.
I can't say I've thought about. But it seems Mr. Elliott has:
Bridging Adapter For Power Amps
Simplest Ever Bridging Adapter for Amplifiers
.
I can't say I've thought about. But it seems Mr. Elliott has:
Bridging Adapter For Power Amps
Simplest Ever Bridging Adapter for Amplifiers
.
working with power is quite confusing for the question you have asked.
Work instead with Voltage.
eg
PSU is +-28Vdc when maximum output is passing and +-31Vdc when the output is open.
The loss through the amplifier is stated in the datasheet. Let's assume a value of 3Vloss.
The maximum unclipped peak voltage (when passing current through the rated load resistance) is therefore 25Vpk (28Vdc-3Vloss).
The maximum unclipped peak voltage when the load is open is 31Vdc-~1Vloss, ~ 30Vpk.
Now you have two amplifiers in parallel.
The maximum output is still 25Vpk.
Does the power output change?
No, because the maximum output voltage has not changed.
If you need those voltage converted to power use the standard formulae:
P=IV=I²R=V²/R (this applies to DC voltages and also to rms values of AC voltages).
If you use peak voltage of a sinewave, then the formulae need to be modified for the non rms/DC values and become
P=IpkVpk/2=Ipk²R/2=Vpk²/R/2 (this only applies to undistorted sinewave).
That 25Vpk is equivalent to 25*25/8/2 = 39W, pretty close to that 38W figure in the earlier post.
A bridged pair of amplifiers is capable of delivering double the power into double the load impedance. The total power of the two amplifiers is EXACTLY the same as the output from the bridged pair. For the above case of two amps with 25Vpk output, then they are capable of delivering 39W+39W into 8r0+8r0, i.e 78W into 16r.
Work instead with Voltage.
eg
PSU is +-28Vdc when maximum output is passing and +-31Vdc when the output is open.
The loss through the amplifier is stated in the datasheet. Let's assume a value of 3Vloss.
The maximum unclipped peak voltage (when passing current through the rated load resistance) is therefore 25Vpk (28Vdc-3Vloss).
The maximum unclipped peak voltage when the load is open is 31Vdc-~1Vloss, ~ 30Vpk.
Now you have two amplifiers in parallel.
The maximum output is still 25Vpk.
Does the power output change?
No, because the maximum output voltage has not changed.
If you need those voltage converted to power use the standard formulae:
P=IV=I²R=V²/R (this applies to DC voltages and also to rms values of AC voltages).
If you use peak voltage of a sinewave, then the formulae need to be modified for the non rms/DC values and become
P=IpkVpk/2=Ipk²R/2=Vpk²/R/2 (this only applies to undistorted sinewave).
That 25Vpk is equivalent to 25*25/8/2 = 39W, pretty close to that 38W figure in the earlier post.
A bridged pair of amplifiers is capable of delivering double the power into double the load impedance. The total power of the two amplifiers is EXACTLY the same as the output from the bridged pair. For the above case of two amps with 25Vpk output, then they are capable of delivering 39W+39W into 8r0+8r0, i.e 78W into 16r.
Last edited:
correct.
if the amplifier can not drive impedance less than 8 ohm than its true.
if it can drive a 4 ohm load, then things get a bit more complicated.
68W--4 Ohms--28V
38W--8 Ohms --28V
let us assume it stands true.
and let us assume you have an 8 ohm load.
if you bridge the amplifier,
you can not drive a load under 8 ohm. as both amps will see an 4 ohm load.
and will deliver the 68 -ish watt as they should.
in real life something around 120 watt is what you get.
just like andrewT wrote.
double the power into double the load.
if you have an 8 ohm speaker and use
LM3886 , you get 38W--8 Ohms --28V .
if you build it in bridge mode, you get about 120 watt into 8 ohm.
if you paralell 2 lm3886 chips into the same load, you get no extra power.
but you can use 2 ohm speakers.
and there you can get the same close to 120 watt as with bridge config.
if the amplifier can not drive impedance less than 8 ohm than its true.
if it can drive a 4 ohm load, then things get a bit more complicated.
68W--4 Ohms--28V
38W--8 Ohms --28V
let us assume it stands true.
and let us assume you have an 8 ohm load.
if you bridge the amplifier,
you can not drive a load under 8 ohm. as both amps will see an 4 ohm load.
and will deliver the 68 -ish watt as they should.
in real life something around 120 watt is what you get.
just like andrewT wrote.
double the power into double the load.
if you have an 8 ohm speaker and use
LM3886 , you get 38W--8 Ohms --28V .
if you build it in bridge mode, you get about 120 watt into 8 ohm.
if you paralell 2 lm3886 chips into the same load, you get no extra power.
but you can use 2 ohm speakers.
and there you can get the same close to 120 watt as with bridge config.
Hi that's why many people confuses (including me) regarding output power.
Datasheet shows this in page 6.
LM3886 in a two-device parallel solution driving a 4Ω load will typically provide 110W of output power.
Even ebay sells parallel amps using LM3886 mentioning 120W per channel into 4 ohms load.
Confusing.
Attachments
Pdmax on page 6 = dissipation.
Pmax,rms = Vpeak^2/(2*R)
With 35v rails and a 4 ohm load, that's 153 watts. Given losses, rail sag, etc, they're saying 110w. With 28v rails and a 4 ohm load, you're getting 98 watts--with losses that's your 68W.
Chips in parallel simply mean you can drive the current and manage the heat.
Pmax,rms = Vpeak^2/(2*R)
With 35v rails and a 4 ohm load, that's 153 watts. Given losses, rail sag, etc, they're saying 110w. With 28v rails and a 4 ohm load, you're getting 98 watts--with losses that's your 68W.
Chips in parallel simply mean you can drive the current and manage the heat.
What they are describing is 2x68 Watt with a 2 ohm load, not a 4 ohm load.
If you half the impedance then you can theoretically double the power, it also means doubling the current output capabilities of the amp and PSU.
If you wanted 4 times the power of a single LM3886 into 4 ohm, you would need a 1 ohm load and 4 lm3886 and a PSU that can supply 4 times the amount of current.
If you half the impedance then you can theoretically double the power, it also means doubling the current output capabilities of the amp and PSU.
If you wanted 4 times the power of a single LM3886 into 4 ohm, you would need a 1 ohm load and 4 lm3886 and a PSU that can supply 4 times the amount of current.
With two chips in PARALLEL you'll get the same output voltage as with a single chip if operating on the same power supply, hence, the output power will be the same: 38 W, 68 W in 8 Ω, 4 Ω, respectively for ±28 V supply.
However, two chips in parallel support more current, hence, can drive a lower impedance. An amplifier with two LM3886es in parallel will be able to drive a 4 Ω load to ±35 V rails. This allows you to build an amp that can drive a 4 Ω load and operate on a ±35 V supply. In this case, you'll get about 60 W into 8 Ω*and 120 W into 4 Ω.
~Tom
Hi thanks for the details.
Now I got it, the whole idea based on the graph on page 12 in LM3886 datasheet.
National has stopped the curve in 4ohms load upto 85Watts on 30V supply.So it is clear the chip can go more than that.The graph will move about 120W when feeding 35V on 4 ohms.
So better to parallel them and feed a 35V supply to share the current.So I can go even for 2 ohms sub as well in future.
Now I got it, the whole idea based on the graph on page 12 in LM3886 datasheet.
National has stopped the curve in 4ohms load upto 85Watts on 30V supply.So it is clear the chip can go more than that.The graph will move about 120W when feeding 35V on 4 ohms.
So better to parallel them and feed a 35V supply to share the current.So I can go even for 2 ohms sub as well in future.
Attachments
There are three things that can limit the output power:
- Output current limit
- Output voltage limit
- Dissipated power (temperature limit)
Current Limit:
While the LM3886 can typically provide 11 A of output current, it is only guaranteed to be able to provide 7 A across all operating conditions. This means a single LM3886 will only be able to provide 7 A. The resulting output power can be calculated as: Pmax = Ipeak^2 * R/2
So: 7^2 * 8/2 = 196 W into 8 Ω
7^2 * 4/2 = 98 W into 4 Ω
7^2 * 2/2 = 49 W into 2 Ω
7^2 * 1/2 = 24.5 W into 1 Ω
For a bridged amp, the 1 Ω number is relevant as each half of the amp "sees" half the load impedance. Hence a bridged LM3886 driving a 2 Ω*load (each half "seeing" 1 Ω) can only provide 24.5 W due to the current limit of the LM3886.
Voltage Limit:
The power supply voltage limits the output swing. In addition, there is also a small voltage drop across the LM3886 itself. This drop is about 2.5 V on the negative rail and slight smaller on the positive rail. I've used Vod = 2.5 V in my math.
If the amp can supply enough current to drive the load to the rails, you can calculate the output power as: Pout = (Vsupply - Vod)^2/(2*R).
±35 V, 8 Ω: Pout = (35-2.5)^2/(2*8) = 66 W
±35 V, 4 Ω: Pout = (35-2.5)^2/(4*2) = 132 W (this exceeds the output current limit!)
±28 V, 8 Ω: Pout = (28-2.5)^2/(2*8) = 41 W
±28 V, 4 Ω: Pout = (28-2.5)^2/(2*4) = 81 W
±28 V, 2 Ω: Pout = (28-2.5)^2/(2*2) = 162 W (this exceeds the output current limit).
Thermal limit:
When current flows through the LM3886, power is dissipated in the LM3886. That's why it gets hot during operation... Calculating the dissipated power gets a bit involved. The math is shown below (grabbed from my website).
An externally hosted image should be here but it was not working when we last tested it.
In order for the IC to survive, it needs to operate at a die temperature below 150 ºC. You need to design your heat sink to ensure the die temperature does not exceed this limit. If you can keep the heat sink temperature below 60-70 ºC, and have a good thermal interface between the LM3886 and the heat sink (so a good SilPad or thermal grease), this means you should aim for a peak dissipated power (as calculated by the equation above) of 40 W or less.
I've made an Excel spreadsheet to make the math easier. You can find it near the bottom of my Taming the LM3886 - Power Supply Design page. I suspect you'll also find the Thermal Design section informative. In addition, National Semiconductor AN-1192 (TI SNAA021B) is a good read: http://www.ti.com/lit/an/snaa021b/snaa021b.pdf
~Tom
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.