Is this the right way to Calculate an Transfomer? - diyAudio
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Old 15th February 2013, 10:42 PM   #1
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Default Is this the right way to Calculate an Transfomer?

Example 1.

Chipamp 70W por channel, if i build an stereo Amp, i need to use something like...

Voltage 24-0-24 = 48V AC => 68 VDC - 2 Volt from diodes = 66VDC
140W Power

To be safe i need to use 140W x 1.4 = 200VA

Amperage = 4.15A

So for an 2 channels amp with 200VA will be OK, right?


If use one channel instead two (to run mono) can use the half of that values

Voltage 24-0-24 = 48V AC => 68 VDC - 2 Volt from diodes = 66VDC
70W Power

To be safe i need to use 70W x 1.4 = 100VA

Amperage = 2.1A

Right?


What is the diference in the maths if i use 8ohms, 6ohms, 4 ohms or 2 ohms load?
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Old 16th February 2013, 12:29 AM   #2
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If the amplifier is 100 watts then you need 140VA as 40% of power is lost in the heatsink and 60% goes to the speaker.
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Old 16th February 2013, 01:44 AM   #3
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I created an Excel spreadsheet to do these kind of calculations. It is useful for designing a power supply, checking out how a given transformer will perform, and to estimate the maximum power that a class AB amplifier (e.g. a chip amp) will put out for a given power supply.

Get it at this link:
Free Audio Software!
It's called "Power Supply Capability and Design".

If you have questions about it, send me a PM or (better) use my web site to contact me.

-Charlie
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Old 18th February 2013, 10:41 PM   #4
gootee is offline gootee  United States
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Quote:
Originally Posted by samsagaz View Post
Example 1.

Chipamp 70W por channel, if i build an stereo Amp, i need to use something like...

Voltage 24-0-24 = 48V AC => 68 VDC - 2 Volt from diodes = 66VDC
140W Power

To be safe i need to use 140W x 1.4 = 200VA

Amperage = 4.15A

So for an 2 channels amp with 200VA will be OK, right?


If use one channel instead two (to run mono) can use the half of that values

Voltage 24-0-24 = 48V AC => 68 VDC - 2 Volt from diodes = 66VDC
70W Power

To be safe i need to use 70W x 1.4 = 100VA

Amperage = 2.1A

Right?


What is the diference in the maths if i use 8ohms, 6ohms, 4 ohms or 2 ohms load?
First, your peak signal voltage level across the load will be:

peak signal voltage = peak rail voltage - p-p ripple voltage - about 4 volts

The "about 4 volts" is for the amplifier itself, usually a Vce voltage plus the voltage across a low-value resistor. Usually 3 to 4 Volts at max power. It might be more than 4 V, depending on amplifier topology. But it probably won't be much less than 3 V.

So assuming 1V p-p ripple, that leaves about 27 V for the output signal peak. The RMS voltage of a maximum-amplitude sine would then be 19 V. RMS output power would be V/Rload, or iRload.

Max RMS Output Power would be 19/8 = 45 W into 8 Ohms, or 60 W into 6 Ohms, or 90 W into 4 Ohms, or 180 W into 2 Ohms.

To get 70 W into 8 Ohms,
V/8 = 70 W
Vrms = 23.7 V
Vpeak = 33.5 V
Vrail_pk = 33.5 + 4 + Vripple_p-p = 38.5 V (if Vripple = 1 V p-p)
Vrms _xfrmr = (38.5 + 2) / √2 = 28.6 VMRS

Rather than taking 1.4 times the needed VA, it would be better to use a factor of at least 3 or more.
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Old 19th February 2013, 01:09 AM   #5
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Quote:
Originally Posted by gootee View Post
Rather than taking 1.4 times the needed VA, it would be better to use a factor of at least 3 or more.
What does "better" mean in this context? Would an infinitely stiff power supply be "better"? It would also (likely) be infinitely expensive, too. I think the point is to estimate the draw on the PS, and then design one that will sag only as much as can be tolerated while still meeting the necessary rail voltage for the desired power output.

For instance, you can choose a higher rail voltage and allow more sag under load, or you can choose to use a lower rail voltage and demand a stiffer supply. The choice might depend on the max rail voltage that your amp can survive, or thermal dissipation concerns, cost of PS, etc. There can be lots of tradeoffs, so I don't think that there is a good ROT for this, except to model the transformer and PS performance under load.

I know you (Bob) put together some spreadsheets for calculating this kind of thing, and my spreadsheet (linked in my post above) also does this kind of calculation. I have found that in most cases, with a typical "music" signal the transformer can be as low as 1.5 times the maximum power output capability of the amp, given that with music signals the amplifier will mostly be operating at a much lower average power level and demand for high power levels is very short in duration and can be (mostly) supplied by the bank of caps in the PS. Do you find something else with your work?

-Charlie

Last edited by CharlieLaub; 19th February 2013 at 01:15 AM.
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Old 19th February 2013, 01:16 AM   #6
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Its quite difficult to rate a transformer for an amplifier because the current draw isn't constant.
The smoothing capacitors take in surges of current at the peak of the sine output from the transformer.
This means you can easily take full current from the transformer for short periods.

So its not as simple as an amp requiring 2amps taking two amps constantly.
Its more likely to take 6+ amps in a surge.
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Old 19th February 2013, 01:57 AM   #7
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wow, thanks too much!

I have an PCB from an rectifier that used with an CT transformer. I have another proyect in mind that dont need +- i just need 0-60V, want to know if is possible to convert or need to use a new PCB, i think that the schematic is similar to this 6*10000uFߵŵԴ - Ű - ŵԴ - ζ

sorry for the offtopic question, but think that the answer will be not enough hard to create a new post
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Old 19th February 2013, 02:00 AM   #8
gootee is offline gootee  United States
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Charlie and Nigel,

Well, "better" means better.

I agree that there isn't a very good rule of thumb for this. I wanted to try to have the OP make sure that he left a generous margin, something about double the 1.4X he was going to use, since that leaves no margin at all. (And I really like the idea of having rail-voltage margin, mainly for headroom for transients, but didn't want to complicate it too much.)

Probably the only way to really do it right is to account for the effects of transformer windings' leakage inductances, and their resistances, and the diodes' on-to-off-time ratio, etc etc. My spreadsheet does solve and simulate the circuit's differential equations (2nd order when diodes are on and 1st order when diodes are off) and also figures out exactly when the diodes turn on and off, and takes into account the capacitor ESR (but not ESL) and plots everything as functions of time. But I guess it does not take into account the heatsink losses.

Also, possibly the actual transformer being used would need to be modeled. The one in my spreadsheet is modeled from a real transformer, and is scalable in terms of VA rating, Vrms output voltage, input frequency, and input voltage. But transformers vary and may need their own model, for accurate results. The spreadsheet does include the instructions for measuring a new transformer, and will automatically create a new model from the measurements.

And, I used a particular Schottky diode, for the rectifier diodes. That could be easily changed, too, by changing the diode's equation in the VBA code. The model and the procedure I used to obtain the model are in the PDF file describing the spreadsheet, in the post at Power Supply Resevoir Size .

Also note that my spreadsheet performs a worst-case analysis, in terms of the output signal. It assumes a constant DC output current that is at the level that the PEAK current would be at, for a sine output at the max rated power.

The spreadsheet's latest version, for Excel 2007 and later, is downloadable from the post at:

Power Supply Resevoir Size

If someone needs the Excel 95-2003 version, it is downloadable from the post at:

Power Supply Resevoir Size

Cheers,

Tom

Last edited by gootee; 19th February 2013 at 02:17 AM.
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Old 19th February 2013, 02:46 AM   #9
gootee is offline gootee  United States
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In post # 4, the Vrms transformer voltage that I calculated, to be able to get 70 W RMS max output power, is not high enough. It only enables about 60 W RMS max, into 8 Ohms. I forgot to account for transformer losses, I think.

It's best to use the spreadsheet.
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