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Power Supply Question
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-minus-



Joined: Oct 26, 2008
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PostPosted: Tue Jan 24, 2012 8:11 am    Post subject: Power Supply Question
Subject description: How many modules can I power?
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Hi all,

A while ago I bought a regulated power supply kit and 2 X 11.5VAC at 1.25 AMP power supply. It all works fine and has liberated me from the 9V battery constraint.

So I have been happily building away making stripboard circuits of various modules and now want to house them into a case.

My question is, how many modules can I reasonably power from this supply? I was hoping that I can build enclosures to house say three rows of five modules per row at 3U depth. This would make 15 modules. Any further expansion would require a second separate supply and enclosure, which is fine by me. 18 modules would be better, but if I have to cut this down to 12, so be it.

How do I work out how many AMPS these circuits will require? They are mainly MFOS and GGS circuits and others. Other than hooking up 15 boards and seeing what happens, is there some kind of formula for working this out? Confused

Thanks!
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marvkaye



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PostPosted: Tue Jan 24, 2012 10:36 am    Post subject: Reply with quote  Mark this post and the followings unread

Ray Wilson generally lists current draw for his modules for both 12v and 15v supplies in each module's documentation on his website. Could be that CGS has the same info available. If all else fails you can just run each module's power through an ammeter, write down all the current draws and you'll know what you need. Hope this helps.

<marv>
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-minus-



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PostPosted: Tue Jan 24, 2012 12:25 pm    Post subject: Reply with quote  Mark this post and the followings unread

Hey thanks <marv> Smile

After I posted I looked at the MFOS site and saw Ray had listed current draw for his current modules (no pun intended!). I've made some of his older circuits where this information is not available. However, I am following your advice regarding the ammeter. I shall power up each module individually and note the power draw. I was never sure how to do this... Confused but hey, youtube to the rescue again:

http://www.youtube.com/watch?v=9Zt3P9bvn8M

Thanks for your help!
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marvkaye



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PostPosted: Tue Jan 24, 2012 12:45 pm    Post subject: Reply with quote  Mark this post and the followings unread

Not a problem, glad I had something to offer. Back when I first started in SDIY I built an MFOS power supply and incorporated ammeters into it for both rails, strictly for use during the building/testing process. I realize it's probably a bit over the top, but it makes doing the kind of thing you're trying to do pretty simple. Go to http://electro-music.com/forum/topic-48135.html and scroll to the last message on the page... several pics of that PS. JIC you're interested.

<marv>
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Osal



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PostPosted: Thu Jan 26, 2012 12:57 pm    Post subject: Re: Power Supply Question
Subject description: How many modules can I power?
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-minus- wrote:

How do I work out how many AMPS these circuits will require?
Hey Minus, You also should know how many AMPs your power supply could deliver.
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PostPosted: Fri Jan 27, 2012 5:49 am    Post subject: Reply with quote  Mark this post and the followings unread

OK... hang on. So apart from the fact the AC wall wart says 1.25 Amps on it, do I need to look at what the +/GROUND/- 12V kit I bought can deliver in Amps? Bit confused here Confused

Thanks!
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Osal



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PostPosted: Fri Jan 27, 2012 6:48 am    Post subject: Reply with quote  Mark this post and the followings unread

Yes, all the components determine the maximum current output of a power supply. See: http://electro-music.com/forum/topic-51694.html
Could you provide more details about the wall wart and the regulated power supply kit? Do you have its schematics?

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PostPosted: Sat Jan 28, 2012 5:23 am    Post subject: Reply with quote  Mark this post and the followings unread

OK. Here's some images of what I have, and a link. Thanks for your help.

The AC power supply:

http://www.jaycar.com.au/productResults.asp?keywords=MP-3055&keyform=KEYWORD&SUBMIT.x=0&SUBMIT.y=0


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Osal



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PostPosted: Sun Jan 29, 2012 7:16 am    Post subject: Reply with quote  Mark this post and the followings unread

OK, there are several points to consider.

1) Current rating of the transformer: As far I understand the wall-wart you have is a an AC/AC wall-wart and they are basically a transformer. You can open it and see it -disconnected-
The current in the secondaries of the transformer doesn’t equals to the current output. Alternating Current in secondaries splits in two components after the bridge rectifier: One alternating and other direct. The alternating component goes to ground through the capacitors leaving the direct current to the regulator. The ratio to calculate is Isec/1.8=Iout. So, if your wall-wart is specified 1.25AC, 1.25/1.8=0.69A, and this is -regarding transformer's current rating- the maximum current you should output.

2) Voltage rating of this transformer:
For standard regulators at +/-12V would be better a 2X15VAC transformer in order to stand voltage drops in the mains electricity (usually you calculate with a 10% of voltage drop). This transformer that you have is 2x11.5V. Normally the voltage rating of a transformer is specified under load. This means that it is 2x11.5V when 1.25A are flowing through the secondaries.
This is not enough.
11.5VAC -10% of mains voltage drop =10.3VAC; to know the peak voltage we should multiply this by its crest factor, now as an approximation be use 1.4; 10.3*1.4=14.4V; We subtract one voltage drop of a rectifier diode, which is approx 1V, 14.4-1V=13.4; Still, we must subtract the regulator voltage drop (about 2V) and the ripple. As you see there is no room for that.
you could try to use this transformer considering the following:
A) Depending where you are located you can take advantage of a more high mains voltage like for-example 240VAC in Australia, thus you can expect smaller voltage drops (related to 230VAC)
B) Output less current than the rated for the transformer. This is relevant because a)Voltage in secondaries will increase due the transformer regulation, b)Ripple will be smaller, and c) Regulator's voltage drop will be also smaller.
C) Increase size of the filter capacitors as much as possible. Think that to output for example 0.5A with 1000uF capacitors the ripple could be as big as 4.4Vp-p!

3) Power dissipation.
Heat-sink in the picture is too small. It would be advised to increase its size. (or/and add a fan)

So, we could estimate closest how much current you can output, but we would need some measurements.
Otherwise you can try to use it up say 0.3A, this or more than this, you can expect -considering worst case situations- 1) Insufficient Voltage at regulator's input, resulting in regulation failure, generating hum. 2) The regulator working at more temperature than its maximum rating (125C). This will decrease its life and degrade its performance. If temperature increases up to regulator thermal protection, it will turn off/on generating noise.

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PostPosted: Sun Jan 29, 2012 8:38 am    Post subject: Reply with quote  Mark this post and the followings unread

Thanks. There is clearly a lot to consider here. I was just comparing the MFOS Wall Wart Bipolar Supply with what I have. Ray has 3x 3300uF caps on each rail(?)... what I have has just 1000uf on each. Ray appears to have no heat sinks on the regulators. Question

I could make up a stripboard of the MFOS circuit and use the AC transformer I have. I am in Australia...

I have hooked up a few modules with alligator leads from this supply before. Not sure how many... two VCO's, AR EG, VCA, and a VCF. I was just curious how many modules I could hook up to this.

I was initially thinking I could make up small enclosures/racks of 12 or 15 modules and run each rack off its own power supply. If this power supply issue becomes problematic I shall just have to get hold of a Power One supply or something.
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Osal



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PostPosted: Sun Jan 29, 2012 7:02 pm    Post subject: Reply with quote  Mark this post and the followings unread

-minus- wrote:

I could make up a stripboard of the MFOS circuit and use the AC transformer I have. I am in Australia...


You could use your AC transformer with your kit also. But yes capacitors should be increased at least up to 4700uF. I would like to know how we can take benefit of the transformer regulation, but this implies measurements, you would need power resistors.

There is the advantage you are located in Australia, so you could calculate the power supply to stand 6% of mains voltage drop, (instead -10%)

And I didn't thought it first, but low voltage in the regulators input has difficulties but also the benefit of less power to dissipate! so more little heat-sink.

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PostPosted: Wed Feb 01, 2012 12:13 pm    Post subject: Reply with quote  Mark this post and the followings unread

I've begun some measurements. Bare in mind some of these circuits I'm using are very small, often just one IC and a few components.

I just measured a dual triangle/square wave oscillator based around a single 4069 IC. I'm getting a reading of no more than 0.04A. That doesn't seem like much at all. Confused (EDIT: Actually, that is quite a bit. The current MFOS VCO consumes 28mA... with 7 IC's! I'm going to have to do some careful accounting here. Might be worth investing in a POWER-ONE HCC15-3-A POWER SUPPLY before this all gets out of hand Laughing )

I'm trying to come up with an enclosure design at the moment. It may be 4 panels/modules wide by 3 deep... 12 modules. Or even 15 modules if I go for 5 modules wide. I'm then figuring that as the synth progresses, I'll make more standard enclosures and run them from similar separate supplies. I'll stack the enclosures. Eventually I'll invest in a serious power supply and run say 6 of the enclosures from that.

I'll have to do some more current measuring here. I'll make note of what circuit is pulling how many amps and report back.
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Osal



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PostPosted: Fri Feb 03, 2012 4:13 am    Post subject: Reply with quote  Mark this post and the followings unread

Hello Minus,

The ideal would be to design a power supply for a determined load. However for modular synthesizer's applications it could be unknown. For example, you don't know still what modules you will use, or how many modules you will fit in the case, etc...

The 40mA that you measured could be an average module consumption, although that easily other modules will consume 80mA or 100mA. In the site of Dave Brown you have some data about: http://modularsynthesis.com/modules/parts/DJB-Power.pdf . You could calculate an average.

However the best would be to finish one of your cases and then measure the consumption. Ah! be careful when you measure currents with modules connected , because if a rail fails due a bad connection some parts could be damaged. So, connect properly the multimeter.

------

About to use one big power supply or several smaller, I would prefer to use a little power supply for each case -or several if the case is big- than one big power supply to feed several cases. Mainly because the power distribution and the heat's dissipation are much more simple.

------

I thought about the power supply you already have, and it could be a good start meanwhile you decide other details.
I will try to explain it. The idea is that mains electricity is something wild from which we need to obtain an accurate DC Voltage. Mains electricity voltage is not stable, it has fluctuations and the voltage regulator needs a minimum input voltage to warranty regulation, so an accurate voltage output. Therefore we need room for those fluctuations happen without affect voltage output.
The example is a 11.5VAC-0-11.5VAC to output +12V/0/-12V. This is too tight. However you are located in Australia where mains voltage can be 240V.
There are several points from where we can scratch some volts. Let's put the example that we want to output 0.5A from your supply.

1) Transformer regulation. Transformers normally are specified under load. So, if they have no load, voltage in secondaries raises due the transformer regulation . For example, if this transformer you have had 15% of regulation, it means that voltage with no load will be 15% higher than the specified. But... how to know what voltage you will have at half load? You can measure it. You place two 24 Ω 50W power resistors from the rails to ground, so 0.5A is flowing. You measure mains voltage (Vp) and voltage in secondaries (Vs), then (Vs*230)/Vp=Vs Nominal.
OK, for example, let's say that you measured it and estimated 12V-0-12V instead 11.5-0-11.5

2) Bridge rectifier. In a split or complementary dual power supply as this one of the example, after the bridge diode there is one rectifier diode voltage drop to the peak voltage. Standard p-n rectifiers like the 1N4002 have 1.1V of voltage drop.
You could use a Schottky rectifier that have about 0.5V of voltage drop. Look for repetitive peak reverse voltage 60V. For example STPS2L60

3) Ripple. You can use a big capacitor to restrain the ripple under the volt. You could try to fit a 25V 4700μF capacitor in that PCB.
Electrolytic capacitors have 20% tolerance, so the worst case will be a value of 3760μF
Ripple=(0.5A*0.007s)/0.00376F=~0.9Vp-p

4) Regulator voltage drop: the voltage drop of a regulator depends on current through itself. Looking LM7812 data-sheet it would be 1.5V (at 0.5A) . Adding a little margin, lets calculate with 2V. (The manufacturer do the tests with more big (Vin-Vout) so maybe here we are out of manufacturer's specifications, and you can expect, for example, less ripple rejection. However it happens also with other parameters, Ideally we should measure all parameters under our operating conditions to be sure of the specifications)

The immunity to mains voltage drops:
Now we calculate from the output, 12V,
Add the regulator's voltage drop, 12+2=14V
Add Vripple p-p, 14+0.9=14.9V
Add the Rectifier voltage drop, 14.9+0.5=15.4V
Divide this peak voltage by the crest factor of the wave in secondaries, for this example CF=1.4 (measure it if you want to be more accurate), 15.4/1.4=11VAC.
This, 11VAC, is the minimum voltage needed in each secondary to warranty regulation.
To know how much % of immunity to mains voltage drops it has: (11*100/Vs nominal)-100, we estimated before Vs nominal as 12V, so (11*100/12V)-100=-8.3% which would be acceptable for Australia.

Now we should calculate the power to dissipate and the heat sink's thermal resistance. The low Vin that was a problem before now is a an advantage:
Power to dissipate (Pd):
Iout=0.5A
Vout=12V
Max Vin = 12+10%=13.2V
13.2*1.4=18.5V
18.5-0.5=18V
18-(0.9/2)=~17.6V

Pd=~(Vin-Vout)*Iout
Pd=(17.6-12)*0.5=2.8W Very Happy

Heat-sink's thermal resistance (Θsa):
Θja=Θjc+Θcs+Θsa
max operating junction temperature Tj=125C (datasheet)
max operating ambient temperature Ta=40C
Pd=2.8W
Θjc=5CW (data-sheet)

Θca=((Tj-Ta)/pd)-Θjc
Θca=((125-40)/2.8 )-5=27.1C/W
You could estimate Θcs=1C/W, so
Θsa=<25.4CW

This seems the heat-sync you are using and is rated as 20C/W so seems that is enough. However the specified thermal resistance of a heat-sink is true for the test conditions used by the manufacturer which could vary from your operating conditions. If you can and is easy I would try to replace the heat-sinks by bigger ones.

As a Summary: Take measurements if possible to know your power supply. Replace the capacitors by 4700μF and try the Schottky STPS2L60 rectifiers. Easily will be possible to output 0.5A
Another advise, specially if you are going to make tests, is to place at least the protection diodes 1N4002 from rails to ground, protecting against accidentally shorts within negative and positive outputs.

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PostPosted: Fri Feb 03, 2012 7:56 am    Post subject: Reply with quote  Mark this post and the followings unread

Thanks Osal for your VERY detailed reply!

There is a lot more to this than I thought. I have to confess that equations are not my strong point at all. Embarassed

I'll reread everything several times and see if I can understand this fully. Interesting what you say about several smaller supplies over one large supply. I shall take heed of this! I'll try and be conservative in my estimations on how many modules can run from a supply this size.

Once again, I appreciate the time you have spent explaining this. You are clearly very knowledgeable in this area.
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Osal



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PostPosted: Sat Feb 04, 2012 7:14 am    Post subject: Reply with quote  Mark this post and the followings unread

Hey -minus- Let me know if I can explain anything better.

I would like to clarify that when I said "little power supply" I was referring up about 0.8A~1A dc output. Just wanted to quantify it Smile

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