# Why does Qts increase with added mass?

#### ghettoblasterfanatic

Can someone help me understand the physics behind why adding mass to a driver increases Qts? In my mind, it makes more sense that adding mass would lower Fs, and smoothen out the resonance peak. This is indicative of dampening being increased right? Which is the opposite of what increasing Qts does? So why would adding mass increase Qts?

#### Davor D

In simple words, Q represents voice coil ability to control the movement of the cone. Lower Q means better control and vice versa.

Now, we add some extra moving mass. The voice coil will have harder job to control the cone, meaning Q gets higher.

Same thing happens when Vas is lowered, beacuse voice coil have to "fight" against more spring action.

1 user

#### markbakk

In addition, the Qms (due to mechanical losses in the suspension) increases too when you add weight to the cone.

#### lcsaszar

Generally speaking, lower Qts means "better" loudspeaker (there is an optimum between 0.3 and 0.5). It can be achieved by lightweight cone, soft suspension and strong magnet, many turns of voice coil, but low DC resistance. In real life each involves some compromise.

2 users

#### VoxCelestial

Q is a measure of resonator "quality" - its ability to, once excited, oscillate for a continued time. Simply speaking, it is a ratio of amount of energy the resonator can store to amount of energy it loses with every oscillation. Compare tuning fork, which has Q of ~1000, can ring for a several seconds when excited, and loudspeaker cone with a Q of 0.5, which doesn't ring at all.
Q can be increased by reducing losses or increasing amount of stored energy. More mass means more stored energy, so Q increases.

2 users

#### Mister Audio

To get the same Qts after adding mass to a moving diaphragm you need to proportionally increase the amount
'electromagnetic energy' to move it ... a more powerful 'motor'.
This is also how low Q speakers are required to overcome smaller 'box volume' for decent bass - powerful motor.

1 user