Controversy regarding Bridged Amplifier power output

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
Administrator
Joined 2004
Paid Member
I spent a lot of time working with the 7294.. The 7295 as you say will operate with 26V rails into a 4 ohm load, but that is the outer limit, and you haven't accounted for what might happen with 10% high line - one of the newbie gotchas I didn't mention in my last post.

Depending on cost I would probably go with the 7294 instead and make sure that the supply voltage under high line conditions does not exceed 27V or so if you intend to drive 8 ohm loads in bridge mode.

It's a big juggling act, but you really do not want to exceed SOA under any foreseeable operating conditions. You need a fairly stiff supply that does not sag severely under load, and one that meets the design criteria for SOA at nominal line. Under load it may sag as much as 10% which would permit you to meet the SOA requirement under 10% high line conditions.

I would download Duncan Amps PSUD and do some basic measurements on your power transformer. (Unloaded AC voltage at nominal line voltage and primary and secondary DCR) With sufficient information you can model the behavior of your transformer under load and know with relative certainty that you will not fry the chips. Antek has a great line of inexpensive high performance toroids if yours does not suit. (I use and recommend them)

A note on heatsinking these. Use an aluminum bar across the body of the chip (not the tab) bolted to the heatsink. You can use a single bar for two devices with three 8-32 screws to clamp it to the heat sink. The standard arrangement does not work well due to the stiff leads and the difficulty of getting uniform contact across the entire tab. Use mica with grease for the lowest real thermal resistance - silpads are nice but never performed as well as claimed.
 
I spent a lot of time working with the 7294.. The 7295 as you say will operate with 26V rails into a 4 ohm load, but that is the outer limit, and you haven't accounted for what might happen with 10% high line - one of the newbie gotchas I didn't mention in my last post.

Depending on cost I would probably go with the 7294 instead and make sure that the supply voltage under high line conditions does not exceed 27V or so if you intend to drive 8 ohm loads in bridge mode.

It's a big juggling act, but you really do not want to exceed SOA under any foreseeable operating conditions. You need a fairly stiff supply that does not sag severely under load, and one that meets the design criteria for SOA at nominal line. Under load it may sag as much as 10% which would permit you to meet the SOA requirement under 10% high line conditions.

I would download Duncan Amps PSUD and do some basic measurements on your power transformer. (Unloaded AC voltage at nominal line voltage and primary and secondary DCR) With sufficient information you can model the behavior of your transformer under load and know with relative certainty that you will not fry the chips. Antek has a great line of inexpensive high performance toroids if yours does not suit. (I use and recommend them)

A note on heatsinking these. Use an aluminum bar across the body of the chip (not the tab) bolted to the heatsink. You can use a single bar for two devices with three 8-32 screws to clamp it to the heat sink. The standard arrangement does not work well due to the stiff leads and the difficulty of getting uniform contact across the entire tab. Use mica with grease for the lowest real thermal resistance - silpads are nice but never performed as well as claimed.

Thanks man, you are already my king!!!

The toroidal transformer is rated at:

150.9VA --> 20.1V X 2, 3.75A

An externally hosted image should be here but it was not working when we last tested it.


So power transformer is rated to have an output of two times 20.1VAC when it has its rated load. So its resistance has already dropped the open circuit voltage and its peak will be 1.414 times higher which is 28.4V peak. The single rectifier drops it to 27.4V and the ripple drops it to 26VDC.

Are the calculations right?
 
Last edited:
20.1Vac is an unusual transformer voltage specification.
Did you measure 20.1Vac?

If you have a 115/230:20+20Vac dual secondary 6% regulation transformer, then it's output voltage is 20Vac+20Vac when the output current is 3.75Aac (two 5r333 loads) and the mains input voltage at the primary tappings is 115/230Vac.

If mains voltage varies the output voltage varies.
If the load current varies the output voltage varies.
If the load is a capacitor input filter the output voltage varies.
You need to be able to allow for ALL these variations to make the informed decision on which secondary voltage you can use on your amplifier.

edit.
the pic shows a 230:20.1-0-20.1Vac centre tapped secondary without a current rating and without a VA rating.
 
Last edited:
Administrator
Joined 2004
Paid Member
Thanks man, you are already my king!!!

The toroidal transformer is rated at:

150.9VA --> 20.1V X 2, 3.75A



So power transformer is rated to have an output of two times 20.1VAC when it has its rated load. So its resistance has already dropped the open circuit voltage and its peak will be 1.414 times higher which is 28.4V peak. The single rectifier drops it to 27.4V and the ripple drops it to 26VDC.

Are the calculations right?

The calculations are reasonable based on the information you currently have. I would still however do the suggested transformer measurements and bear in mind that the drop across a standard silicon bridge might be 2.4V or more at full load which helps. Don't forget the effect of high line in your calculations, and be aware that the voltage at the peak of your ripple may also exceed the SOA of the device momentarily. The transformer rating is reasonable for hifi use with a single pair of amp chips for 100W output into 8 ohms on a noncontinuous basis.

Should the voltage prove to be too high in use you can use a set of pi-filters on the rails with 0.33 - 0.47 ohm 5W resistors (Caps: 4 x 4700uF or greater @ 35V - 50V) which will get your ripple down and drop the supply output voltage sufficiently I think to stay inside your SOA requirement. There is another benefit that shows up at clip where the amp goes open loop and lots of ripple shows up on the output - the ripple will be reduced and clipping while still very obnoxious will be somewhat less so without so much 100Hz ripple reaching the output..
 
Hi is this the first amplifier you have decided to build ? . If you like this hobby and want to keep liking it i suggest you dont build bridged amplifier as your first project.
Regards Mark. Ps if done right you will be amazed how much volume level you can get from a lower powered amplifier. Spl not watts
 
20.1Vac is an unusual transformer voltage specification.
Did you measure 20.1Vac?

If you have a 115/230:20+20Vac dual secondary 6% regulation transformer, then it's output voltage is 20Vac+20Vac when the output current is 3.75Aac (two 5r333 loads) and the mains input voltage at the primary tappings is 115/230Vac.

If mains voltage varies the output voltage varies.
If the load current varies the output voltage varies.
If the load is a capacitor input filter the output voltage varies.
You need to be able to allow for ALL these variations to make the informed decision on which secondary voltage you can use on your amplifier.

edit.
the pic shows a 230:20.1-0-20.1Vac centre tapped secondary without a current rating and without a VA rating.

This toroid came with my Logitech Z-2300. I want to use this toroid. I specifically contacted the maker Ten Pao to know its VA rating. I provided them with all the details labeled on the transformer. After a dozen request they replied it is a manufacture specific 150.9VA transformer.

My request to Ten Pao -->

Kindly inform me about be the VA rating of this toroidal transformer

Model No: TOG433028F0
Input: AC 230V/50Hz
Output: AC 20.1V x 2

This transformer is used by my Logitech Z-2300 and had gone dead. Since my Z-2300 is outside warranty & thus no support from Logitech, I need to buy one from the market. So, I need the VA rating of this toroidal transformer. I have repeatedly asked for help in the "Online Inquiry" section of you website with no reply.

Kindly please help me.

Thanking You,

RishiGuru
Country : India

Their answer after a dozen request :

Dear RishiGuru,

After confirming with our engineer, the VA rating for this model is 150.9, and our company didn't set any retail place in India, so please contact with your Logitech Z-2300 supplier.

Best Regards
Sam Su
Marketing Department
TEN PAO INTERNATIONAL LTD.
Tel: 86-752-2312309
Fax: 86-752-2312813
Skype: sam_su85
Email: sam@tenpao.com
MSN: sushan851028@hotmail.com
website: Ten Pao International Ltd

From: RishiGuru
Sent: Sunday, December 12, 2010 9:39 PM
To: mkt@tenpao.com
Subject: 線上查詢

----------------------------------------------------------------------------------------------------

The internal fuse of the toroid was blown and I rectified it in a local repairing shop. Now the toroid is as good as new.

The toroid is 150.9VA
So, it seems 150.9 VA --> 2 x 20.1 V, 3.75A

I have two 10,000uF, 35V capacitors that I want to use as ripple filters.

Now coming to the chip TDA7295, its spec sheet says:

The Output Peak Current of TDA7295 is 6A.

VS = ± 34V, RL = 8Ω --> 80W MAX
(***) VS = ± 26V, RL = 4Ω --> 80W MAX

Note (***): Limited by the max. allowable out current

What does the above sentence mean?

Does this in any way mean that if the current is less then 6A, say 3.75A, then it can handle much higher voltage?
 
Last edited:
Administrator
Joined 2004
Paid Member
This toroid came with my Logitech Z-2300. I want to use this toroid. I specifically contacted the maker Ten Pao to know its VA rating. I provided them with all the details labeled on the transformer. After a dozen request they replied it is a manufacture specific 150.9VA transformer.

<snip>
Kindly please help me.

Thanking You,

RishiGuru
Country : India

Their answer after a dozen request :

Dear RishiGuru,

After confirming with our engineer, the VA rating for this model is 150.9, and our company didn't set any retail place in India, so please contact with your Logitech Z-2300 supplier.

<snip>

The internal fuse of the toroid was blown and I rectified it in a local repairing shop. Now the toroid is as good as new.
The toroid is 150.9VA
So, it seems 150.9 VA --> 2 x 20.1 V, 3.75A

I have two 10,000uF, 35V capacitors that I want to use as ripple filters.

Now coming to the chip TDA7295, its spec sheet says:

The Output Peak Current of TDA7295 is 6A.

VS = ± 34V, RL = 8Ω --> 80W MAX
(***) VS = ± 26V, RL = 4Ω --> 80W MAX

Note (***): Limited by the max. allowable out current

What does the above sentence mean?

Does this in any way mean that if the current is less then 6A, say 3.75A, then it can handle much higher voltage?

On peaks your capacitors are more than capable of delivering the 6A required so the short answer is no they cannot handle any higher voltage.

I'm also concerned that the thermal fuse in that transformer blew as that is usually the last line of defense before a transformer heats to the point where insulation breaks down. Depending on insulation class this transformer may have been stressed to the limit, and the fuse only senses the immediate locale, it is quite possible for there to be hotter spots in the insulation. Call me conservative, but I think the best place for that transformer is in the trash!! (dustbin)

A bridge will only give you 3dB more output as explained in several previous posts so unless you really need 100W you might want to consider this.
 
Last edited:
a 230:20.-0-20.1Vac 6% regulation transformer when fed with 230Vac +10% will have a maximum output after rectification and smoothing of ~
230 *1.1 / 230 * 20.1 * {1+6%} -400mV ~32.7Vdc
when very lightly loaded. The 35V capacitors are safe.

When quiescent current is drawn, the voltage will drop. Expect about a 1Vdc to 2Vdc reduction, i.e. ~30 to 32Vdc.
This is way above what is recommended for your proposal.

When the voltage is at a more normal 230+3% (during times of lower power demand) the expected voltage being fed to the amplifier to be between 28Vdc to 30Vdc.
This is still way above what you propose.
 
Last edited:
Bridging gives you twice the power into twice the load.

Sure, but what sense does that make?

You've got 8 ohm speakers. So you bridge your amp and then go out and buy 16 ohm speakers?

Sorry, but I just don't get the point of the phrase "bridging gives you twice the power into twice the load."

If you want to know what you would get into the same load, you first have to discover what one amp will give you into half the load (then double the power for bridging).

Assuming it can deliver the current and dissipate the power, and also depending on how the input signal is handled, bridging essentially gives you four times the power into the same load.

If you want to get the right answer, you have to ask the right question.

Ok. Don't recall seeing anyone asking how much power you get from bridging into twice the load.

se
 
If the amp can deliver the current without overheating or PSU droop then you can get four times the power. The issue is the 'if' at the start of the sentence. The factor of 4 comes from two factors of 2:
2 comes from bridging, and is always equal to 2
2 comes from halving the impedance seen by each amp, and can be less than 2

So the question boils down to what the amp does with half the load impedance. This is the question to ask. Ask the right question; get the right answer.
 
If the amp can deliver the current without overheating or PSU droop then you can get four times the power. The issue is the 'if' at the start of the sentence. The factor of 4 comes from two factors of 2:
2 comes from bridging, and is always equal to 2

Not always. It depends on how the input signal is handled in bridged mode. If you're simply splitting the signal to get an inverted signal to feed to the second channel, then you end up with a factor of 1, not 2.

2 comes from halving the impedance seen by each amp, and can be less than 2

The impedance isn't halved. The voltage across the load is doubled, again depending on how the input signal is handled.

So the question boils down to what the amp does with half the load impedance.

Well, if you're using 8 ohm speakers, and if the amp is capable of driving 4 ohm speakers, then just double the four ohm power rating, which for a decent amp will be roughly four times the 8 ohm rating when bridged.

se
 
Now you are confusing gain with power delivery. In the unlikely event of splitting the signal between the amps you get reduced gain, but the same maximum power output as I have been saying.

I will say it once more: to discover what power output you can get from N ohms in bridge mode, first find what you get for N/2 ohms with one amp then double it. That's all. Anything else is pure confusion. The reason this works is that each amp sees half the load; it can't distinguish between N/2 ohms connected to ground, and N ohms connected to an antiphase signal. Think of it as the opposite of bootstrapping, because that is exactly what it is.
 
Now you are confusing gain with power delivery.

No, I'm not.

In the unlikely event of splitting the signal between the amps you get reduced gain, but the same maximum power output as I have been saying.

If you simply split the input signal, say with a simple resistor divider or a 1:1 input transformer, the gain would remain the same as if you were driving just a single channel.

Each channel would get half the signal so when they're bridged, you're back to square one as far as gain goes.

I will say it once more: to discover what power output you can get from N ohms in bridge mode, first find what you get for N/2 ohms with one amp then double it.

Which, if it's a decent amp, would be roughly eight times the power into N.

The reason this works is that each amp sees half the load; it can't distinguish between N/2 ohms connected to ground, and N ohms connected to an antiphase signal.

This "sees half the load" is a source of confusion as well.

They don't see half the load. The reason the current doubles is because the voltage across the load has doubled. Which may or may not actually be the case depending on how the input signal is split.

se
 
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.