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Old 4th October 2012, 08:25 PM   #1
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Default Tool for sizing linear unregulated power supplies

In order to determine what transformer power (VA) rating and total capacitance I will need for the power supplies that I am building, I put together a handy Excel spreadsheet that I would like to share with the community. This is for linear, unregulated power supplies using a single bridge rectifier often employed in class AB amplifiers. I have included a screen shot, below.

Please take a look at it and comment if you see anything that is incorrect, etc. It can be found at:
Amplifier Power Supply Design Calculations.xls

The one thing that I feel is missing is a specification for the transformer in terms of the no load secondary voltage and VA rating. Unfortunately, it looks like it is necessary to get the transformer under consideration "in hand" and make some measurements on it in order to see how it will perform, if it will meet the design spec, and how the secondary voltage will sag under load (thus limiting output power).

-Charlie


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Old 4th October 2012, 11:41 PM   #2
WSJ is offline WSJ  United States
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Did you include the amplifier efficiency and transformer derating factor?

Derating transformers for rectifier use.

Last edited by WSJ; 4th October 2012 at 11:44 PM.
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Old 5th October 2012, 03:32 AM   #3
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Quote:
Originally Posted by WSJ View Post
Did you include the amplifier efficiency...
Yes. The amplifier efficiency is accounted for in this way: power is either delivered to the load, or dissipated as heat. I believe that these two fates of input power should account for the efficiency of the amplifier for the most part. Often the information on power dissipation can be gained from calculations, or from datasheet info. The user enters values for these in watts.

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Originally Posted by WSJ View Post
Did you include the ... transformer derating factor?
I have not included a "derating factor" although I do make some mention of this in the notes section. The link you cited has some good information, but it's a complex subject that involved various aspects of the rectification, the avereage to peak power ratio, and so on. It's true that the pulses of power drawn through the diode bridge make for a much different type of load for the transformer compared to a full sine wave. I have not seen how much derating should be applied as a result of it, and I did not (yet) include any factor for it. If you have more info on this, please post about it here.

You might find this section in a web page on power supplies to be of interest:
Elliott Sound Products - Linear Power Supply Design
I also stumbled across an old but good article but Nelson Pass that is written for a lay audience:
Product Review -

-Charlie
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Old 5th October 2012, 09:17 PM   #4
WSJ is offline WSJ  United States
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The required RMS secondary current rating of a transformer is 1.8 x DC current for a full wave bridge with a capacitor filter. Amplifier efficiency is 63% with a sign wave at the full output level.

References:
1982 National Semiconductor Voltage Regulator Handbook, page 8-4.
PowerVolt Application Notes
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Old 5th October 2012, 11:53 PM   #5
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Quote:
Originally Posted by WSJ View Post
The required RMS secondary current rating of a transformer is 1.8 x DC current for a full wave bridge with a capacitor filter. Amplifier efficiency is 63% with a sign wave at the full output level.

References:
1982 National Semiconductor Voltage Regulator Handbook, page 8-4.
PowerVolt Application Notes
I don't see the factor of 1.8 in practice. Here is real world data:

I have a power supply next to me right now, operating, but with little to no demand on it. It consists of a center tapped transformer secondary feeding a KBPC2504 bridge rectifier and two caps, one per rail, each with 46k uF (-10/+75%). It's exactly what is shown as one of the figures in the web page you cited:
Click the image to open in full size.

Let's do a "sanity check" - the secondary voltage on the transformer of my PS measures 48VAC end-to end. The DC voltage across both rails at the caps, measures 66V. Math check of my calculation method: 48V*1.4-1.4V=65.8V (calculated) versus 66V (measured). That's pretty close if you ask me. In fact the behavior of my power supply exactly follows the behavior predicted by my spreadsheet calculations.

Your figure of 1.8 times for a full bridge is something that I have seen before. I have no idea where it comes from, but I always observe the 1.414 factor.

Thanks for mentioning the 63% efficiency factor. I went back and checked my calculation and there was an error. I had changed a voltage listed in a cell used in the calculation from the DC rail voltage to the AC secondary transformer voltage, so the dissipated power value was incorrect. I updated the formula, and indeed the power deliver to the load is about 65% of the total consumed power. Another way to look at it is that another 50% of the amount of power delivered to the load is waste heat. But this is not always the case - it depends on the load and the rail voltage (although not strongly). See the formula in my "note 3".

I uploaded an updated version of the spreadsheet. The screen shot now shows the correct value for the dissipated power as well.

-Charlie

Last edited by CharlieLaub; 6th October 2012 at 12:06 AM.
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Old 6th October 2012, 03:14 AM   #6
WSJ is offline WSJ  United States
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The transformer derating is from transformer manufacturers specification for FWB capacitor filter.
electro-music.com :: View topic - Linear Power Supply For Modular Synth Application.

This Defence Systems document indicates an additional transformer derating of 80%.
http://www.sars.org.uk/old-site-arch...-77)/p3c07.pdf
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Old 6th October 2012, 03:40 AM   #7
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Quote:
Originally Posted by WSJ View Post
The transformer derating is from transformer manufacturers specification for FWB capacitor filter.
electro-music.com :: View topic - Linear Power Supply For Modular Synth Application.

This Defence Systems document indicates an additional transformer derating of 80%.
http://www.sars.org.uk/old-site-arch...-77)/p3c07.pdf
I'm not sure why you are giving much merit to these documents. From what I could tell, the arguments for transformer derating provided in the links you posted are very un-scientific. Ah yes, let's just derate the equipment by xx% so that is lasts longer/runs cooler/has more room for error, etc. etc. I didn't find much substance behind the approaches, so I don't really find them to be very compelling. I am not building a missile or military aircraft. I don't need the level of reliability that is argued in the defense document. There are no "unknown unknowns". This is just an amplifier power supply... something that is pretty well understood.

On the other hand, if there is a reason behind a proposed derating factor, then it is more believable (to me). For instance "...because the AC waveform at the secondary is altered by the strong pulses of current drawn by the power supply through the diode bridge, component YY must be derated by the factor Z" would be something that would be more believable to me. I could go and research this waveform issue and check the physics, the math, etc.

If you correctly account for the demands that will be placed on your power supply, there really is no need to include any "derating" fudge factors... that being said, if there are any aspects of the power supply that are not accounted for in the design calculations, please let me know. I did mention that the transformer sag is not explicitly included (see notes in spreadsheet). In itself that is not a motivation to derate, but rather to calculate the actual sag that a transformer in question will experience, which is possible after a couple of simple measurements. See this thread for more info:
How to estimate transformer sag under load

-Charlie

Last edited by CharlieLaub; 6th October 2012 at 03:44 AM.
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Old 26th August 2013, 01:55 PM   #8
AndrewT is offline AndrewT  Scotland
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Quote:
Originally Posted by WSJ View Post
Did you include the amplifier efficiency and transformer derating factor?
Quote:
Originally Posted by CharlieLaub View Post
..........I have not included a "derating factor" although I do make some mention of this in the notes section. ...........
The VA of a transformer is the maximum continuous value given by the manufacturer for a resistor load ONLY.

Those same manufacturers also tell us, the users, that when feeding a capacitor input filter that the continuous VA rating MUST be derated to take account of the extra heating effect of the IČR losses that are very different from the sinusoidal current into a resistive load.

This derating has nothing to do with the type of load being fed by the capacitor input filter. It could be resistive or it could be varying with a music signal. What matters is the shape of the current pulses into the capacitor input filter and the different heating effect that those pulses have on the transformer windings and thence the core and insulations.
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