus chip losses) inside 30V. That can't be true.

The sim is totally unexplained. Where does the 14V number come from? *In Audio* we normally supply a Sine input and watch to be sure the output remains Sine. In crude power systems (say an incandescent lamp) we can push the amp into heavy clipping, high THD, and yes get 14V rms with less than 40V peak to peak. But that's not "clean".

The 6V "on speaker" needs more detail. Older voltmeters would read "an average". On speech/music signals there are transients far above average. So it can be clipping the heck out of those transients, sounding bad, yet not come over a few V average. Looks like your meter reads more transients. 6V rms would be 17.5V p-p. If we add 6 or 8 volts for chip losses, 25V. So you are not getting all the 30V supply that you paid for. (Or your real supply sags below the 30V infinite battery on the sim.) Or you just were not pushing ALL the way to the edge of clipping.

"Speaker load" is also dubious. Speakers are "never" the rated impedance. Trying to document all possible loads would drive the amplifier designer crazy (or crazier). The usual thing is to test in a likely Resistor, verify that nothing unexpected happens at other impedances (usually infinite up and 50% down), and quote that as "output".
 
Yeah I agree, I just ordered a 8Ohm Power resistor to see my actual RMS.

Interesting I didn't think of Pk-Pk, I measured my iphone, it spits out 0.963RMS so its 2.724Vpp? I mean the 14V rms makes sense no? If my gain in the picture is around ~11 then you have Using my laptop so I measured around 1.320RMS 1.320*11 = 14.52V RMS. Then theoretically I am clipping going full volume should be as simple as that? Let see here with the simulations and it looks like it is Imgur: The magic of the Internet. Side question what is that R2 doing? Cause I am pretty sure the R3 and R4 is forming the voltage divider and not the R2?
 
If the gain of your amp is 21 (using 20k and 1k) and your input is 1V then the output will be 21V.

How high does the PSU voltage have to be to output this?

It isn't 21, its 11. It will simply just clip then, if the Output Voltage is over the supply voltage, like shown above. When using my Laptop the RMS Voltage was 1.320V so * 11 = 14.52V Rms, this then translates to 14.52*sqrt(2) *2 = 41.07Vpp. This is not possible as I have a power supply of 30V so it will clip hard. This is shown in the simulation even. I tested my RMS output with transient input which isnt accurate at all, which the gentlemen above said it can peak at highs for a second. I just ordered a 8 Ohm resistor so I can test my output without killing my ears with a 1Khz hum. I just find it weird that my output voltage wasn't even close to 14.52 it was like 7V RMS or something. So either my gain is weird, or my power supply is sagging hard, but its a bench power supply with a 150Watts rating, so I am sure thats fine. I dont see the voltage collapsing either. It pulls like 300mA.
 
I see above a ton of unjustified suppositions, such as inputting 1V to get 21V RMS or lowering it slightly to get 14V RMS .

Alll of them unrealistic because they fully ignore what the PSU can actually supply.

Real World calcukation:

1) starting with a 30V standard supply, not a regulated one, so voltage can and will sag: typically 15% drop; even worse if small transformer or small capacitance (increased ripple substracts fromavailable voltage)
So 30V*0.85=25.5V.

2) power transistors still need some voltage across them to pass maximum current, usually as much as 4V ; let´s be optimistioc and think designer applied all his tricks and got down to 2V loss ... at each peak (positive or negative) so we must substract b4 volts from supply:
25.5-4=21.5V peak to peak.

Whicgh amounts to: 21.5/2.82=7.72 V RMS .

Now, what did you say you read? ;)
 
I see above a ton of unjustified suppositions, such as inputting 1V to get 21V RMS or lowering it slightly to get 14V RMS .

Alll of them unrealistic because they fully ignore what the PSU can actually supply.

Real World calcukation:

1) starting with a 30V standard supply, not a regulated one, so voltage can and will sag: typically 15% drop; even worse if small transformer or small capacitance (increased ripple substracts fromavailable voltage)
So 30V*0.85=25.5V.

2) power transistors still need some voltage across them to pass maximum current, usually as much as 4V ; let´s be optimistioc and think designer applied all his tricks and got down to 2V loss ... at each peak (positive or negative) so we must substract b4 volts from supply:
25.5-4=21.5V peak to peak.

Whicgh amounts to: 21.5/2.82=7.72 V RMS .

Now, what did you say you read? ;)


hahah wow! Really close.

Some questions for you though, you are talking about a transformer and I get that however I am using a DC bench power supply which I am pretty sure is a SMPS. Does those sag as well? I wouldnt assume so as they are close loop systems?


What you calculated I think is the maximum voltage RMS I can have at my output, so either that means I am clipping hard and thats why it isnt going over 7.72V RMS or something is wrong with my gain loop.
 
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