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Old 21st October 2015, 05:58 AM   #1
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Default LM1875 BTL from single supply

so I have been using a LM1875 for a while now and decided to switch to a BTL setup to boost the ouput from the same voltage.

here is how I have it wired, however I am having an issue that the gain on the inverting amplifier (green) is very frequency dependant, so at 100Hz for instance its output would be 2V p-p, when the non-inverting one is at 10v p-p on a test sine wave.

but at 1k Hz they are both 10v p-p, and at 10k, the inverting amp is clipping a square wave at 22v p-p when the non-inverting one is still happily humming along at 10v p-p output.

the other thing i noticed is the output is not out of phase on the inverting one, it is only like 60 degrees shifted rather than 180 deg. hopefully this makes sense.

this is how In have it wired right now. (forgive the crude print, its 2 am and I have to work tomorrow
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Old 21st October 2015, 06:52 AM   #2
Mooly is offline Mooly  United Kingdom
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The green 1k needs to be in series with the green 1uf. The unmarked cap at the bottom isn't needed. The green 22k feedback resistor should be 23k to make the gain of the inverting stage equal the gain of the non inverting.


Downsides... the inverting stage has a 1k input impedance. This means it loads the source heavily. The green 1uf input cap needs raising to 100uf to equalise the input cut off frequency between the two stages. Upper left blue 100k needs a cap across it to remove all audio, say 47uf.

Its not a great set up tbh unless you just experimenting.
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Old 21st October 2015, 06:58 AM   #3
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You can simplify it a bit - remove the green 1uF. Remove the green cap just above GND and route the dangling end of the 1k to the bottom end of the blue cap above the higher GND. Disconnect said cap from GND.

As Mooly pointed out, the feedback resistors need to be nudged a little to get equal gain from the two halves. The bottom one needs to be 23k when the top's 22k.
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Old 21st October 2015, 02:21 PM   #4
johnr66 is offline johnr66  United States
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Look up the datasheets for the TDA2040. IIRC, they show a bridge configuration which will work for the 1875. I don't recommend your config without op amp buffers/inverters.

With 8 ohm load, you are safe to operate the 1875 up to 30v supply which includes some headroom for reactive loads. If you decide to try a 4 ohm load, 18v is about as far as you should go.
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Old 21st October 2015, 03:21 PM   #5
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i guess since my question is if the lm1875 is basically an op amp at its core it shouldn't draw nearly any current into its input so what is the difference between running it into a power op amp -vs- running into a buffer op amp into a power op amp?
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Old 21st October 2015, 04:40 PM   #6
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Variations of this circuit have been around for a long time. Its performance is dependent on an array of precision matched resistors and oddball values. Any error in the resistor network will degrade performance. I haven't seen any commercial applications of this circuit (common in cheap car radio "power boosters" of yesteryear) that worked very well.

An easy way out would be to use a differential line driver like DRV134. DRV134PA Texas Instruments | DRV134PA-ND | DigiKey The whole resistor network is integrated into the chip and laser trimmed for highest precision and lowest distortion. It provides a fixed gain of 6 dB in differential mode. No external feedback network is required. It does a lot of heavy lifting for its $6.00 price tag.

This could drive non-inverting chip amps. Since it's designed to drive 600 ohm loads it could even drive low impedance input inverting amps if you like. It frees up a lot of wiggle room for your design.

There is a circuit floating around out there that uses this chip in a composite differential chip amplifier, but this greatly complicates your design.

No matter how you go, this chip will make your design easier and better.
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Old 21st October 2015, 05:55 PM   #7
Mooly is offline Mooly  United Kingdom
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Quote:
Originally Posted by JWoodrell View Post
i guess since my question is if the lm1875 is basically an op amp at its core it shouldn't draw nearly any current into its input so what is the difference between running it into a power op amp -vs- running into a buffer op amp into a power op amp?
The inputs don't draw any current (well only microscopic amounts). What draws the current is the configuration the device is used in.

In inverting mode the input signal is applied through a series resistor to the inverting input. That in itself draws no current. What does cause the current flow is the fact the inverting input has another resistor tying it to the output of the chip. This means the inverting input always sees zero volts at the input pin, and as a consequence, the value of the series input resistor now becomes the input impedance which is seen by the driving source component.
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Old 21st October 2015, 06:24 PM   #8
AndrewT is offline AndrewT  Scotland
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the 1875 is current crippled into 4ohms loading.
A bridged pair of 1875 feeding an 8ohms speaker will also be current crippled.

Don't do it.
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Old 22nd October 2015, 04:49 AM   #9
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ok, thank you mooly. when / if I use this in my project i will bump the values of that divider so the proportion is the same but the values are higher.

I am using this in an industrial environment. so I only have +24V DC power as my supply. with a single 24V rail I was putting out only around 7W to the speaker but for what they were used for it worked well and was simple. the new ones I have to make now have to be significantly louder but I can't completely gut the design for these.

I am also considering having a switch mode inverter to make me a -24V rail and using the single 1875 as well but the it dumps 19W of heat like that and I am running into limitation of a convection mode heatsink to handle that. I may end up doing the BTL and maybe the inverter as well (tuned down to -5 to -12v probably) depending on how the thermal loading ends up with the two 1875s.

the next version I have decided to switch up to a class D amp (aiming for 80W - 100W and the speaker is rated for 125W RMS) but that is a future project. (with a single 30V rail and 90% efficiency it should be around there)
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Old 23rd October 2015, 01:05 PM   #10
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https://www.sparkfun.com/datasheets/...ral/STA540.pdf

why not leave the lm1875 and use something more suited ?
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