Crown Audio MA-2402 operation manual AC Power Draw and Thermal Dissipation

MA 2402 Power Amplifier

9 AC Power Draw and Thermal Dissipation

This section provides detailed information about the amount of power and current drawn from the AC mains by the Macro-Tech 2402 amplifier and the amount of heat produced under various conditions. The calcula- tions presented here are intended to provide a realistic and reliable depiction of the amplifier. The following assumptions or approximations were made:

•The amplifier’s available channels are loaded, and full power is being delivered.

•Amplifier efficiency at standard 1 kHz power is estimated to be 65%.

•Quiescent power draw is 90 watts (an almost negligible amount for full-power calculations).

•Quiescent thermal dissipation equals 105btu/hr at 90 watts.

•The estimated duty cycles take into account the typical crest factor for each type of source material.

•Duty cycle of pink noise is 50%.

•Duty cycle of highly compressed rock ‘n’ roll midrange is 40%.

•Duty cycle of rock ‘n’ roll is 30%.

•Duty cycle of background music is 20%.

•Duty cycle of continuous speech is 10%.

•Duty cycle of infrequent, short duration paging is 1%.

Here are the equations used to calculate the data presented in Figure 9.1:

		Total output power with all	x	Duty
AC Mains Power	=	channels driven (watts)		Cycle	+	Quiescent Power
Draw (watts)		Amplifier Efficiency (.65)				Draw (watts)

The quiescent power draw of 90 watts is a maximum value and includes power drawn by the fan. The following equation converts power draw in watts to current draw in amperes:

					AC Mains Power
		Current Draw	=		Draw (watts)
		(amperes)	AC Mains			x		Power
				Voltage			Factor (.83)
The power factor of 0.83 is needed to compensate for the difference in
phase between the AC mains voltage and current. The following equa-
tion is used to calculate thermal dissipation:
Thermal		Total output power with all			x Duty	x .35			Quiescent Power
		channels driven (watts)			Cycle
Dissipation =	(							+	Draw (watts)	)	x 3.415
(btu/hr)		Amplifier Efficiency (.65)

The constant 0.35 is inefficiency (1.00–0.65) and the

factor 3.415 converts watts to btu/hr. Thermal dissipation in btu is divided by the constant 3.968 to get kcal. If you plan to measure output power under real-world conditions, the following equation may also be helpful:

Thermal		Total measured output power x .35		Quiescent Power
Dissipation =		from all channels (watts)				x	3.415
			+	Draw (watts)
(btu/hr)	(				)
		Amplifier Efficiency (.65)

Operation Manual page 25