Ok, so I've finally had the time to try to build an 1875T chipamp for a PC subwoofer, and I'm having a bit of an issue, I think it may be ground related, I'm still pretty new to this, but it's distorted quite a bit, and I can't seem to figure out what it could be. I'm using this schematic, I can't seem to figure it out, I'm using a 15v supply, as I don't want too much bass(in an apartment).
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It might be clipping.
What is your maximum peak input voltage? Your gain is about 33.4 [(R2 + R3) / R2], which seems a little high. Even if the amp could swing to the +/-15v rails (which it can't), your max input level could only be +/- 0.45 Volts.
What is your load (speaker) impedance?
With +/-15v supply, you can only expect to get a maximum of about 9.5 Watts output. Watts = V squared over R. So, for an 8 Ohm speaker, 10 Watts would mean 8.71 V RMS max output which for a sine wave would be +/- 12.33 V peak, so your gain of 33.4 would imply a max input peak of roughly +/- 0.37 Volts.
If your speaker is 4 Ohms, 9.5 Watts would imply 6.16 V RMS max output or roughly +/- 0.26 Volts max input peak.
If you can't measure your input level's peak, you could raise R2 to 2.2k in order to lower the gain to 11, which is near the minimum stable value of 10, and see if that helps. Alternatively, you could put a pot or voltage divider on the input, to lower the input voltage level.
Check my arithmetic.
After that is all sorted out, you should probably also add the Zobel network, as shown in the datasheet at national.com, which is a 1 Ohm resistor in series with a 0.22 uF capacitor, from output to ground.
Tom
What is your maximum peak input voltage? Your gain is about 33.4 [(R2 + R3) / R2], which seems a little high. Even if the amp could swing to the +/-15v rails (which it can't), your max input level could only be +/- 0.45 Volts.
What is your load (speaker) impedance?
With +/-15v supply, you can only expect to get a maximum of about 9.5 Watts output. Watts = V squared over R. So, for an 8 Ohm speaker, 10 Watts would mean 8.71 V RMS max output which for a sine wave would be +/- 12.33 V peak, so your gain of 33.4 would imply a max input peak of roughly +/- 0.37 Volts.
If your speaker is 4 Ohms, 9.5 Watts would imply 6.16 V RMS max output or roughly +/- 0.26 Volts max input peak.
If you can't measure your input level's peak, you could raise R2 to 2.2k in order to lower the gain to 11, which is near the minimum stable value of 10, and see if that helps. Alternatively, you could put a pot or voltage divider on the input, to lower the input voltage level.
Check my arithmetic.
After that is all sorted out, you should probably also add the Zobel network, as shown in the datasheet at national.com, which is a 1 Ohm resistor in series with a 0.22 uF capacitor, from output to ground.
Tom
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Good point. If he's using a single supply, 0-15 V, then he's using the wrong schematic.
Correct. No filter needed.
What about your input signal voltage level? Sounds like you have a multimeter. See if you can tell what the highest peak voltage is. You could also measure as AC volts which would probably be in RMS Volts. You can't quite calculate the real peak voltage from RMS unless the signal is a pure sine wave, in which case you could just multiply the RMS value by the square root of two, i.e. 1.414. But you can multiply by that anyway and probably see if you're in the ballpark. OR, if you can change the input signal level, just try turning it down and see if the distortion goes away. But I still think your gain looks a little high.
If you have the specs for your system, just try to find out what the signal level range is, for that subwoofer output.
To be clear, do you have two wires or three wires for DC power to the chipamp circuit?
You should have three wires, if you have +15 V, Gnd, and -15 V.
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OK. You need to be using the schematic for the "single supply" version of an LM1875 amplifier circuit.
Go to National Semiconductor | High-performance Analog and download the datasheet for the LM1875. The circuit you posted is the dual-supply one, from page 1. The circuit you need is the single-supply one, from page 2 (at least in my version of the datasheet).
Note that C1 and probably C5 should be film caps, polypropylene. Also, connect the 0.1uF and 100uF decoupling caps as close to the chip's power pins as you can.
You could increase the values of the C1 1.0uF cap and the 100uF cap, if you wanted to.
Tom
Go to National Semiconductor | High-performance Analog and download the datasheet for the LM1875. The circuit you posted is the dual-supply one, from page 1. The circuit you need is the single-supply one, from page 2 (at least in my version of the datasheet).
Note that C1 and probably C5 should be film caps, polypropylene. Also, connect the 0.1uF and 100uF decoupling caps as close to the chip's power pins as you can.
You could increase the values of the C1 1.0uF cap and the 100uF cap, if you wanted to.
Tom
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With a TDA2003 he probably could, but the LM1875 just won't "swing" that. The TDA2050 has a bit more output swing at a given Vs, but these higher supply voltage "Hifi" chips weren't designed with optimized output swing in mind.
. I uh, made the mistake of buying a grab-bag of resistors so I'm spending my day Iding and organizing them 
Toroids are the preferred choice in power-supplys due to their high output power and little magnetic radiation.
First you should decide how much output power you want. So look at Page 3 on the datasheet. Lets assume 20W, thats around +-22V
One page forward look at the power dissipation vs. output power. At 20W with +-22V we get a disipation at around 15W
As a last step we want to decide which heat sink to use. We know we dissipate 15W at maximum power, and given a really hot summer day on which we incidently have a party we look at 40°C ambient temperature. The 5°C/K won't do since it will drop just below 15W at 40°C. 2°C/K is plenty so 3°C/K will work, 4°C/K might also.
Now back to the power supply, the chipamp wants 15W heat + 20W output power. so you need at least 35W or 35VA output power. be safe and add ~30% for loss in the various components so above 45VA is a safe choice.
The voltage of the toroidal: after rectification and filtering you get: (initial AC Voltage) * (square root(2)) = (DC Voltage)
We want +-22V DC so thats: around 16V AC.
So the toroidal you want is 2 * 16V AC @ 45VA or something like that.
keep in mind that you need a PSU for rectification and such..
cheers
First you should decide how much output power you want. So look at Page 3 on the datasheet. Lets assume 20W, thats around +-22V
One page forward look at the power dissipation vs. output power. At 20W with +-22V we get a disipation at around 15W
As a last step we want to decide which heat sink to use. We know we dissipate 15W at maximum power, and given a really hot summer day on which we incidently have a party we look at 40°C ambient temperature. The 5°C/K won't do since it will drop just below 15W at 40°C. 2°C/K is plenty so 3°C/K will work, 4°C/K might also.
Now back to the power supply, the chipamp wants 15W heat + 20W output power. so you need at least 35W or 35VA output power. be safe and add ~30% for loss in the various components so above 45VA is a safe choice.
The voltage of the toroidal: after rectification and filtering you get: (initial AC Voltage) * (square root(2)) = (DC Voltage)
We want +-22V DC so thats: around 16V AC.
So the toroidal you want is 2 * 16V AC @ 45VA or something like that.
keep in mind that you need a PSU for rectification and such..
cheers
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