Quote:

Originally Posted by

**3ll3d00d** /t/1528032/understanding-input-sensitivity-specs#post_24623875

Yes sorry. The manual is here -

http://www.focux.us/support-doc/manual/D_Series_Amp_Manual_V6.2_FOCUX.pdf

That manual has pretty detailed information there. Here are some details of a few relationships between gain, input sensitivity, and maximum output power at a given load resistance.

- Given the input sensitivity, the maximum output power and the resistance at which it is measured, one can calculate the amp's gain.
- Given the gain, the maximum output power and the resistance at which it is measured, one can calculate the amp's input sensitivity.

To do this, let's arm ourselves with some formulas. The output power P out for a load resistance R is given by:

P out = V RMS(out) 2 / R

If we know the power and load resistance and want the RMS output voltage, we must solve for V RMS(out) . We get:

V RMS(out) = sqrt(P out * R)

The voltage gain, call it A R , expressed as a ratio is:

A R = V RMS(out) / V RMS(in)

The voltage gain in dB, call it A dB is defined as:

A dB = 20 * log 10 (A R )

= 20 * log 10 (V RMS(out) / V RMS(in) )

If we know the gain in dB A dB and want to find the gain as a plain ratio A R , we must solve for A R . We get:

A R = 10 (A dB / 20)

Let's look at the D12 maximum output power into 8 Ohms. It is 1200 W. We solve for the RMS output voltage

V RMS(out) = sqrt(P out * R) = sqrt(1200 * 8) = 97.98 Volts RMS

This occurs by definition when the RMS input voltage is equal to the input sensitivity (the RMS input voltage required to reach maximum power). From the spec sheet, this is 4.90 Volts RMS. The voltage gain as a ratio A R is:

A R = 97.98 / 4.90 = 20.0

We convert this to dB by taking 20 times the base 10 log of the ratio to get A dB . This gives us:

A dB = 20 * log 10 (A R ) = 20 * log 10 (20.0) = 26.0 dB.

This is exactly what the spec sheet says for the gain just underneath the sensitivity specs. But they don't tell you the sensitivity when the gain is set to 32 dB. Let's do that as an example. First we convert the 32 dB gain number (A dB ) from dB to the voltage ratio A R .

A R = 10 (32 / 20) = 39.81

Now that we know the voltage gain as a ratio, we can take the output voltage at full power, which is 97.98 Volts RMS, and divide it by this value to get the input voltage, which by definition is the input sensitivity.

Input sensitivity at 32 dB gain for D12 = 97.98 / 39.81 = 2.46 Volts RMS

Similar calculations can be done for the other amplifier models.

They do have an interesting, and in my view useful way of controlling the gain. For two of the three settings (26 dB and 32 dB), the gain is independent of which amplifier model you're using. To get a 1 Volt sensitivity for each model, the gain must be higher for the higher-powered models, since they will have a higher output power (therefore a higher output voltage) for a fixed input voltage (1 Volt RMS).

Whew! Sorry for such a long post.