Average DC Voltage:Where does the 2/π factor come from?
Vavg = 2 x Vmax / π
or
Vavg = Vmax x (2/π)
2/π = 0.636
Your Math is perfect, only missing transformer internal drop, both resistive and magnetic, both much larger than expected on first sight, because of the very low duty cycle charging peaks, so HUGE current must pass during them to get needed average/RMS load demands.Actually, it's (68-1.4) * √2 = 94.2 V, peak, rectified, assuming 700 mV across each rectifier diode. Obviously the voltage across the diodes is temperature dependent, so it may be more like (68-1.0) * √2 = 94.8 V, peak, rectified once the diode bridge has reached operating temperature.
This does not account for the ripple voltage, however, so the DC voltage at the output of the power supply will be lower than the 94 V, peak, rectified.
Typically you'll have around 1.30-1.35 times the applied AC voltage at the output of the supply, under load, so you're looking at 68*1.30 = 88.4 V to 68*1.35 = 91.8 V, under load, depending on the amount of supply capacitance. If you're into solving differential equations you can probably work out an analytic solution. If not, I suggest simulating the circuit to determine the exact output voltage.
Above numbers assume that the transformer provides 68.000 VAC and has zero DC resistance (i.e., is superconductive at room temperature). In reality, production tolerances, mains variation, etc. will result in quite a bit of variability. It's reasonable to assume at least ±5 % variation just in the mains voltage (assuming you're in North America or most places in Europe), so ±10 % variation once all the other factors are considered is not out of line. So now you're looking at something like:
0.9*1.30*68 = 80 V, minimum
1.1*1.35*68 = 101 V, maximum
So expecting ±90 V at nominal mains for a dual supply is pretty reasonable. Specifying the output voltage with five significant digits is silly as you really don't know more than 1-2 digits.
Where does the 2/π factor come from?
Tom
On the other side, you know I always try to "put some numbers into it [tm]" 😉 , but at the same time am kinda obsessive about measuring 🙄
That´s why on post #6 I suggested:
which made me estimate :Raw/Peak voltage drops between 10% (oversized supply) to 20% (skimpy one).
Different paths to roughly reach the same place 🙂Since this one looks like a beancounter approved amp, let´s consider 20% so we have +/-76V rails, under full load.
Substract 4V each side for transistor drop, we have "useful" 72V peak, so 51V RMS, so possible 325Winto 8 ohm, and about 50% more into 4 ohm.
In a nutshell: home Hi Fi use?
Or relatively "light" PA use? (Church, large supermarket PA and background Music, Jazz/Chill/Lounge music Club)?
Perfect, not stressed, will last a long time.
Headbanger/Reggae/Punk band or Club PA or DJ use? ..... not so sure.
The calculated amperage requirements (6.2 A at 8 ohm and 12.4 A at 4 ohm) still seem to be greater than the secondary windings amperage ratings (4 5 A).
The only way I can understand the maximum continuous power ratings is if the amplifier is actually driven into curent overload.
Those dual secondaries being wired in series 34-0-34 V at 4.5 A for an output of 68 V x 4.5 A = 306 VA, are actually capable of withstanding over 9 A of current as if they were wired in parallel for 34 V x 9 A = 306 VA.
One last question: the secondary windings voltage ratings shouldn't be proportional with the primary winding voltage rating (120 V)?
The only way I can understand the maximum continuous power ratings is if the amplifier is actually driven into curent overload.
Those dual secondaries being wired in series 34-0-34 V at 4.5 A for an output of 68 V x 4.5 A = 306 VA, are actually capable of withstanding over 9 A of current as if they were wired in parallel for 34 V x 9 A = 306 VA.
One last question: the secondary windings voltage ratings shouldn't be proportional with the primary winding voltage rating (120 V)?
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As we’ve tried to tell you 15 times already, you CAN draw more current from a transformer that it’s “rated” for. It won’t hurt anything as long as the long term average is reasonable. The amplifier portion itself will have its own output current limiting which happens very quickly, and at a MUCH higher current. Probably in the neighborhood of 30 amps. It will also have a temperature sensor on the heat sink that will shut it down WHEN it gets too hot. The required heat sink for full power continuous sine wave operation is the size of the ones in the 1978 original Peavey CS800, with a fan that sounds like a low flying plane. They dont make them like that anymore.
Want to see what any amp is really made of? Drive it to full output with a sine wave and time it with a stopwatch. Or give it to a DJ and do the same.
When a modern amp is rated for “X” watts “continuous” - it just means a steady state signal As opposed to a pulsed, low duty or noise signal. It does NOT mean it will do it “indefinitely”. At one time it used to. Of course you’re free to build them that way if you want. You would need a 12 or 15 amp 4x35 volt transformer (Antek 15635 would do nicely) and a LOT more heat sink.
Want to see what any amp is really made of? Drive it to full output with a sine wave and time it with a stopwatch. Or give it to a DJ and do the same.
When a modern amp is rated for “X” watts “continuous” - it just means a steady state signal As opposed to a pulsed, low duty or noise signal. It does NOT mean it will do it “indefinitely”. At one time it used to. Of course you’re free to build them that way if you want. You would need a 12 or 15 amp 4x35 volt transformer (Antek 15635 would do nicely) and a LOT more heat sink.
The primary and secondary voltages are proportional in theory, without considering the internal losses due to loading.One last question: the secondary windings voltage ratings shouldn't be proportional with the primary winding voltage rating (120 V)?
I am amazed by degree of cheapness. By my calculations, that transformer is not good for more than 200 W/channel.
Ed
Ed
If you read a bit further down in my post, I do point out the resistive losses in the transformer with the sentence, "Above numbers assume ..." I did not account for the magnetic losses. That's true. Also true that the current pulses in the rectifier diodes can get surprisingly high (which is why you keep the rectifier away from sensitive nodes in your circuit).Your Math is perfect, only missing transformer internal drop, both resistive and magnetic, both much larger than expected on first sight, because of the very low duty cycle charging peaks, so HUGE current must pass during them to get needed average/RMS load demands.
The bottom line is still that if you want five digits of precision you're in for a bit of "fun" with differential equations. And even then you'll likely only get close. All our math so far has assumed that the mains voltage is a perfect sine wave. It isn't. It has significant distortion. So there is that.
The power rating of the transformer is another one of those 'fuzzies' for the exact reasons you point out.
Some people think engineering is about arriving at one correct and exact answer. In reality engineering is about choosing one answer out of many possible and sometimes we don't even know the value of something with one digit of precision (take the thermal voltage, VT, for example).
Tom
Ah. Thanks. I'm curious how that's relevant, though.Average DC Voltage:
Vavg = 2 x Vmax / π
Tom
I'm not. Transformers are expensive. Shaving a few percent off of the cost of the transformer improves the profit margin of the product significantly. Besides, as pointed out above, most of us listen to music rather than sine waves. And even when we do listen to music, we often don't do so at ear-splitting levels. Now, if the amplifier was intended to drive a shaker table at the full rated power for days with a sine wave I'd definitely recommend a larger transformer. 🙂I am amazed by degree of cheapness. By my calculations, that transformer is not good for more than 200 W/channel.
Tom
Well, I am a DJ and I do use a PA reinforcement system, but with much higher power capabilities.
This particular amplifier (XLI1500) is used to drive a set of loudspeakers rated for 165 W continous at a nominal impedance of 6 ohm.
I never use the amplifiers to their full rated power ratings or close to their limits, because I want to maintain a good headroom for peaks and ensure the termal stability of the amplifier.
I want to understand the workings and limitations of this amplifier and if, possible, to improve its performance.
This particular amplifier (XLI1500) is used to drive a set of loudspeakers rated for 165 W continous at a nominal impedance of 6 ohm.
I never use the amplifiers to their full rated power ratings or close to their limits, because I want to maintain a good headroom for peaks and ensure the termal stability of the amplifier.
I want to understand the workings and limitations of this amplifier and if, possible, to improve its performance.
calculations are good when designing amps, but when you have the amp up ang running, then full power FTC testing validates all of the design assumptions taken....
i once asked an audio guru here oh more than 16 years or so ago, he advised me, "get the biggest size traffo that will fit in your box....."
i once asked an audio guru here oh more than 16 years or so ago, he advised me, "get the biggest size traffo that will fit in your box....."
They're the same amp layout. Difference in power transformer sizeMaybe, but then why the 34V taps?
Do you have a schematic?
Attachments
these amps delivers massive power.i own 4 xli 800 and 1 xli 3500.tested with any compination on pro neo subwoofers and side to side with my dynacords 5 times more expensive amps they are amazing and not underated. i mean look at the price the only thing i found is crap capacitors inside (ok cheap amp) change it with high quality and the amp keep going for meny years
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