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LetterBeacon
Joined: Mar 18, 2008 Posts: 454 Location: London, UK

Posted: Sat Aug 25, 2012 5:37 pm Post subject:
Transformer for 2x adjustable power supply PCB 


I'm running out of power and space in my modular cabinet. At the moment I have one power supply supplying 1A to my modules.
I have two of the MFOS 1.5A adjustable power supply PCBs but unfortunately, not enough room to run them off two separate transformers, so the plan is to run them from one large transformer.
Using this formulas in this thread: http://electromusic.com/forum/post351747.html#351747
I think I need to get a 160VA 30CT transformer which can supply 2.96A at the regulators' outputs. Meaning each PCB will be supplying approx 1.5A.
Does that sound right? Would someone be able to check please?
Thank you! 

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Osal
Joined: Aug 16, 2011 Posts: 145 Location: Here

Posted: Sun Aug 26, 2012 2:41 am Post subject:



May I ask what voltage you want in the supply output?
Also keep in mind that the maximum output current should be 0.8A if:
The power dissipation of the regulator is passive.
Considering worst case situations like 40C of ambient temperature and 10% of high mains fluctuations.
This is accordingly my temperature measurements of regulator's cases and estimations of the junction's temperature; in a +/15V supply with a 36VCT transformer using LM317/337.
The difficult part of a power supply is the power dissipation of the regulator if it is passive. _________________ electronicsea.net 

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LetterBeacon
Joined: Mar 18, 2008 Posts: 454 Location: London, UK

Posted: Sun Aug 26, 2012 3:11 am Post subject:



Sorry, I didn't mention that I'll be running +/15V. I thought I'd go for 30VCT rather than 36VCT to lessen the amount of power to be dissipated caused by too high voltages on the secondaries when not loaded.
Also, can you clarify something for me: When a +/15V supply is touted as '1.5A' (as is the case with these PCBS), does that mean 750mA per rail?
Thanks a lot. 

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Osal
Joined: Aug 16, 2011 Posts: 145 Location: Here

Posted: Sun Aug 26, 2012 4:36 am Post subject:



Quote:  Sorry, I didn't mention that I'll be running +/15V. I thought I'd go for 30VCT rather than 36VCT to lessen the amount of power to be dissipated caused by too high voltages on the secondaries when not loaded. 
This is not advisable for standard regulators. Mains voltage has fluctuations, it is not stable. Usually power supplies are calculated to stand up at least a 10% of mains voltage drop.
Quote:  Also, can you clarify something for me: When a +/15V supply is touted as '1.5A' (as is the case with these PCBS), does that mean 750mA per rail? 
A 1A power supply is 1A for each rail. Actually is the same current that goes from the positive rail to the negative rail.
Regarding your initial question, your calculus is correct; 160/30/1.8=2.96 _________________ electronicsea.net 

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LetterBeacon
Joined: Mar 18, 2008 Posts: 454 Location: London, UK

Posted: Sun Aug 26, 2012 4:48 am Post subject:



So if there's 1.5A on each rail, I'll need a total of 3A per PCB, correct? So I'm actually after a transformer that can provide 6A if I want to run two PSU PCBs from it, right? 

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Osal
Joined: Aug 16, 2011 Posts: 145 Location: Here

Posted: Sun Aug 26, 2012 4:52 am Post subject:



No no. To say that 1.5 ampere flows from the +15 to the 15 V is the same as to say that each rail provides 1.5A. _________________ electronicsea.net 

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LetterBeacon
Joined: Mar 18, 2008 Posts: 454 Location: London, UK

Posted: Sun Aug 26, 2012 5:21 am Post subject:



Great, thank you.
So I should go back up to a 36VCT transformer then. Bearing that in mind I'm looking at a 225VA 36VCT transformer that can provide:
225/36/1.8 = 3.47A.
Hooked up to my two PSU PCBs, I should be able to get 1.5A from the output of all four rails (2x +/15V). So my modules will now be able to draw 3A in total.
If the PSU PCBs only need to draw, say 2A. Does this mean that my regulators will need to dissipate more power as there will be more volts present at the regulator's input? 

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Osal
Joined: Aug 16, 2011 Posts: 145 Location: Here

Posted: Mon Aug 27, 2012 6:48 am Post subject:



You're welcome,
For a standard regulator we simplify the power dissipation with:
Pd=(VinVout)*Iout.
Then, to estimate the voltage at the regulator's input (Vin) for power dissipation:
Vin= ((Vsec +%HMV)*CF)Rvd(Ripple/2)
Where:
(Vsec) It is the AC voltage in the secondaries and as you say, to be accurate, it should include the voltage regulation of the transformer. For such big transformer that you want to use 225VA, the typical regulation could be something like 7%. Thus, under no load and nominal mains voltage, Vsec could be 18+1.26~=19.3. Using the transformer at half of its capacity we could expect something like 18.6 for each secondary. (measure it)
(%HMV) It is the percentage of high mains voltage respect its nominal value. Usually it is calculated with +10%
(CF) It is the crest factor of the wave in the secondaries. We need it in order to know the peak value of the wave after the rectification. For a sine wave it is sqrt(2). But this is when there is not load. Under load the sine wave is distorted. If more load, more distorted. Make measurements if you have an oscilloscope. If not, use CF=1.35
(Rvd) It is the rectifier voltage drop. For these power supplies with LM317/337 and one rectifier bridge, there is one rectifier voltage drop and it is about 1 volt.
(Ripple) It is the ripple in volts peak to peak. RippleVpp=(0.007*Iout)/C
(C) It is capacitance. Consider the 20% of tolerance of the electrolytic capacitors. Let's say that for this example is 3 V.
So lets put the example:
Vin= ((18.6+%10)*1.35)1(3/2)≈25.1V
So Pd=(25.115)*1≈10.1W
Now, let's calculate the thermal resistance of the heat sink:
Θsa=((maxjtmaxta)/pd)  (Θjc+Θcs)
where:
(maxjt)max junction temperature LM317=125C
(maxta)max worst case ambient temperature=40C
Pd=10.1W
Θjc=5C/W
Θcs=1C/W
Θsa=((12540)/10.1)6≤2.4C/W
It is good to know that heatsinks' manufacturers' specifications respond to a determined test conditions that can not coincide with your working conditions. In any case it needs a big heat sink. You can play with several aspects to reduce Vin, but you will lose immunity to mains voltage drops. You can calculate it now with 1.5A output; you will see negative thermal resistance. Also you can calculate it for a half of ampere.
It is good to know that when smaller junction operating temperature, more durability and reliability of the regulator is achieved.
edit: where I wrote power factor i meant crest factor. It is already corrected. _________________ electronicsea.net Last edited by Osal on Mon Aug 27, 2012 1:03 pm; edited 2 times in total 

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LetterBeacon
Joined: Mar 18, 2008 Posts: 454 Location: London, UK

Posted: Mon Aug 27, 2012 6:58 am Post subject:



Wow, thank you for taking the time to type all that out!
So from that I take it we can estimate we need to dissipate 10.1W per heatsink and, if the ambient temperature is 40C, I'll need heatsinks rated at least 2.4C/W.
EDIT  I see your calculation is with 1A. Let me try it with 1.5A...
So if I pulling 1.5A from my regulators, I need to dissipate 15.15W. That's going to require a fairly large heatsink!
That's extremely helpful thank you very much! 

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LetterBeacon
Joined: Mar 18, 2008 Posts: 454 Location: London, UK

Posted: Mon Aug 27, 2012 7:15 am Post subject:



So what's the solution? I'm not sure I have the space to put a large heatsink on four regulators. Is it worth me just trying to find the space to put in a separate transformers for each of the two PSU PCBs?
Perhaps get two 100VA 36VCT transformer which will provide 1.54A at the regulators' secondaries...? 

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Osal
Joined: Aug 16, 2011 Posts: 145 Location: Here

Posted: Mon Aug 27, 2012 8:55 am Post subject:



Hey LetterBeacon! The example that I wrote was more to explain better the subject than to give you a exact solution to your specific case. If you calculate with 1.5A the result will be negative resistance. Which means that you need active dissipation with a fan.
Also see that I put as example a 3Vpp ripple... I should edit it... Lets say that we use 4700uF filter capacitors. Due the 20% of tolerance of the electrolytic capacitors, the value can be from 3760uF to 5640uF. For one ampere output this means that the ripple can be from 1.2Vpp to 1.9Vpp. To calculate worst case power dissipation we use the smaller possible ripple and for calculate the immunity to the mains voltage drop we use the bigger.
So lets put the example again with this new value:
Vin= ((18.6+%10)*1.35)1(1.2/2)≈26V
So Pd=(2615)*1=11W
Now, let's calculate the thermal resistance of the heat sink:
Θsa=((maxjtmaxta)/pd)  (Θjc+Θcs)
where:
(maxjt)max junction temperature LM317=125C
(maxta)max worst case ambient temperature=40C
Pd=11W
Θjc=5C/W
Θcs=1C/W
Θsa=((12540)/11)6≤1.7C/W
That it is a big heatsink. _________________ electronicsea.net 

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Osal
Joined: Aug 16, 2011 Posts: 145 Location: Here

Posted: Mon Aug 27, 2012 9:12 am Post subject:



LetterBeacon wrote:  So what's the solution? I'm not sure I have the space to put a large heatsink on four regulators. Is it worth me just trying to find the space to put in a separate transformers for each of the two PSU PCBs?
Perhaps get two 100VA 36VCT transformer which will provide 1.54A at the regulators' secondaries...? 
You could use active cooling, or calculate it for 35C of maximum temperature. Measure the temperature inside of your cabinet. Subtract the current ambient temperature and add the worst case temperature in your country.
To me the solution using these regulators is to make little power supplies like 0.5A that you can make with 50VA transformers. The maximum I would make is 0.8A with large heatsinks. _________________ electronicsea.net 

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Osal
Joined: Aug 16, 2011 Posts: 145 Location: Here

Posted: Tue Aug 28, 2012 8:24 am Post subject:



To me the solution for these chips is using 36VCT transformers as I already said. However coming back to your initial idea to use 30VCT transformer and for a local solution you can be interested in this. http://electromusic.com/forum/topic517880.html
Regards. _________________ electronicsea.net 

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LetterBeacon
Joined: Mar 18, 2008 Posts: 454 Location: London, UK

Posted: Tue Aug 28, 2012 8:32 am Post subject:



Thanks for your reply and the link to the thread.
I think I've decided I'm going to try and it two PSUs in the case each supplying a smaller amount of current. It seems a lot simpler that way! 

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Osal
Joined: Aug 16, 2011 Posts: 145 Location: Here

Posted: Fri Aug 31, 2012 1:39 pm Post subject:



Yes! Small power supplies are much more simple. _________________ electronicsea.net 

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