As in Bob C's last post... He was discussing the Power Supply C values and the rail "droop" due to a 20Hz square, or sine wave. I thought that is what prompted you to put up the raw data?
No doubt, a continous, full power signal of any kind is a rigorous test, however unlike music...
But, You may not have been subjecting the amp to less than enough C, as Bob C was suggesting ??? You may have only been driving the amp closer to the rail, by lowering it? You did not explain that part...
No doubt, a continous, full power signal of any kind is a rigorous test, however unlike music...
But, You may not have been subjecting the amp to less than enough C, as Bob C was suggesting ??? You may have only been driving the amp closer to the rail, by lowering it? You did not explain that part...
I confess to having blurred vision after the last few pages, but I'd like to offer a few quick observations (forgive me if these were already covered):
1) The proper way to compare power supplies is not uF/W, but Joules per Watt. It misses the point entirely to compare a 25W amplifier using 20000uF with a 100W amp with the same amount of capacitance. Why? Because the rails of the 100W amp are higher, and the energy storage goes up with the square of the voltage.
2) Regarding the use of large VA transformers...put yourself in the position of the circuit, looking at the power supply. What do you want to see? Constant voltage, with current on demand. This is the very definition of a voltage source (as opposed to a current source, which supplies constant current, voltage according to demand). What is the primary characteristic looking into a voltage source? Low impedance. Given that the transformer's secondary represents a large proportion of the impedance of the power supply, it then becomes obvious that anything that lowers that impedance will bring the power supply closer to being a theoretically perfect voltage source. Granted, it'll be a diminishing returns curve, but that doesn't mean there's no benefit to be gained.
I rather missed the point of the poster who wanted his rails to collapse. Setting aside the fact that the operating points for the gain devices will change dynamically (bad idea), this behavior begins to approximate a current source. The only case where I can see this sort of power supply being at all tolerable is a circuit whose current draw sums to DC; a subset of class A circuits.
Grey
1) The proper way to compare power supplies is not uF/W, but Joules per Watt. It misses the point entirely to compare a 25W amplifier using 20000uF with a 100W amp with the same amount of capacitance. Why? Because the rails of the 100W amp are higher, and the energy storage goes up with the square of the voltage.
2) Regarding the use of large VA transformers...put yourself in the position of the circuit, looking at the power supply. What do you want to see? Constant voltage, with current on demand. This is the very definition of a voltage source (as opposed to a current source, which supplies constant current, voltage according to demand). What is the primary characteristic looking into a voltage source? Low impedance. Given that the transformer's secondary represents a large proportion of the impedance of the power supply, it then becomes obvious that anything that lowers that impedance will bring the power supply closer to being a theoretically perfect voltage source. Granted, it'll be a diminishing returns curve, but that doesn't mean there's no benefit to be gained.
I rather missed the point of the poster who wanted his rails to collapse. Setting aside the fact that the operating points for the gain devices will change dynamically (bad idea), this behavior begins to approximate a current source. The only case where I can see this sort of power supply being at all tolerable is a circuit whose current draw sums to DC; a subset of class A circuits.
Grey
If in a load test, the rail voltages just drop (for example in no load +/-45V, in 100W test, it drops to +/-37V), what is this tell us? Is this test means the VA of the transformer is just not big enough?
Even very big VA rating power transformer will drop the rail voltages on load test. Just how many % is considered normal?
Even very big VA rating power transformer will drop the rail voltages on load test. Just how many % is considered normal?
I ran the test with a 20 Hz signal, 100 watts into 8 ohms -- at the rail voltage of 51V THD+N is 0.006%, dialing down the power supply voltage to 46VDC THD+N is 1.1 % and it really blooms when you go lower.
I would include some pix but the DIYAUDIO website is a bit wonky today.
I would include some pix but the DIYAUDIO website is a bit wonky today.
Hi Lumanauw,lumanauw said:
I will not answer directly, but remind you what B Cordell said.
the amp should deliver at least 180% power into half load impedance. This implies that the output peak voltage remains within 95% of the reference maximum output.
I use a slightly different rule, the maximum power output into double impedance should triple into half impedance. i.e.60W into 16r and 180W into 4r. This results in 87% of output voltage when the load is quartered. This is very similar to BC's rule.
To achieve either will require that the supply rails move from about 50Vdc to 52Vdc when quiescent to about 45Vdc @100W into 8r and fall further to 42Vdc or so @180W into 4r. That 3V drop is about 6% or 7% rail droop into the more severe load.
Note however, this is for an 8ohm capable amplifier.
If you are designing a 4ohm capable amplifier all the test load resistances/impedances should be halved.
jackinnj said:
Nice work.
As you dialed down the supply rail voltage, at what voltage did you begin to see visible onset of clipping on tips of the sinusoid?
Also, under the conditions you described, I assume that when you mention supply voltage you are referring to the average supply voltage in the presence of 120 Hz ripple. If you put a scope on a rail, did you see amplitude modulation of the 120 Hz ripple by the 20 Hz sinusoid?
Thanks,
Bob
I didn't have time to put it up -- this is 20Hz 100W into 8R showing both the output amplitude and the ripple on the positive supply rail -- 1% THD
An externally hosted image should be here but it was not working when we last tested it.
Is there any practical guidance to test a certain power transformer, to know how many A, V, VA it delivers? If the power transformer is 500VA, is there any way to do the testing without having the 500VA dummy load?
Generally there's a pretty tight correlation between core cross-section area and the square root of VA, although of course that will vary based on core material. For the usual mains power transformers, the constant of proportionality if cross-section is in cm^2 would be close to one.
David,
the only way to find out what the top speed of a car really is, is to drive it.
If you are talking toroids, the weight is often a pretty good parameter to determine the VA rating.
Comparing the cross section and the secondary wire diameter with those of a transformer with a known VA rating narrows it down.
The nominal secondaries voltage is derived by measuring the zero load voltage and correcting it for the regulation factor.
The nominal A that comes out can be compared with the secondary wire diameter.
To know for sure, you'll need a dummy load.
But the dummy load has to have a higher rating than 500 watts to find the VA rating the transformer is designed for.
P-A stated that the most accurate way of determining it is to measure the temperature of the core.
the only way to find out what the top speed of a car really is, is to drive it.
If you are talking toroids, the weight is often a pretty good parameter to determine the VA rating.
Comparing the cross section and the secondary wire diameter with those of a transformer with a known VA rating narrows it down.
The nominal secondaries voltage is derived by measuring the zero load voltage and correcting it for the regulation factor.
The nominal A that comes out can be compared with the secondary wire diameter.
To know for sure, you'll need a dummy load.
But the dummy load has to have a higher rating than 500 watts to find the VA rating the transformer is designed for.
P-A stated that the most accurate way of determining it is to measure the temperature of the core.
I disagree, as that would be variable depending on how much of the core is exposed, whether they skimped on copper (a bit lower wire gauge giving more resistive heating), insulation, and so on. The proper way would be to determine how much magnetic flux the core crossection can take from the usual formula (assuming standard core material), derate it appropriately, and get your VA number from there.jacco vermeulen said:
Nixie,
it's P-A's statement, not mine.
An unknown transformer could be a custom job, i gathered from transformer manufacturers info that custom transformers can be designed for a different insulation class.
it's P-A's statement, not mine.
An unknown transformer could be a custom job, i gathered from transformer manufacturers info that custom transformers can be designed for a different insulation class.
Nixie said:
Hi that core ampere turns is one criteria, the others (that I know about) are core/insulation temperature and regulation.
The three specification values are decided and the current output will hit one or more of these limits. That then determines the VA.
I'm thinking about this, don't know if it is right or wrong. It could be dangerous too (never done it before).
To know how many VA a certain power transformer can put into the speaker.
I assume the audio amp cct already has low impedance (to be neglected in the test), then the determining factor is actually the power transformer itself.
Connect the secondaries of the power transformer to dioda bridge rectifier (with its capacitor bank, to make +/-rail DC).
Then connect this +/-rail to dummy load (4 ohm or 8ohm, but I will have to provide about 1000W dummy for this transformer test).
Then I can measure the voltage drop on this 4ohm load, how many volts. From the equation P=VxV/R, if we know the V and R value, we can know the P output into 4ohm or 8ohm.
This is to get DC wattage. To get AC wattage (audio), what is the conversion factor?
More important : Is this dangerous to do?
To know how many VA a certain power transformer can put into the speaker.
I assume the audio amp cct already has low impedance (to be neglected in the test), then the determining factor is actually the power transformer itself.
Connect the secondaries of the power transformer to dioda bridge rectifier (with its capacitor bank, to make +/-rail DC).
Then connect this +/-rail to dummy load (4 ohm or 8ohm, but I will have to provide about 1000W dummy for this transformer test).
Then I can measure the voltage drop on this 4ohm load, how many volts. From the equation P=VxV/R, if we know the V and R value, we can know the P output into 4ohm or 8ohm.
This is to get DC wattage. To get AC wattage (audio), what is the conversion factor?
More important : Is this dangerous to do?
lumanauw said:
Yep, because you may have a 99 % chance of overloading the transformer and a 1 % chance of getting the right VA rating.
A toroidal transformer can deliver much more VA than is written on it, a 500VA transformer does not limit it's output power at 500VA.
An overloaded transformer will deliver much more VA, efficiency will drop like a fly, and the transformer will seriously start to overheat.
imo, a better approach would be to hook up a sizeable variac inbetween the transformer and the dummy load.
Measure the incoming VA at the primary side and the outgoing VA on the secondaries to determine the efficiency rate of the transformer with the unknown data.
Chart the efficiency rates, and compare the outcome with transformer efficiency graphs you can download from several toroid manufacturers.
hi,
you don't even have to employ rectifiers and caps, just put the dummy load accross the transformer secondary and see how fast the temperature rises, or how low the secondary voltage drops. that will give you a clear picture of the transformer's capacity...
you don't even have to employ rectifiers and caps, just put the dummy load accross the transformer secondary and see how fast the temperature rises, or how low the secondary voltage drops. that will give you a clear picture of the transformer's capacity...
Hi, Jacco,
Yes, it will be dangerous. I won't try it.
Hi, Tony,
If I want to know the transformer capability of delivering to 4ohm load, what is the R dummy size (ohm) that must be used for the test like you proposed?
In real amplifier (with -speaker connected to 0/gnd), the energy that is delivered to the speaker is just 1/8 of DC wattage, because it is sinusoidal area (shadded area).
Yes, it will be dangerous. I won't try it.
Hi, Tony,
If I want to know the transformer capability of delivering to 4ohm load, what is the R dummy size (ohm) that must be used for the test like you proposed?
In real amplifier (with -speaker connected to 0/gnd), the energy that is delivered to the speaker is just 1/8 of DC wattage, because it is sinusoidal area (shadded area).
Attachments
David,
most toroid efficiency graphs show the highest efficiency rate at around 25%=> Pout/Pnominal.
Increase the output load and the efficiency rate starts to go down.
At a point beyond Pout/Pnominal = 1, the efficiency rate will collapse.
The nominal VA rating often is 90% of the output of the breaking point in the efficiency line, or the point where d(Pout)/d(Pnominal) suddenly changes is at 110% of the nominal VA rating.
90% is a common continuous rating number for a lot of appliances.
Engines are designed for a maximum continuous rating of 90%, diesel engines pollute the least operated at 90%, which lengthenes the overhaul interval.
Funny enough, even humans perform best at 90% of the maximum output they can deliver continuously.
Somehow there's a universal truth.
most toroid efficiency graphs show the highest efficiency rate at around 25%=> Pout/Pnominal.
Increase the output load and the efficiency rate starts to go down.
At a point beyond Pout/Pnominal = 1, the efficiency rate will collapse.
The nominal VA rating often is 90% of the output of the breaking point in the efficiency line, or the point where d(Pout)/d(Pnominal) suddenly changes is at 110% of the nominal VA rating.
90% is a common continuous rating number for a lot of appliances.
Engines are designed for a maximum continuous rating of 90%, diesel engines pollute the least operated at 90%, which lengthenes the overhaul interval.
Funny enough, even humans perform best at 90% of the maximum output they can deliver continuously.
Somehow there's a universal truth.
Easy, take a process management class.
These numbers have been thoroughly examined at manufacturing plants in the past, the (US) automobile industry in particular.
It's how the belt speed for continuous processes is set, whether you manufacture cars or make glass bottles. (Me worked at a plant that made Coca Cola PET bottles and glass works)
For humans, both the 90%, 100%, and 110% also apply.
Tony right, why bother with all the strenuous stuff ? 9 of 10 toroids will have similar regulation numbers.
These numbers have been thoroughly examined at manufacturing plants in the past, the (US) automobile industry in particular.
It's how the belt speed for continuous processes is set, whether you manufacture cars or make glass bottles. (Me worked at a plant that made Coca Cola PET bottles and glass works)
For humans, both the 90%, 100%, and 110% also apply.
Tony right, why bother with all the strenuous stuff ? 9 of 10 toroids will have similar regulation numbers.
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