The original on page 1 has a typical error of assumption. A resistively terminated MOSFET SE is not optimally biassed when Vds equals one half of the power supply voltage. This is only true if there is a current source instead of the 15 ohm resitor, but then, there is nothing to set the gain except the loudspeaker, which means it varies at the impedance characteristic of the loudspeaker varies with ffrequency. Normally this will be considerable variation.
Things are better but not a wjole lot with the 15 ohm resistor, the outpit impedance of the amp is roughly equal to the 15 ohms, which means the damping factor is considerably less than 1 with an 8 ohm load.
The outout power you get is considerably reduced by the wrong choice of operating point (Vds). The load current can swing positive up to (Vdd - Vds)/(15+8) but negative by Vds/8. Vds = (Vdd-Vds) means the positive half period of the signal will produce a current of maximum 12/23 A and negative 12/8A, clearly asymetrical. It will clip on the positive long before it will on the negative, and the maximum output power at onset of clipping will be ~1.1W.
Optimum is achieved when both currents (positive and negative) are the same. in this case (24-Vds)/23 = Vds/8
A bit of re-arranging gives the optimum result for Vds=6.2V, at which point thie maximum output power before clipping will be ~2.4W. However, there will be twice as much heat generated on the 15 ohm resistor (and half on the transistor) at idle.
This calculation however assumes that the voltage acros the MOSFET can drop to 0V which in realiti it can approach but with a lot of distortion. Because of this the actual optimum Vds is more (7-8V) and the maximum power less. With the given values of the power supply and 15 ohm resistor, about 2W maximum can be expected wit reasonable distortion levels.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.