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Many, many choices. And this is just Microchip's offerings.

I'm going to be amplifying by a maximum of 5x using an op-amp. So I choose an op-amp with a GBWP of 50 kHz * 5 = 250 kHz. That still leaves many op-amps. What slew rate should I be looking for? (with low distortion on the output.) And are there any other parameters to look out for, like input noise and offset?

EDIT: It seems I need to clarify. I'm only looking for a single channel op-amp for amplifying a mono signal with reasonable levels of distortion. I'm not looking for high quality prosumer gear audio.

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    \$\begingroup\$ ....Microchip? Go with Linear tech or Burr-Brown if you're looking for ultra-high sound quality. \$\endgroup\$
    – Connor Wolf
    Commented Nov 15, 2010 at 0:53
  • \$\begingroup\$ Not ultra-high sound quality, just acceptable THD. \$\endgroup\$
    – Thomas O
    Commented Nov 15, 2010 at 7:58
  • \$\begingroup\$ Most important aspect unstated so far is output power and load impedance. However Op AMps are not known for this, so I hope this is for a pre-amp. Otherwise. you r question is incorrectly stated. \$\endgroup\$
    – D.A.S.
    Commented Nov 30, 2012 at 17:58

4 Answers 4

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the slew rate needs to be \$\geq 2\pi f V_{\text{peak}}\$ to not impart distortion on a sinusoidal signal.

\$f\$ = frequency of the sinusoid and \$V_{\text{peak}}\$ = peak voltage.

So in your case you'd probably use 20kHz for the frequency and whatever your \$V_{\text{peak}}\$ is to find your minimum slew rate.

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    \$\begingroup\$ of course you don't listen to sine waves, however music, as with any remotely periodic signal, can be devolved into a sum of sinusoids, so looking at the highest band-limited frequency is all thats needed. If your trying to say that you need to consider the impulse response, i disagree, any wave form with content outside ~20-20khz won't be in the source material anyway. \$\endgroup\$
    – Mark
    Commented Nov 15, 2010 at 4:18
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    \$\begingroup\$ @XTL - Not approximately, exact! Or, alternatively, you could create a sine wave as an infinite sum of square waves. \$\endgroup\$
    – stevenvh
    Commented Jun 7, 2011 at 13:17
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    \$\begingroup\$ @stevenvh, can you draw a circuit for that? :) \$\endgroup\$
    – Kortuk
    Commented May 18, 2012 at 22:03
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    \$\begingroup\$ @supercat - It does too! The infinite sum gives an exact square wave! (BTW, it's "Fourier") \$\endgroup\$
    – stevenvh
    Commented May 21, 2012 at 4:26
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    \$\begingroup\$ @supercat - Thanks for the link. But It is important to put emphasis on the word finite because even though every partial sum of the Fourier series overshoots the function it is approximating, the limit of the partial sums does not. I see no problem here; the infinite series doesn't seem to have the overshoot. And also, CMIIW, it seems to be about the overshoot, not the edge's steepness. \$\endgroup\$
    – stevenvh
    Commented May 21, 2012 at 15:29
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As Mark said, if you want to look at the distortion of a 20kHz sine wave with 1Vpp you get about 0.125V/us every Vpp. It depends how much your output peak-peak is. Because your amplification is only 5 times, I don't think you're going for 30Vpp but more something like 10Vpp?

Because you're amplifying audio, you could use an AC coupling so you don't have to worry about the offset very much. The noise can be significant, but not with these amplifications. You can calculate the RMS noise RTI according to your bandwidth:

e_noise = sqrt(Bandwidth) * NoiseFigure (nV/sqrt(Hz))

So your Bandwidth = 20kHz (ideal situation, a simple low pass will add penalties), your noise figure is for example 50nV/rt Hz (not counting 1/f noise), which means your RTI (input) noise RMS value is 7.071uV. If you amplify the signal by 5 times, you RTO (Output) noise, taking away external noise sources (resistors, etc.) you get 35.4uV rms noise added. At this gain and bandwidth, that isn't very much, but please note that your input noise (from the source) may already be 120uV, which means you would get 0.6mV RMS noise at the output.

Noise becomes a significant factor at high gain and/or higher bandwidths.

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  • \$\begingroup\$ Good point on the LPF ( or any filter ). Depending on the tolerance for non-constant group delay the designed for bandwidth may need to be higher. However if he's not after 'hifi' performance non-constant group delay, on the high end especially, may not matter. \$\endgroup\$
    – Mark
    Commented Nov 15, 2010 at 19:43
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Sine_Max_slew = 2*pi*f* Vpeak

so "V peak" which should not to be confused with V peak-to-peak (Vpp).

However, the equation can be reduced further though when using Vpp because Vp = Vpp/2 which means:

Sine_slew = 2*pi*f*Vpp / 2

and the 2's cancel, which leaves:

Sine_Max_slew = pi*f*Vpp

There's also a linear slew (triangle waveform) and sine waveform max slew amplitude relationship occurring under equal frequency. To understand this relationship, a reduced amplitude sine waveform will fit inside a triangle waveform of higher amplitude but same frequency when the sine waveforms max slew equals the linear slew rate. At that point, the sine waves amplitude is:

Sine(Vpp) = 2/pi * Triangle(Vpp)

There's also a relationship for equal slew rates between linear and sine when amplitudes are the same but frequency is different:

Sine(Hz) = 2/pi * Triangle(Hz)

Just a little noteworthy background on the whole slew subject...

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Two AmpOps in the same package could be a good choise if you're planning to design stereo sound amplifier.

With the two of them in the same package the distortion from temperature variation would be the same.

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  • \$\begingroup\$ Stereo support is not going to be used. Thanks for the suggestion though. \$\endgroup\$
    – Thomas O
    Commented Nov 15, 2010 at 13:06

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