Calculating a Hybrid Bridge Tied Load

To many ads? Support ODJT and see no ads!
My point well taken merely pointing out that through the muck in the end if you cant add or subtract you will disagree with calculations presented in this thread. No matter how you dumbed down the Theory.
 
Here is the conclusion I've come to:

Each channel of the hybrid configuration is seeing a 2.67 ohm load at 300 watts. If the amplifer is capable of 2 ohm loads per channel there should be no problem with this configuration.
 
MIXSIRS said:
Here is the conclusion I've come to:

Each channel of the hybrid configuration is seeing a 2.67 ohm load at 300 watts. If the amplifer is capable of 2 ohm loads per channel there should be no problem with this configuration.
Click to expand...

In the words of an Algebra teach I used to have....

Can you show your work? ;)

The value of the resistance shouldn't change so the difference is how we're going about calculating it. I think there's a fault in the math or theory somewhere along the way.. and I should probably print out the original post so I can review it more closely.
 
Rebuttal.

Thunder said:
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!
Click to expand...


It didn't take long to get nasty. Imply what you will but do not imply I'm a thief or a plagiarist. That work is mine. All of it. My text, my schematics. If I reference some one's work they are credited for it. Normally I post the URL. If I quote someone, it will be attributed. You may also be assured it [the math] was checked by two other people. I'm not saying I couldn't make a mistake, simply that none as of yet have been found. I believe it to be good accurate work.


Thunder said:
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!
Click to expand...


I assume you are referencing my figure 3 above?
Thunder, why did you halve the voltage? You can't do that. The load is across the whole supply. Therefor you have to apply the whole voltage. This is fundamental stuff.


R = E / I
R = 56.56/ 7.07 = 8 ohms
That matches the specified load. What's the problem?


Thunder said:
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!
Click to expand...


The analysis was made on the circuit network you suggested. Why would I change the values? All the loads were supposed to be 8 ohms. If I did[change them] everyone can be sure you'd complain about that. There was scrupulous attention paid to detail and fairness. Whatever numbers I'd use you'd complain. Thunder, they're your numbers. You'd complain if they hung ya with a new rope.


Thunder said:
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!
Click to expand...


You shouldn't concern yourself. Be assured I took it as sarcasm. We've come to expect it from you.


There is real life outside and I must answer its call.
 
Amplifier

100 watt into 8 ohms per channel
200 watts into 4 ohms per channel
400 watts into 2 ohms per channel

400 watts into 8 ohms bridged
800 watts into 4 ohms bridged


Each 8 ohm speaker connected in stereo is receiving 100 watts into 8 ohms from its channel

The 8 ohm bridged speaker is receiving 200 watts into 4 ohms from each channel

The load on each channel is 2.67 ohms, 4 ohms from the bridged speaker and 8 ohms from the non bridged speaker.
 
Balanced Line said:
It didn't take long to get nasty. Imply what you will but do not imply I'm a thief or a plagiarist. That work is mine. All of it. My text, my schematics. If I reference some one's work they are credited for it. Normally I post the URL. If I quote someone, it will be attributed. You may also be assured it [the math] was checked by two other people. I'm not saying I couldn't make a mistake, simply that none as of yet have been found. I believe it to be good accurate work.





I assume you are referencing my figure 3 above?
Thunder, why did you halve the voltage? You can't do that. The load is across the whole supply. Therefor you have to apply the whole voltage. This is fundamental stuff.


R = E / I
R = 56.56/ 7.07 = 8 ohms
That matches the specified load. What's the problem?





The analysis was made on the circuit network you suggested. Why would I change the values? All the loads were supposed to be 8 ohms. If I did[change them] everyone can be sure you'd complain about that. There was scrupulous attention paid to detail and fairness. Whatever numbers I'd use you'd complain. Thunder, they're your numbers. You'd complain if they hung ya with a new rope.





You shouldn't concern yourself. Be assured I took it as sarcasm. We've come to expect it from you.


There is real life outside and I must answer its call.
Click to expand...

I apologize, I didn't know you had actually authored the document you posted, I also did not know that you wrote the Peavey Pro Sound hand book! Because that is exactly the same diagram that is in that book for hybrid setup (I know because that is where I learned it from back in the early 80's)

I am proud of my 8th grade education! LOL

Since I was using your figures how can I possibly be wrong?
 
Please apply the correct voltage in your formula.

If you apply the numbers incorrectly, you get the wrong answer. I answered what you asked in my rebuttal.

You applied the wrong voltage.

Apply the right voltage, you'll get the right answer.

As I said this is pretty fundamental.



Never read the 1980's Peavey sound manual, never saw it. You described the network I drew it and then analyzed it.

Now I really do have to go.
 
Balanced Line said:
It didn't take long to get nasty. Imply what you will but do not imply I'm a thief or a plagiarist. That work is mine. All of it. My text, my schematics. If I reference some one's work they are credited for it. Normally I post the URL. If I quote someone, it will be attributed. You may also be assured it [the math] was checked by two other people. I'm not saying I couldn't make a mistake, simply that none as of yet have been found. I believe it to be good accurate work.





I assume you are referencing my figure 3 above?
Thunder, why did you halve the voltage? You can't do that. The load is across the whole supply. Therefor you have to apply the whole voltage. This is fundamental stuff.


R = E / I
R = 56.56/ 7.07 = 8 ohms
That matches the specified load. What's the problem?





The analysis was made on the circuit network you suggested. Why would I change the values? All the loads were supposed to be 8 ohms. If I did[change them] everyone can be sure you'd complain about that. There was scrupulous attention paid to detail and fairness. Whatever numbers I'd use you'd complain. Thunder, they're your numbers. You'd complain if they hung ya with a new rope.





You shouldn't concern yourself. Be assured I took it as sarcasm. We've come to expect it from you.


There is real life outside and I must answer its call.
Click to expand...

I apologize, I didn't know you had actually authored the document you posted, I also did not know that you wrote the Peavey Pro Sound hand book! Because that is exactly the same diagram that is in that book for hybrid setup (I know because that is where I learned it from back in the early 80's)

I am proud of my 8th grade education! LOL

Since I was using your figures how can I possibly be wrong?

Seems you are using two different formulas when it comes to hybrid versus a single 8 ohm!


56.56/ 7.07 = 8 ohms
That matches the specified load. What's the problem?

Versus

28.28/ 10.61 = 2.67

Again if we are using your formula them we end up with this.

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

28.28/ 14.14 = 2 ohm ****

Again I am using your own figures and figuring it for both setups using the same formula just using them consistantly for both setups.

However, I can post it the other way as well!

56.56/ 14.14 = 4 ohms for the conventional bridged setup

Versus

56.56/ 10.61 = 5.33 ohms for the hybrid setup

Opps it seems like you're wrong again, these are based on your numbers and your formulas!
 
I'll answer later, I don't have time now. In the meantime, it is all laid out in my original post.
 
When you experts finally figure out what you are doing wrong I will explain it in detail! LOL
 
If I can find the Peavey hand book I will copy and paste it here for you!

The figure is almost a perfect copy (from my recollection) not the numbers!
 
You must log in or register to reply here.
Top Bottom
Back