<p>This is a partial answer because I'm not an expert and because I don't know what the dynamic range of 137 dB means. Hopefully you can add a little more information.</p>

<p><strong>tl;dr:</strong> if 137 dB is the dynamic range in power, then it's 68.5 dB in voltage and velocity which sounds more plausible, and makes the velocity sensitivity well below 1 micron per second. However we don't yet know what the noise and bandwidth of your signal are yet so we can't evaluate that.</p>

<blockquote>
  <p>I have a few specs from the datasheet...</p>
</blockquote>

<p><em>The more information you share from the data sheet the better</em> although I've now just noticed that the question is about three years old.</p>

<p>Also, there may be some helpful insight at <a href="https://earthscience.stackexchange.com/q/16178/6031">How sensitive are typical seismometers?</a></p>

<hr>

<p>137 dB is a factor of <span class="math-container">$10^{-137/10} = 10^{-13.7} =$</span> 2E-14.</p>

<p><strong>Let's assume</strong> <strong><em>just for a moment</em></strong> that the dynamic range applies to voltage and velocity, rather than power.</p>

<p>If the full scale is 20 Volts then the minimum resolution (ignoring noise) is 4E-13 Volts.</p>

<p>If the sensitivity is 2400 V per m/s then it's </p>

<p><span class="math-container">$$\frac{4 \times 10^{-13} \text{ V}}{2400 \text{V per m/s}} \approx 1.7 \times 10^{-16} \text{m/s} $$</span> </p>

<p>which is <span class="math-container">$1 \times 10^{-10}$</span> microns per second, and absurdly low number.</p>

<p><strong>On the other hand if the 137 dB is power</strong> then we multiply the 20 V by <span class="math-container">$10^{-137/\mathbf{20}} = 10^{-6.85} =$</span> 1.4E-7 to get a minimum resolution (again ignoring noise) is 2.8E-6 Volts, and</p>

<p><span class="math-container">$$\frac{2.8 \times 10^{-6} \text{ V}}{2400 \text{V per m/s}} \approx 1.2 \times 10^{-9} \text{m/s} $$</span> </p>

<p>which is <span class="math-container">$1.2 \times 10^{-3}$</span> microns per second or about 1 nanometer per second, which is indeed a realistic number for some <a href="https://en.wikipedia.org/wiki/Capacitive_displacement_sensor#Precision_positioning" rel="nofollow noreferrer">capacitive</a> and <a href="https://electronics.stackexchange.com/a/460877/102305">optical</a> displacement sensors. It sounds <em>plausible</em>.</p>

CC BY-SA 4.0
2020-04-30T08:41:05.433Z
2020-04-30T08:47:43.673Z
2020-04-30T08:47:43.673Z
uhoh
<p>create a bench test using an acceleramentor and a transducer since their values are known then equate the results to your sesimo</p>

CC BY-SA 4.0
2018-05-11T06:44:37.390Z
Hi Chuck, welcome to stackexchange. This is a rather terse answer and, while it does offer a way forward, I think the asker of the question was hoping it was possible to tell from the datasheet. Would you consider expanding it and, if the answer cannot be determined from the datasheet, explaining why not?
Semidiurnal Simon
CC BY-SA 4.0
2018-05-10T18:44:37.273Z
2018-05-10T18:44:37.273Z
chuck
<p>I would like to know if a specific seismometer can measure 1 micron/sec velocity. I have a few specs from the datasheet but I'm not a seismologist and am trying to figure out how to relate the specs to one another.</p>

<p>I have:</p>

<pre><code>Velocity output band: 30s (0.03Hz) to 100 Hz
Output Sensitivity: 2400 V/m/s
Peak/Full scale output: Differential: +- 20V 
Sensor dynamic range: 137 dB @ 5 Hz
</code></pre>

<p>Thanks in advance!</p>

CC BY-SA 4.0
2020-04-30T08:23:27.940Z
There may be some helpful insight at [How sensitive are typical seismometers?](https://earthscience.stackexchange.com/q/16178/6031)
uhoh
<geophysics><seismology><instrumentation>
How can I calculate the sensitivity of a seismometer?
NaN
# How can I calculate the sensitivity of a seismometer?

(Asked by: username_0 on 2017-08-25)

I would like to know if a specific seismometer can measure 1 micron/sec velocity. I have a few specs from the datasheet but I'm not a seismologist and am trying to figure out how to relate the specs to one another.

I have:

```
Velocity output band: 30s (0.03Hz) to 100 Hz
Output Sensitivity: 2400 V/m/s
Peak/Full scale output: Differential: +- 20V 
Sensor dynamic range: 137 dB @ 5 Hz

```

Thanks in advance!

**Tags:** geophysics, seismology, instrumentation

> **Comments:**
> **username_1**: There may be some helpful insight at [How sensitive are typical seismometers?](https://earthscience.stackexchange.com/q/16178/6031)  

---

## 2 answers

### Answer by username_1 on 2020-04-30. Score: 1 | Accepted answer

This is a partial answer because I'm not an expert and because I don't know what the dynamic range of 137 dB means. Hopefully you can add a little more information.

**tl;dr:** if 137 dB is the dynamic range in power, then it's 68.5 dB in voltage and velocity which sounds more plausible, and makes the velocity sensitivity well below 1 micron per second. However we don't yet know what the noise and bandwidth of your signal are yet so we can't evaluate that.

> I have a few specs from the datasheet...

*The more information you share from the data sheet the better* although I've now just noticed that the question is about three years old.

Also, there may be some helpful insight at [How sensitive are typical seismometers?](https://earthscience.stackexchange.com/q/16178/6031)

---

137 dB is a factor of $10^{-137/10} = 10^{-13.7} =$ 2E-14.

**Let's assume** ***just for a moment*** that the dynamic range applies to voltage and velocity, rather than power.

If the full scale is 20 Volts then the minimum resolution (ignoring noise) is 4E-13 Volts.

If the sensitivity is 2400 V per m/s then it's

$$\frac{4 \times 10^{-13} \text{ V}}{2400 \text{V per m/s}} \approx 1.7 \times 10^{-16} \text{m/s} $$

which is $1 \times 10^{-10}$ microns per second, and absurdly low number.

**On the other hand if the 137 dB is power** then we multiply the 20 V by $10^{-137/\mathbf{20}} = 10^{-6.85} =$ 1.4E-7 to get a minimum resolution (again ignoring noise) is 2.8E-6 Volts, and

$$\frac{2.8 \times 10^{-6} \text{ V}}{2400 \text{V per m/s}} \approx 1.2 \times 10^{-9} \text{m/s} $$

which is $1.2 \times 10^{-3}$ microns per second or about 1 nanometer per second, which is indeed a realistic number for some [capacitive](https://en.wikipedia.org/wiki/Capacitive_displacement_sensor#Precision_positioning) and [optical](https://electronics.stackexchange.com/a/460877/102305) displacement sensors. It sounds *plausible*.


---

### Answer by username_2 on 2018-05-10. Score: 0

create a bench test using an acceleramentor and a transducer since their values are known then equate the results to your sesimo

> **Comments:**
> **username_3**: Hi username_2, welcome to stackexchange. This is a rather terse answer and, while it does offer a way forward, I think the asker of the question was hoping it was possible to tell from the datasheet. Would you consider expanding it and, if the answer cannot be determined from the datasheet, explaining why not?  

---

