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Is this value useful in calculating signal to noise ratio? Or what fun calculations can I do with this number?

To convert the spectral density \$\tilde v\$ (in nV/√Hz) to a voltage (in V_RMS), you need to multiply it by the square root of the bandwidth: $$ v_\mathrm{RMS}=\tilde v \cdot \sqrt{\Delta f} $$ For example, if the op-amp is a TLC071, with equivalent input noise voltage density of 7 nV/√Hz, and audio bandwidth, the total equivalent input noise is:

• 7 nV/√Hz ⋅ √(20000 Hz - 20 Hz) = 0.99 μVrms

Assuming this is the dominant noise source, if the gain of your amp is 10× (= +20 dB) the output noise is then:

• 0.99 μVrms ⋅ 10 = 9.9 μVrms

Note that the actual noise curve is not always 7 nV/√Hz, it slopes up at low frequencies:

Turns out that's ok because the X axis is logarithmic and the units of noise are not, so it has very little effect on the total (the non-flat part below 1 kHz is only 5% of our total bandwidth, measured linearly). If you need a more accurate value you can (numerically) integrate and get the area under the (squared) curve: $$ v_\mathrm{RMS}=\sqrt{\int^{f_2}_{f_1} \! \tilde v(f)^2\,df} $$ Or simulate it in SPICE (I get 0.82 μVrms EIN).

Also, real circuits do not have ideal brickwall HPF and LPF filters, so you can compensate for this using "brickwall correction factors" to calculate the "equivalent noise bandwidth".

If your circuit has 1-pole filters, for instance, the total noise would then be

• 7 nV/√Hz ⋅ √(1.57 ⋅ (20000 Hz - 20 Hz)) = 1.24 μVrms

(Sanity check: SPICE with noiseless filters measures at 1.22 μVrms.)

Is this value useful in calculating signal to noise ratio? Or what fun calculations can I do with this number?

To convert the spectral density \$\tilde v\$ (in nV/√Hz) to a voltage (in V_RMS), you need to multiply it by the square root of the bandwidth: $$ v_\mathrm{RMS}=\tilde v \cdot \sqrt{\Delta f} $$ For example, if the op-amp is a TLC071, with equivalent input noise voltage density of 7 nV/√Hz, and audio bandwidth, the total equivalent input noise is:

• 7 nV/√Hz ⋅ √(20000 Hz - 20 Hz) = 0.99 μVrms

Assuming this is the dominant noise source, if the noise gain of your amp is 10× (= +20 dB) the output noise is then:

• 0.99 μVrms ⋅ 10 = 9.9 μVrms

Note that the actual noise curve is not always 7 nV/√Hz, it slopes up at low frequencies:

Turns out that's ok because the X axis is logarithmic and the units of noise are not, so it has very little effect on the total (the non-flat part below 1 kHz is only 5% of our total bandwidth, measured linearly). If you need a more accurate value you can (numerically) integrate and get the area under the (squared) curve: $$ v_\mathrm{RMS}=\sqrt{\int^{f_2}_{f_1} \! \tilde v(f)^2\,df} $$ Or simulate it in SPICE (I get 0.82 μVrms EIN).

Also, real circuits do not have ideal brickwall HPF and LPF filters, so you can compensate for this using "brickwall correction factors" to calculate the "equivalent noise bandwidth".

If your circuit has 1-pole filters, for instance, the total noise would then be

• 7 nV/√Hz ⋅ √(1.57 ⋅ (20000 Hz - 20 Hz)) = 1.24 μVrms

(Sanity check: SPICE with noiseless filters measures at 1.22 μVrms.)

Is this value useful in calculating signal to noise ratio? Or what fun calculations can I do with this number?

To convert the spectral density \$\tilde v\$ (in nV/√Hz) to a voltage (in V_RMS), you need to multiply it by the square root of the bandwidth: $$ v_\mathrm{RMS}=\tilde v \cdot \sqrt{\Delta f} $$ For example, if the op-amp is a TLC071, with equivalent input noise voltage density of 7 nV/√Hz, and audio bandwidth, the total equivalent input noise is:

• 7 nV/√Hz ⋅ √(20000 Hz - 20 Hz) = 0.99 μVrms

Assuming this is the dominant noise source, if the gain of your amp is 10× (= +20 dB) the output noise is then:

• 0.99 μVrms ⋅ 10 = 9.9 μVrms

Note that the actual noise curve is not always 7 nV/√Hz, it slopes up at low frequencies:

Turns out that's ok because the X axis is logarithmic and the units of noise are not, so it has very little effect on the total (the non-flat part below 1 kHz is only 5% of our total bandwidth, measured linearly). If you need a more accurate value you can (numerically) integrate and get the area under the (squared) curve: $$ v_\mathrm{RMS}=\sqrt{\int^{f_2}_{f_1} \! \tilde v(f)^2\,df} $$ Or simulate it in SPICE (I get 0.82 μVrms EIN).

Also, real circuits do not have ideal brickwall HPF and LPF filters, so you can compensate for this using "brickwall correction factors" to calculate the "equivalent noise bandwidth".

If your circuit has 1-pole filters, for instance, the total noise would then be

• 7 nV/√Hz ⋅ √(1.57 ⋅ (20000 Hz - 20 Hz)) = 1.24 μVrms

(Sanity check: SPICE with noiseless filters measures at 1.22 μVrms.)

Is this value useful in calculating signal to noise ratio? Or what fun calculations can I do with this number?

To convert the spectral density \$\tilde v\$ (in nV/√Hz) to a voltage (in V_RMS), you need to multiply it by the square root of the bandwidth: $$ v_\mathrm{RMS}=\tilde v \cdot \sqrt{\Delta f} $$ For example, if the op-amp is a TLC071, with equivalent input noise voltage density of 7 nV/√Hz, and audio bandwidth, the total equivalent input noise is:

• 7 nV/√Hz ⋅ √(20000 Hz - 20 Hz) = 0.99 μVrms

Assuming this is the dominant noise source, if the gain of your amp is 10× (= +20 dB) the output noise is then:

• 0.99 μVrms ⋅ 10 = 9.9 μVrms

Note that the actual noise curve is not always 7 nV/√Hz, it slopes up at low frequencies:

Turns out that's ok because the X axis is logarithmic and the units of noise are not, so it has very little effect on the total (the non-flat part below 1 kHz is only 5% of our total bandwidth, measured linearly). If you need a more accurate value you can (numerically) integrate and get the area under the (squared) curve: $$ v_\mathrm{RMS}=\sqrt{\int^{f_2}_{f_1} \! \tilde v(f)^2\,df} $$ Or simulate it in SPICE (I get 0.82 μVrms EIN).

Also, real circuits do not have ideal brickwall HPF and LPF filters, so you can compensate for this using "brickwall correction factors" to calculate the "equivalent noise bandwidth".

If your circuit has 1-pole filters, for instance, the total noise would then be

• 7 nV/√Hz ⋅ √(1.57 ⋅ (20000 Hz - 20 Hz)) = 1.24 μVrms

(Sanity check: SPICE with noiseless filters measures at 1.22 μVrms.)

Is this value useful in calculating signal to noise ratio? Or what fun calculations can I do with this number?

To convert to a voltage, you need to combine it with the bandwidth. For example, a TLC071 has 7 nV/√Hz voltage noise, so for audio bandwidth, the total equivalent input noise is:

• 7 nV/√Hz ⋅ √(20000 Hz - 20 Hz) = 0.99 μVrms

Assuming this is the dominant noise source, if the gain of your amp is 10× (= +20 dB) the output noise is then:

• 0.99 μVrms ⋅ 10 = 9.9 μVrms

Note that the actual noise curve is not always 7 nV/√Hz, it slopes up at low frequencies:

Turns out that's ok because the X axis is logarithmic and the units of noise are not, so it has very little effect on the total (the non-flat part below 1 kHz is only 5% of our total bandwidth, measured linearly). If you need a more accurate value you can simulate it in SPICE (I get 0.82 μVrms EIN) or calculate it with a spreadsheet.

Also, real circuits do not have ideal brickwall HPF and LPF filters, so you can compensate for this using "brickwall correction factors" to calculate the "equivalent noise bandwidth".

If your circuit has 1-pole filters, for instance, the total noise would then be

• 7 nV/√Hz ⋅ √(1.57 ⋅ (20000 Hz - 20 Hz)) = 1.24 μVrms

(Sanity check: SPICE with noiseless filters measures at 1.22 μVrms.)

Is this value useful in calculating signal to noise ratio? Or what fun calculations can I do with this number?

To convert the spectral density \$\tilde v\$ (in nV/√Hz) to a voltage (in V_RMS), you need to multiply it by the square root of the bandwidth: $$ v_\mathrm{RMS}=\tilde v \cdot \sqrt{\Delta f} $$ For example, if the op-amp is a TLC071, with equivalent input noise voltage density of 7 nV/√Hz, and audio bandwidth, the total equivalent input noise is:

• 7 nV/√Hz ⋅ √(20000 Hz - 20 Hz) = 0.99 μVrms

Assuming this is the dominant noise source, if the gain of your amp is 10× (= +20 dB) the output noise is then:

• 0.99 μVrms ⋅ 10 = 9.9 μVrms

Note that the actual noise curve is not always 7 nV/√Hz, it slopes up at low frequencies:

Turns out that's ok because the X axis is logarithmic and the units of noise are not, so it has very little effect on the total (the non-flat part below 1 kHz is only 5% of our total bandwidth, measured linearly). If you need a more accurate value you can (numerically) integrate and get the area under the (squared) curve: $$ v_\mathrm{RMS}=\sqrt{\int^{f_2}_{f_1} \! \tilde v(f)^2\,df} $$ Or simulate it in SPICE (I get 0.82 μVrms EIN).

Also, real circuits do not have ideal brickwall HPF and LPF filters, so you can compensate for this using "brickwall correction factors" to calculate the "equivalent noise bandwidth".

If your circuit has 1-pole filters, for instance, the total noise would then be

• 7 nV/√Hz ⋅ √(1.57 ⋅ (20000 Hz - 20 Hz)) = 1.24 μVrms

(Sanity check: SPICE with noiseless filters measures at 1.22 μVrms.)