A Simple opamp question?

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An externally hosted image should be here but it was not working when we last tested it.

So, this is the last version :) of my DAC LPF circuit..
Features;
1.5X gain (1,65V RMS output)
Single supply, it shares DAC's analog supply (+5v)
Capable to direct drive 32 to 250R headphones/speakers.. (85mW @32R, 11mW @250R), may be lower loads..
Based on BB's reference design..
...
Any suggestion on this?
 
Finally,

This is the last status of my PCM2702 DAC;
An externally hosted image should be here but it was not working when we last tested it.

And already prepared a PCB for it;
An externally hosted image should be here but it was not working when we last tested it.

At the outputs, I use AD8397 as mentioned before. It is used for both for LPF (@45Khz) and for headphone drive.. After the opamp, Elna Cerafine type high value caps used (1,000uF) bypassed with 470nF Evox-Rifa PHE426 type MKP caps.
For power, I used seperated regulators for digital and analog section.. For digital supply 78L33 and for analog LM317 used..
This DAC can directly drive 85mW in to 32R headhones and also 42mW for 64R ones.
I expect to have really good measurements and sounding results with this circuit :) Because I am working for over one month just for it!
So I am waiting for your last comments on the circuit and PCB design pls.. I'll purchase the PCB in oe or two days..
Thanks in advance.
 
Regarding my last circuit,

Its actually AC coupled and there is no cap on inputs.. However if you consider the opamps NI inputs are biased with 5v/2 (common) point then I think there is no need for a capacitor on inputs (like TI's reference design)..
And about the input resistor, I was expecting to have 1.5X gain and this resistor is obtaining that gain. So it was not possible to use equal value with feedback resistor..
Am I right?
 
I mean put a resistor in series with the non-inverting input which is currently connected directly to +5V/2. This is a common method of removing the offset due to input bias currents (having equal resistance seen by both inputs causes the input offset voltage to be equal, and thus cancel, resulting in no output offset voltage, ideally).

Since it is DC coupled, the resistance needed will be 3.9K+(18K||12K) ~= 11K, ignoring the output impedance of whatever the preceding stage is.
 
Dxvideo said:
By the way, my simulator says 8K2 will cause the best dc offset value.
That is dependent on the particular parameters of the model used. You will find that the perfect value is different in real life, and different for each op-amp you try. It depends on input offset current and other things.

If you want the perfect offset for one single build of the circuit, then by all means try different values (in real life) and pick the best, but if you want the circuit to work well with any op-amp, then stick to a calculated value.
 
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