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Small Op-Amp Circuit Question

mike7877

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So we've got the 0.1uF'ers right on the pins, and you can't see them here, but there's 1,000uF bulks on each rail, too (to match the resistors lol)

If this circuit's purpose is to be a very basic headphone output (to power a 1/4" jack like you'd see on old Sony 5 disc changers - static level, same voltage as line out)


Should there be a 5-20 ohm resistor on the output, and if so, where should it be located?
1) between the rightmost arrow and black dot?
or
2) between the black dot and the op a?


and

Do I have to enable low pass filters with resistor/capacitor couples in the feedback if the DAC's output is already low passed and 99.99% under 20kHz?
 
You are thinking of an inverting version of the CMoy headphone amp.

There is a write up here:
Yes, an output resistor on this non-inverting amp of around 50 Ohms is beneficial for some aspects of operation. Specifically, will help taming output DC offset, at the expense of the the already modest output.
Amir measured the CMoy amp here:

I think you will find the very inverting configuration has serious issues with DC offset, and maybe noise too, being that they are going to be multiplied by the gain. There are fine solutions, but add complexity.
 
View attachment 454031

So we've got the 0.1uF'ers right on the pins, and you can't see them here, but there's 1,000uF bulks on each rail, too (to match the resistors lol)

If this circuit's purpose is to be a very basic headphone output (to power a 1/4" jack like you'd see on old Sony 5 disc changers - static level, same voltage as line out)


Should there be a 5-20 ohm resistor on the output, and if so, where should it be located?
1) between the rightmost arrow and black dot?
or
2) between the black dot and the op a?


and

Do I have to enable low pass filters with resistor/capacitor couples in the feedback if the DAC's output is already low passed and 99.99% under 20kHz?
The 0.1uF (10nF is probably better) must be close to the power pins directly to a ground plane and is there to prevent oscillations by locally decoupling the power pins.
This means for HF only the power supply pins have a low 'resistance'.

In cases where it is also essential to have a low resistance of the power lines for audio frequencies (preventing the power line voltages to sag) when say battery fed or fed from a rail-splitter you might also need some reservoir caps. These do not have to be as close to the pins as decoupling caps.
How big (capacitance) those reservoir caps need to be depends on circumstances.

The circuit above is inverting so only use this when one needs to have the polaritiy inverted, otherwise use a follower circuit.

The output resistor might be required for certain op-amps and is there to either limit output current or, more likely, is there because the chip manufacturer recommends using one to ensure a high capacitance load will not destabilize the feedback loop and oscillate.
The value thus depends on several factors.


1) between the rightmost arrow and black dot?
or
2) between the black dot and the op a?
That would depend on the goal of the design.
Both options can be used.
1: to prevent a capacitive load to cause oscillation but increases the output R
2: to create a current limiter and to ensure a low output resistance. The circuit might act weird with certain non resistive loads.
 
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