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Old 19th April 2007, 03:19 AM   #21
Clipped is offline Clipped  Thailand
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Quote:
Originally posted by Eva



Double voice coil speakers have different T/S parameters depending on whether the coils are wired in series or parallel. Enclosure requirements are different too, and frequency response is not the same.
but if i wire the coils in parrallel then series between the two subs it sounds the same.

as

series then parrallel between the two subs.

but i get what you mean, the 'q' will change, but i didnt think it would have this much of an impact on system performance.
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Old 19th April 2007, 03:22 AM   #22
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Quote:
Originally posted by janneman

If the amps have the same gain, and one speaker has 6dB less sensitivity, it will sound half as loud.
If one amp has 3dB less gain but the speaker has 3dB more sensitivity, they would still sound the same loudness-wise.
I think you get the point I'm trying to make?

Jan Didden


ok i think im getting this, as voltage goes up current goes down , which relates to a higher dc current gain.

the gain of the 36v rails is higher than the 22 v rails.

the higher the voltage the higher the gain?

for figure 8. which operating temperature would you go by?

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Old 19th April 2007, 04:09 AM   #23
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That's DC current gain for the transistor. It has nothing to do with the audio signal gain (RCA input level vs speaker output level).

For BJTs, the collector current is directly related to the base current. The DC beta tells you the ratio of base current (drive current) to collector current (output current).

For the curves of the 2n6488, at 1 amp of collector current the curves show that the transistor would require a drive current of 1/150 of the collector current. For a collector current of 5 amps, the drive current would have to be 1/25 of the collector current.

None of this affects the overall gain of the amp. If there were no negative feedback and marginal drive current, the lower DC gain could result in a slightly lower signal (a somewhat distorted signal) at higher current output conditions. Since the amp uses negative feedback, the output will not vary significantly (no audible difference) as long as the amp is operated within its design parameters. The drive circuit simply drives the output transistors harder (when necessary) to compensate for the lower DC gain.
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Old 19th April 2007, 04:21 AM   #24
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***bangs head against wall.

mmmm just wondering...i was playing with the numbers
say:

36v x 1.7current x 10 transistors = 612 watts
22v x 3.5current x 10 transistors = 770 watts

it seems to be inline with the power ratings.

could these numbers be used to deduce power output?

or am i just screwing up again? :bawling
---------------------------------------------------

im guessing that signal gain is done at the opamps via resistors?
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Old 19th April 2007, 04:59 AM   #25
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The easiest way to get real world power output is to drive the amp into a dummy load. Using the output voltage just before clipping and the load resistance, you can get reliable, repeatable power ratings.

If the rails hold 'precisely' to the regulated voltage, you can use the rail voltage and the load impedance to calculate the power output but it won't be exact due to losses in some of the components. The emitter resistors will drop a bit of voltage. The transistors will not be able to drive all of the way to the rail. The Vce(sat) curves will tell you approximately how much voltage you'll lose at various current levels. These are not going to cause audible differences in power output levels between the amps.

If you use the rail voltage (assuming that it can be held) and the load impedance, you will get 18 amps for the 2150 and 44 amps for the 250. Those values would be divided by 5 (the number of output transistors in parallel) to determine the peak current through the output transistors.

The overall gain of the amp is determined by the various preamp stages and the gain of the power amplifier section (inside the feedback loop). All of the various stages of gain are determined by resistors (voltage dividers).
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Old 19th April 2007, 07:04 AM   #26
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Default Predicted maximum output power

Quote:
Originally posted by Clipped
[B36v x 1.7current x 10 transistors = 612 watts
22v x 3.5current x 10 transistors = 770 watts

it seems to be inline with the power ratings.

could these numbers be used to deduce power output?[/B]
yes, a designer could use the cold transistor power specification to get a handle on whether his chosen transistors can supply the desired output power.

An Example.
5pair of 150W devices gives 750W per half side of the push pull ClassAB amplifier.
For max power into a 45degree phase angle load divide by 2 resulting in maximum output power of about 375W into a 45degree load.
Into a severe load i.e. 60degree phase angle, divide by 3.
Resulting in about 250W into a 60degree load.
This method works fairly closely for 4 and 8ohm reactive loads. For the typical voltages used in CAR amplifiers and the extremely low load values you can probably squeeze a bit more out of the transistors, maybe +10% to +20%.

If the amp uses 5 pair of 70W devices operating into low phase angle, low value loads then [350/2]*1.2 might get into the ballpark. i.e. about 210W into 2ohms. When bridged this equates to 420W into 4ohms. Note that when bridged the amp is effectively a 10pair arrangement and 10*70/2*1.2=420W (same prediction).

If CAR amplifier manufacturers are specifying more power they are telling lies about the amplifier's suitability to drive reactive loads.
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Old 19th April 2007, 04:53 PM   #27
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im contemplating whether or not its a good idea to ask you what a phase angle load is
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Old 19th April 2007, 05:41 PM   #28
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Hi,
Quote:
what is a phase angle load
= a speaker.

The more severe the speaker phase angles, the more likely an amplifier will damage itself trying to drive current into the load or go into self protection mode. Neither of which are condusive to good sound.
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Old 19th April 2007, 06:06 PM   #29
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What he means is that the relation of the voltage and current being drawn is not exactly aligned at zero degrees. If you imagine two sinewaves, one is current and one is voltage, and one is shifted left or right from the other a bit, that bit can be described by phase angle. Thus it should be apparant that once phase angle of a load happens, the effective power draw becomes heavier.
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