Calculating a Hybrid Bridge Tied Load

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Aug 14, 2006
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A recent post re-introduced the idea of connecting speaker loads using both standard (stereo/dual) connection and bridge tied load in a single implementation. The given elements are:
all three speaker loads are 8 ohms
one of the speakers will receive 6db greater signal
the amplifier will “see” a 4 ohm load as the amp is bridgeable to 4 ohms


If you haven't practiced doing these computations, the steps are all broken out and the work is shown.


Here is the original proposed schematic solution:


View attachment 16149


If you don't have your Ohm's law and power calculations memorized find a power wheel cheat sheet here:


View attachment 16153



In order to analyze the current, power and load on an amplifier we will make some assumptions. The original solution stipulates a 6db greater amplitude for the bridged tied load. To provide an easy to understand power relationship the normally connected speaker pair will be set at a very typical 100 watts each. Now let's calculate the current and voltage.


View attachment 16150


Piece of cake right? Drop in the numbers and we find 3.54 amps of current will flow when 100 watts are dissipated in the load. The amplifier will have 28.28 volts at it's terminal (let's call it V term) when doing so.


View attachment 16151




Under the same conditions, the same V term, what are the conditions in a bridged load (BTL or Bridge Tied Load). The amplifier must have the capability of being bridged. To do so the amplifier must be able to drive both sides with the same signal and invert the output of one side (usually #2). What this means is that whatever signal is output on one side, it's mirror image is produced on the other. If one side outputs +2 volts the mirror outputs -2 volts. As one terminal swings up in voltage the other swings down and vice versa. Each terminal of the BTL puts out half the voltage [differential] to the load. As a result, the voltage across the load is also doubled. If voltage is doubled and resistance remains the same, the current is doubled as well. With voltage and current doubled the power dissipated in the load is now 4X higher than in a single load run off one side of the amplifier.


Remember that connecting an amplifier in BTL extracts no more power than two normal loads sized for maximum power transfer. The maximum BTL load is half that ( two times more resistance) of a normally connected load.


It may help to imagine an electrical center or balance point in the BTL load. At the center point electrically mid-way between the two terminals, the voltage is always zero as either side's voltage swings up or down in unison with its respective terminal. Think of one side pushing and the other pulling. When the current reverses again, the same side is pulling and other pushing.


The current in all parts of a series circuit is the same. While each terminal of the BTL supplies only half the voltage, each terminal supplies the same current. When only the BTL load is present, the current at both terminals must be the same. Under normal BTL conditions this is the case. Standard textbook and instruction manual descriptions depict the load connected only to the two hot terminals.


View attachment 16152
This is what you've been waiting for, the final load calculation. Kirchoff says that the total current is equal to the sum of the branch currents. Both amplifiers have two paths for current. One path is shared with the other positive terminal (7.07 Amps) and the second path returns to the low terminal on the same side of the amplifier (3.54 Amps). The total current flowing from each terminal is 10.61 amps. The resistance = the voltage at the terminal / total current = 2.67 ohms.


To double check the figures, square the current and multiply it by the calculated resistance = 112.57 * 2.67 = 300. This is all of the normally connected load and half the BTL.


A final word on BTL and the possible 6db increase of output. In the hybrid configuration shown in fig 1, the BTL load will always be driven with 6db more signal than the normally configured loads. But how does this fit the power curve of your amplifier? The power capability of an amplifier will follow a curve which rises as the load impedance decreases, roughly following ohm's law. Below a certain load impedance, output falls, in some cases steeply. Where this occurs will vary from amp to amp but will most certainly occur below the amplifier's minimum output impedance . Many amplifiers may be capable of driving 2 ohm loads but often at a reduced power rating. The 6db of additional drive you might be expecting will not be available as the output falls attempting to drive a load at an impedance lower than it was designed for. Distortion figures will suffer, the amp may overheat and in the worst case situation, fail completely. Super amplifiers do exist that will drive very low impedance loads but they are more rare than common and certainly more expensive. Be aware of where on the load impedance / power output curve your configuration will be before assuming your configuration can benefit from a BTL configuration.



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Thanks! That is some good reference reading!
 
Which is actually less of a load than a bridged amp would see if it were set up with (2) 8 ohm speakers in the conventional bridged setup!

In a normal bridged 4 ohm setup the amp see 4 ohms but will act as if it has a 2 ohm load!

28.28 X 2 = 56.56/4 = 14.14 X 56.56 = 800 watts
 
I like!!!! :D:D:D:D:D
 
The problem with the "myth buster" posted is that it is comparing a single 8 ohm speaker versus 3 8 ohms in a hybrid setup.

Any 4 ohm bridge capable amp will follow what I posted above with 2 8 ohm or a single 4 ohm load in bridged mode. So even given that you are using 3 8 ohm speakers the actual load on the amp is less than 2 8 ohm speakers in a conventional bridged setup!
 
The nicest thing you ever said

Thunder said:
The problem with the "myth buster" posted is that it is comparing a single 8 ohm speaker versus 3 8 ohms in a hybrid setup...
Click to expand...

Steve, that's the nicest thing you ever said to me. I'm a big fan of Mythbusters and think they do a pretty good job. I can only aspire to do as well as they. Now, on to your myth.


The question has always been what impedance does your speaker network present to either terminal of the power amp?

Do you still hold it is 4 ohms?

How did you calculate it?

May we see the calculations?
 
I never said what the load was only what the amp sees! Only that it sees the same load with three 8s in a hybrid setup as it does with 2 8s in a conventional bridged setup!

But actually I was wrong the amp actually sees less of a load with three 8s in a hybrid setup than it does with a conventional 4 ohm bridged setup.

Are you saying that 2 8s are more of a load than three or is it a fact that 2 8 ohm speakers is a two ohm load?

My calculations using the same amp criteria as in your post for two 8s!

28.28 X 2 = 56.56/4 = 14.14 X 56.56 = 800 watts

28.28/14.14 = 2 ohm
 
What does "sees" mean?

Thunder said:
I never said what the load was only what the amp sees! ...
Click to expand...


Steve, as you well know, amplifiers don't have eyes. Amplifiers are imperfect current sources. When operated within design specifications they respond largely, if somewhat imperfectly, to ohm's law. An eight ohm load will draw roughly half the current of a four ohm load etc. Whenever a co-worker said to me “it sees” that meant it functioned in the same way Ohm's Law would dictate.


When you say the Amplifier “sees”, do you mean effectively? If an amp sees four ohms, then a four ohm capable amplifier can drive it?
 
From Houston's thread - How would you hook up these three components
Thunder said:
Since I thought this was a teaser I want to throw one out to those in the know.

I have an amp that is bridgable to 4 ohms, how can I run three 8ohm speakers on this bridged amp and still only have a 4ohm load?
Click to expand...

If there is a four ohm load then a four ohm capable amplifier should be able to drive it - no?
 
Ok, now you aren't making any sense! I just said yes, I also showed you that the actual load to the amp in a hybrid setup using three 8 ohm speakers is less than the load the same amp would have if it were setup in a conventional bridged mode using 2 8 ohm speakers!

Or did this all go over your head!
 
Ok... Here's the bottom line question.

If the amp is only rated stable to 4 Ohms, and you hook it up as described, which shows the load at only 2.67 Ohms, the potential is that the amp will become unstable, potentially overheat, etc? Am I right in thinking this? Or is there some super secret amplifier magic that's going on here, which violates Ohm's law?
 
Perhaps the same could be said about you

Steve, I've been on the same tack since the beginning. That was the purpose of the calculations. You can't say you didn't know what was being determined, all the work was shown and illustrated. You rejected the conclusion and stated “The problem with the "myth buster" posted is that it is comparing a single 8 ohm speaker versus 3 8 ohms in a hybrid setup.”


The conclusion of the calculations is each terminal of the amplifier “sees”, works into, supplies current is consistent with a 2.66 ohm load. When a network with multiple conditions is analyzed it has to be normalized otherwise how do we know what it does? In this case, the issue was the total load and therefor the current demanded by the load at either terminal. What is the effective load? The amplifier did not “see” four ohms.


My problem from the start has been that someone will connect up that hybrid circuit (valid but obscure as it is) and not understand the effective load it presents. The load cannot be driven by a four ohm capable amplifier. The load the amplifier “sees” is 2.66 ohms.


I'll have to take a break here as I'm already late for an appointment.
 
And as I pointed out using the very same formula that was presented in the material you used, that it is still less of a load than two 8 ohm speakers. And I am refering to an amp that is 4 ohm capable in Bridged mode!

Do the math using the formula you presented! and you will see I am correct "again"!:sqwink:

Yes there is some magic involved!
 
Let's take it one better, let's complete the formula in the authors example for a single 8 ohm speaker bridged in the same amp!

28.28 X 2 = 56.56/8 = 7.07 X 56.56 = 399.89 watts

Now let's double check that using his same formula

28.28/7.07 = 4

Opps was there a mistake here or has the 8 ohm speaker magically changed to a 4 ohm speaker!

Now do you get the idea that 2.67 is actually less of a load on the amp than 2 ohms!

Again using the same formula for two 8 ohm speakers conventionally bridged


28.28 X 2 = 56.56/(8/2) = 14.14 X 56.56 = 800 watts

28.28/14.14 = 2 ohm

If the amp is capable of powering two 8 ohm speakers in a conventional bridged mode (showing 2 ohms by his own formula) would not common sense tell you that the same amp would have no problem supporting three in hybrid mode (showing 2.67 ohms again by his own formula)?

The truth is the guy was trying to play myth buster but failed to apply his own math equally across the board! Had he been trying to be honest he would have used two 8 ohm or a single 4 ohm speaker in the conventional bridged example, and he would have completed the formula for the single 8 ohm speaker as well, but he couldn't do that because it would have made that magical number change appear!

BTW I wasn't calling you the myth buster I was refering to the guy who authored the document you posted! You were just taking what you were reading at face value without checking any of the math!
 
Actually, I did a basic circuit, ran calculations and came to the conclusion that the only way to make this understandable and put it to rest is to do something so incredibly simple that the math would be undisputed. In other words, I used really easy to understand numbers and very simple formulas. The only way it would be easier is if I drew pictures.


Here's what I did.

3 Terminals - +24V/Ground/-24V

This, I think we can agree is the 'typical' layout of a 'bridged' amplifier.

I loaded the terminals using the 'hybrid' description, and then with a 4 Ohm load as 2x8Ohm speakers. I can do the math for 2x8ohm, but I opted to do it for a 4Ohm load alone assuming we could at least agree that the resistance formulas aren't in question. I only used Ohms law, and the power formula.

Ohm's Law
E=IR

Power Formula
P=IE



Now on with the Hybrid Circuit:

With an 8 Ohm load across +24V to Ground:

24V/8Ohm=3A
3A*24V=72W

With an 8 Ohm load across -24V to Ground:
24V/8Ohm=3A
3A*24V=72W

Across -24V and +24V:
56V/8Ohm=6A
6A*48V=288W

Total Wattage=432Watts
Total Current 'Drawn' = 12Amps

Solve for Voltage:
E=P/I = 432/12 = 36V (Remember, we're not applying the same voltage across the entire circuit - this is the 'equivilent' volatage, not VoltageApplied)

36V/12A=3Ohm

Nope, seems to me that 'hybrid' circuit is applying the equivilet of a 3Ohm Load.



Now the 'Standard' Circuit:

-24V to +24 V using 1 4Ohm speaker:

48V/4Ohm = 12Amps

12Amps*48Volts=532 Watts

The math comes out the same if using 2 x 8 Ohm loads with 288 Watts Per speaker with 6 amps of current.

E=P/I=532/12=48V

48V/12A=4 Ohms.





While both circuits draw the same 'current' the speakers in circuit 1 are underpowered, applying the equivilent of a 3Ohm load, and being damaged because the DJ is cranking the volume, causing clipping, all while just trying to 'fix' it.


In circuit #2 both speakers are running great, we have a spare in the truck, and we dance and party all night.



I don't care how you do your math, what tricks you try to claim, or anything else. The hyrbid 'looks' like a 3 Ohm load, draws the same current, at less wattage, and is generally a very bad idea.
 
Meister,

Then how do you reconcile your formula with that of the author of the original posted formula?

And your mistake in the math is that you are using the total of the current into the total of both the plus and minus volts, where the total of the ohms should be divided into the positive volts ie (24+ volts)! For you for your 4 ohm speakers to bridged to get the Ohm load it would be 24/12 which would give you an ohm load of 2 ohms (according to the original posted formula)!

Which regardless of how you want to play the numbers is still a higher load on the amp than a Hybrid setup.

And just for your information I used the hybrid setup for years without any problems what so ever but then I was using rock solid Peavey gear! Granted in Gear like Crown, EV, QSC and such it may have been a totally different experience!
 
There isn't a mistake in my math. Each and every calculation is based on the applied voltage to the load. That's how ohm's law works. Voltage is the difference in potential between two points - and that's what was used in every calculation I did. If you think something other than applied voltage should be used, then you might want to review Ohm's laws, and some basic electronics theories.


As for the original post, I did mine completely apart from him using very easy to understand math and standard formulas. The only mistake I see in the original formula is using wattage as a starting point and getting too far into the weeds for most people to keep up. If he had started with voltage and resistance, then calculated using the same method I did he would have come up with the exact same answer. No matter how you change the values - the math I showed will ALWAYS come up with the same answers for a load - 3 ohms for the hybrid, 4 ohms for the 2 speaker setup.

As for your having used the hybrid setup: Just because you can, or did, do something, doesn't mean it's a good idea. With cheaper equipment it would have likely meant a melt down. And as you can see from the math, the wattage at those speakers means 2 of them are likely underpowered - which means they're not efficient. Most people will continue to turn up the volume to compensate - well into clipping in many cases.

Short of firing up skype and running through this on paper, over and over - and getting an electrical engineer to agree with me, or a fleet of electrical engineers - you still tell me that I'm wrong because you did it, and it worked. I'm just saying that you're lucky because you weren't using cheap equipment because the amp was pushing a 3 ohm load.
 
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