Take 1 driver. 1 voice coil, 1 cone, one amplifier. Driven with 2.83 volts rms the driver produces 88dB. As we are assuming the driver has a perfectly flat impedance, that is 8 ohms, the driver also draws 0.35 amps rms from the amplifier.
If you now take a second driver and wire it in series with the first you've got 2 voice coils, 2 cones and one amplifier. The drive level is exactly the same = 2.83 volts rms, . In series the impedance doubles to 16 ohms, 2.83 volts rms is still dropped across the combined voice coils. As we're dealing with 16 ohms now the current halves to 0.175 amps rms.
0.175 amps now flows through each voice coil meaning that each coil has 2.83/2 volts rms across it.
Lets look at this in terms of power.
P = V*I
In the first instance the driver sees P = 2.83*0.35 = 1 watt.
In the second instance each driver sees P = (2.83/2)* 0.175 = 1/4 of a watt. Both drivers together = 1/2 of a watt. So in going from 8 to 16 ohms the total electrical power has halved.
The method of scaling powers against one another =
10log * (P2/P1) = 10log * ( 0.5/1) = -3dB.
So in other words you lose 3dB of electrical power when you wire two drivers in series when compared to a single driver on its own. However this isn't the end of the story. As you've got two cones with the two drivers in series this also means you've got twice the radiating area or twice the acoustic power vs the single driver, the net effect is that you lose 3dB of electrical power, but you gain 3dB of acoustic power so in the end you're still at 88dB.
Now lets look at parallel. Two coils, two cones, one amplifier, but now the coils are wired in parallel with one another.
Each 8 ohm coil sees 2.83 volts RMS and as we saw for the first driver on it's own this = 0.35amps rms or 1 watt. As we've got two coils in parallel this time and because each coil has 2.83 volts rms across it we've now got a total of two watts of electrical power.
If we pop this into the power formula we now get.
10log * (2/1) = +3.
So we've doubled the electrical power and gained 3dB because of it. Now though as we've got two cones vs one we've also doubled the acoustic power, so we've gained another 3dB too, so we're up +6dB vs the single driver on its own.
At least I think I've got that right, it's been a while since I've thought about it.
If you now take a second driver and wire it in series with the first you've got 2 voice coils, 2 cones and one amplifier. The drive level is exactly the same = 2.83 volts rms, . In series the impedance doubles to 16 ohms, 2.83 volts rms is still dropped across the combined voice coils. As we're dealing with 16 ohms now the current halves to 0.175 amps rms.
0.175 amps now flows through each voice coil meaning that each coil has 2.83/2 volts rms across it.
Lets look at this in terms of power.
P = V*I
In the first instance the driver sees P = 2.83*0.35 = 1 watt.
In the second instance each driver sees P = (2.83/2)* 0.175 = 1/4 of a watt. Both drivers together = 1/2 of a watt. So in going from 8 to 16 ohms the total electrical power has halved.
The method of scaling powers against one another =
10log * (P2/P1) = 10log * ( 0.5/1) = -3dB.
So in other words you lose 3dB of electrical power when you wire two drivers in series when compared to a single driver on its own. However this isn't the end of the story. As you've got two cones with the two drivers in series this also means you've got twice the radiating area or twice the acoustic power vs the single driver, the net effect is that you lose 3dB of electrical power, but you gain 3dB of acoustic power so in the end you're still at 88dB.
Now lets look at parallel. Two coils, two cones, one amplifier, but now the coils are wired in parallel with one another.
Each 8 ohm coil sees 2.83 volts RMS and as we saw for the first driver on it's own this = 0.35amps rms or 1 watt. As we've got two coils in parallel this time and because each coil has 2.83 volts rms across it we've now got a total of two watts of electrical power.
If we pop this into the power formula we now get.
10log * (2/1) = +3.
So we've doubled the electrical power and gained 3dB because of it. Now though as we've got two cones vs one we've also doubled the acoustic power, so we've gained another 3dB too, so we're up +6dB vs the single driver on its own.
At least I think I've got that right, it's been a while since I've thought about it.
Yes. I see. Not to difficult and a great explanation. I have copied it for future reference as a reminder. So I keep the ohms for the "cones and domes" between 4 and 8 ohms and this and my SPL are what guide me in this decision. If I am running three drivers could I run two in series and one parallel and it still sound natural? Just curious. Also. Is it better for distortion reasons on the driver to choose series or parallel? Would distortion be a guiding factor?
Thanks 5th element!
Thanks 5th element!
The only reason not to wire drivers in series is the back-EMF created when the coil moves in the magnetic gap. As the drivers are (I assume) the same, this won't be a problem. The only time when it's really a problem is when you're trying to do a series crossover, and the back EMF from the woofer drives the tweeter, giving increased tweeter excursion (therefore distortion) around the crossover point - it's attenuated below the crossover point.
Can't see anything wrong with it from here.
Chris
Years ago that was a constant theme as the increased load was thought to kill the amplifier's Damping Factor. As has been shown, Damping Factor is just about the most worthless spec in an amplifier's design, although THD is another spec that is just about as worthless.
Best Regards,
TerryO
Hi
Just to throw a little spanner in the works . with speakers we are talking impedance and not just resistance. Most drivers will drop below the rated impedance. This depends on the driver in use . The impedance can almost half with some drivers depending on the input frequency.
Regards Ian
Just to throw a little spanner in the works . with speakers we are talking impedance and not just resistance. Most drivers will drop below the rated impedance. This depends on the driver in use . The impedance can almost half with some drivers depending on the input frequency.
Regards Ian
Actually the damping factor doesn't change with series connected drivers. If you had an 8 ohm woofer and put 8 ohm resistor in series then you would seriously degrade damping. But if you take and 8 ohm woofer and put a second, well matched 8 ohm woofer in series, then the damping doesn't change.
The reason is that the two units have identical impedance curves so they always split the input voltage in half. Since the voltage to either woofer's terminals remain flat then the damping factor must be unchanged.
Always hard to get your head around this but it is a fact.
David S.
I was just mentioning what was being said at the time.
The whole Damping Factor thing is pretty much a non-issue anyway.
Best Regards,
TerryO
Damping factor is a non-issue once it gets sufficiently high, so a damping factor of 1000 isn't any better than a damping factor of 100. However if it drops low enough it can be very significant.
I wasn't disagreeing with your comment but did want to clarify a point that commonly causes confusion.
Regards,
David
I wasn't disagreeing with your comment but did want to clarify a point that commonly causes confusion.
Regards,
David
Crown used to market their amplifiers pretty aggressively on the selling point that they had very high damping factors (and with it, the tightest, most awesome bass ever), maybe this is why they don't seem to be doing that anymore.
Seems likely, I know that many McIntosh amps have had a fairly low Damping Factor and both Nelson Pass and Frank Van Alstine aren't bothered by it very much, if at all.
As speaker dave pointed out it can be pretty low. While different designers have their own opinions, the general sense I get from it is that anything below a 12-14 DF may have problems.
Best Regards,
TerryO
In the end it is about the range of variation of the Z of your loudspeaker and how many dB response variation you are willing to allow. The math is straightforward.
For example an 8 ohm speaker and a damping factor of 10 means the amplifier Z is 0.8 (1 tenth of 8 ohms). Lets say the speaker rises to 4 times its rated Z (32 ohms). At frequencies where it is 8 ohms the open circuit voltage that gets to the woofer (assume an open circuit 1 volt) is 8/8.8 = 0.909 volts. At frequencies where it is 32 ohms then the voltage to the woofer is 32/32.8 = 0.976 volts. 20 x log (.976/.909) = 0.62dB. So a damping factor of 10 and an impedance range of 4 to 1 gives 0.62dB response variation.
I can live with that.
David S.
For example an 8 ohm speaker and a damping factor of 10 means the amplifier Z is 0.8 (1 tenth of 8 ohms). Lets say the speaker rises to 4 times its rated Z (32 ohms). At frequencies where it is 8 ohms the open circuit voltage that gets to the woofer (assume an open circuit 1 volt) is 8/8.8 = 0.909 volts. At frequencies where it is 32 ohms then the voltage to the woofer is 32/32.8 = 0.976 volts. 20 x log (.976/.909) = 0.62dB. So a damping factor of 10 and an impedance range of 4 to 1 gives 0.62dB response variation.
I can live with that.
David S.
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Is it only a variance in response that's the product of DF? I always thought there were "speaker control" effects.
Perhaps I should read more about it.
Edit: Yes, from Wikipedia:
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That is one of the great myths of loudspeakers.
Of course electrical damping is happening and the amplifier source impedance (in conjunction with the voice coil DCR) determines how much electrical drag is exerted and so determines the system Q. But... in the end it is all fully covered if you view it as simply how flat are the electrical volts at the woofer and how does this change if the amplifier source impedance changes. The answer to that comes from looking at the woofer impedance curve.
If the amplifier source impedance rises (DF falls) you tend towards more constant current and the woofer drive voltage looks less flat and more and more like its impedance curve. That is, the woofer's normal response under low Z, high DF drive gets equalized by the impedance curve (or some proportion of the impedance curve) as source impedance rises. The calculations I did were just a shorthand look at how the lowest and highest points of the impedance curve would perturb the frequency response with the given 0.8 ohm source impedance.
If you think about woofer "damping" you will tend to get confused. The variation in driving volts due to source impedance is whats really going on.
David
Of course electrical damping is happening and the amplifier source impedance (in conjunction with the voice coil DCR) determines how much electrical drag is exerted and so determines the system Q. But... in the end it is all fully covered if you view it as simply how flat are the electrical volts at the woofer and how does this change if the amplifier source impedance changes. The answer to that comes from looking at the woofer impedance curve.
If the amplifier source impedance rises (DF falls) you tend towards more constant current and the woofer drive voltage looks less flat and more and more like its impedance curve. That is, the woofer's normal response under low Z, high DF drive gets equalized by the impedance curve (or some proportion of the impedance curve) as source impedance rises. The calculations I did were just a shorthand look at how the lowest and highest points of the impedance curve would perturb the frequency response with the given 0.8 ohm source impedance.
If you think about woofer "damping" you will tend to get confused. The variation in driving volts due to source impedance is whats really going on.
David
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