Choosing a transformer for a power supply design

[Nitefly182]

I'm considering building a custom power supply for devices with the following requirements:

1. 12VDC 1200ma
2. 5VDC 150ma
3. 18VDC 150ma

Im looking around at power supply designs and the GGG 4-output isolated seems like it is the perfect solution with an isolated output for each voltage I need.

(http://www.generalguitargadgets.com/effects-projects/power-supplies/four-output-isolated/)

Problem is I'm coming up short on how to calculate the current available from this power supply. My rookie-level calculations suggest the 18VAC 65ma supply from the transformer listed (http://www.mouser.com/ProductDetail/Triad-Magnetics/F36-65?qs=mZMNBqyh7oHqVNDbRmaL%252bQ%3D%3D) is only going to produce 100ma on the output which is way less than I need.

So basically what I'm asking is this - is it unrealistic to expect to make a simple power supply that provides as much current as I need? Would I be better off just rehousing the couple wall warts Im using into a box and daisy chaining them to an IEC input to get rid of the power strip and bulk of all the individual components? That would certainly be cheaper but if Im going to build something I figure its worth building right.

Thanks in advance for any thoughts!

[R.G.]

Power supplies basics at geofex.com

[Nitefly182]

Power supplies basics at geofex.com

I reviewed the articles on your site before posting and didn't come across information addressing how to design to supply a certain amount of current. The article you mention here is focused on providing a specific voltage which I feel reasonably confident about. Thanks though!

[R.G.]

OK. Current capability discussion.

1. Find the max DC current you're going to supply. You list those, so step one done.
2. Decide if you're going to regulate or not. A typical regulator needs a couple of volts more than the regulator output. That happens at the minimum of the DC ripple voltage, which depends on the filter capacitors etc. and figure out what transformer secondary voltage, etc.
OK, you got that, you're ready.

For current: your transformer must supply enough RMS current to get the filter capacitors to provide enough DC average current out. The quirk is that the current going into the caps is in the form of short pulses. The RMS (heating) value of this current  is what has to be supplied out of the transformer, and what the transformer ought to be rated for.

The RMS value depends on how you rectify. The worst for the transformer is to be full wave bridge rectified. This causes the RMS current in the transformer to be 1.6 to 1.8 times the DC out of the filter capacitor.

So, if you're doing a full wave bridge rectified raw supply to a 12V regulator at 100ma, you're going to have to rate the transformer at 160ma to 180ma and make it put out at least 14V into the filter cap under worst case conditions of low AC line.

[PRR]

Please do not get electrocuted or burn down the barn working with wall-wires. We would be sad.

> GGG 4-output isolated seems like it is the perfect solution

No. Look at the demand. {Edit- I see R.G. typed faster.}  Just your 12V 1200mA (1.2A) is over 14 Watts of DC. That generic GGG plan says "2.5VA" transformers. A VA is not a DC Watt, actually less. You need *much* bigger transformer just for this one output. 12VAC 2Amps (24VA) may be barely big enough. Also the suggested diodes and regulator will be working a bit past their specs--- I predict it will work on Thursday and blow-up at the Friday gig.

Read R.G. again. Sharpen your BS detector and wander around Google/Bing--- there is a LOT of power supply info in the web and not "all" of it is crap. If you can find a bootleg of TI's 1970s Power Supply Design book, the early part is great.

[baloubass]

You can go with this guy and send a pm to a custom TR. The delivery time is so long but good price.
I build my power supply with 7x9v 0.5a and 1x18v 0.5a with 1 TR.
https://fr.aliexpress.com/item/6-8-v-R-Transformateur-Noyau-R40-personnalis-transformateur-220-v-entr-e-cuivre-bouclier-sortie/32953005345.html?spm=a2g0s.9042311.0.0.40696c378s4FRs

[Nitefly182]

Thanks for the knowledge folks! Seems more and more reasonable to deal with a bit of a larger size and install the guts of a handful of inexpensive single supplies into a project box that will fit under my pedaltrain and call it a day rather than try to design a supply from the ground up for this one specific purpose.

[Rob Strand]

Thanks for the knowledge folks! Seems more and more reasonable to deal with a bit of a larger size and install the guts of a handful of inexpensive single supplies into a project box that will fit under my pedaltrain and call it a day rather than try to design a supply from the ground up for this one specific purpose.

1. 12VDC 1200ma
2. 5VDC 150ma
3. 18VDC 150ma

You could probably save one power supply by generating the 5V rail using a 5V linear regulator powered off the 12V rail.   The small package 5V regulators will probably get too hot but a TO-220 package should OK.  (See the datasheet for caps you need to add around the regulator.)

[Nitefly182]

Another thing I hadn't thought about is two of the three power supplies I'm currently using are SMPS so trying to make the whole thing linear with that much current was destined to be a giant brick of a box. Building the compiled setup with three rehoused SMPS circuits would mean a MUCH smaller box in the end.

[rankot]

I'll be gratefull if one can check my "homework math"

I need a power transformer for my mic preamp (green pre). It want to make a PS which will provide ±15V DC and 48V DC, so I can use it for my mic preamp. Preamp use 4 opamps and two transistors, so I have calculated that it will need max. 40mA per rail.

I will use this schematic for a regulator part of a PS (not all those different regulators, LM7912/LM7915 actually, but I don't have models for them), so I wander how to choose a suitable power transformer.


I know that I need ±18V unregulated for this to work. I've found few on-line articles which say different things, so I'm definitely not sure how to calculate.

Sowter has this page: http://www.sowter.co.uk/rectifier-transformer-calculation.php, stating that if I need 18V unregulated DC, Vac should be V_DC*0.71, so for 18V DC I need 13V AC transformer. I presume that's RMS value, so multiplying it with 1.41 I will get 18V p-p AC. I tried with that value in LTspice, but it's not sufficient - actually, I need V_DC*1.41 p-p AC (or 18V RMS AC) for this to simulate as expected.

At the other hand, Hammond has a nice guide which says that if I use full rectifier and capacitor input load, Vdc (unregulated) should be Vac(RMS)*0.9, which brings me to 20V RMS AC transformer (or 28.2V p-p AC for simulation).

The most complicated part (and the one I don't understand) is secondary current rating. Both those articles claim that Idc equals 0.62 Iac, but if I set series resistance of my voltage sources in LTspice to V_ac,RMS/I_AC then the simulation can't provide enough current. I'm getting current peaks of 200mA with source internal resistance set to 30 Ohm and V_AC,p-p set to 25.38V (or V_AC,RMS=18V).

So the question is: shall I use the transformer with current ratio as in proposed in those articles, or I shall go for more, especially if I consider the fact that transformer current is peaking up to 800mA when circuit starts?

[PRR]

Your voltages are off.

First: you need *two* 15V rails. One 18V CT winding will not do that (in this form). First figure out ONE side of the "simple" part (the 15VDC), then double that for the +/-15V part.

Yes, ideally a 15V regulator can work from 18V, 3V headroom. Then add 10% for ripple, 10% for wall variations, and 10% for general unknown. (If your "unknowns" were KNOWNs you would not need a regulator.) You need 24V raw DC. Maybe 25V for diode loss.

25V raw DC needs 25V/1.414 = 17V AC RMS.

This is per-side! So you will need 34V AC RMS to get the two 25V DC rails. 36VCT is probably the closest standard winding.

Simulator shows this making +/-23VDC.


The stacked doubler is a real brain-game. Apparently it works, giving 65VDC, which is probably sufficient for 48V regulated.

You show 40mA each output. Yes, the AC RMS current will be like 1.6 to 1.8 times the DC current; I simply pencil "2". So without the Phantom leg we need 73mA-80mA of AC rating.

On the DC side we have 2*23VDC times 0.040ADC or 1.84 Watts DC. The AC rating is 34Vrms times 0.080A AC or 2.72VA. The transformer VA must be 1.5 times the DC Watts.

The Phantom leg is very screwy. Let's figure from Basic Law (no free lunch). The DC current is shown as 14mA DC. This will be at about 65V. So 0.92 Watts of DC. Multiply by 1.5, needs 1.4VA of transformer.

2.72VA + 1.4VA = 4VA. At 34V: 4VA/34V is 0.12 Amps AC. Actually, because 4VA is a very small lump of iron, I would arbitrarily double that to 8 or 12VA, just to get a better, less saggy, part. (Also, 1/3rd of your power comes from 1/2 of your winding, so you need to be a little generous.)

800mA seems to me to be much more than you need.

> transformer current is peaking up to 800mA when circuit starts?

It takes minutes to burn-up a transformer. After 1 Second you have 200mA peaks but they are very narrow, current is *zero* for 80% of the time. If your plotter has an RMS() function, this will approximate the heating effect of the narrow spikes. _IF_ your voltages come out correct.

> Hammond ...says that if I use full rectifier and capacitor input load, Vdc (unregulated) should be Vac(RMS)*0.9

That Hammond "guide" has real errors. However I am sure that is for CHOKE-input rectifier, which does not make sense for us.

[PRR]

Hmmmm... estimating the winding resistance of a 12VA (36V 0.3A) transformer, we see that the winding RMS current is down to ~~120mA as we approach the 20 Second mark (iron won't burn that fast). However the DC voltages (not shown) are very saggy, giving hardly 18V raw DC which leaves no room for wall or part variation.

If you were making a million of these you would want a sharp pencil and a lot of testing of prototypes to get the absolute lowest-cost transformer and avoid going broke too soon.

For DIY, over-specing is cheaper than brain-pain. Go at least 24VA and if you see a 40VCT part use that. (48VCT makes 35V DC which may be your regulators' Max.)

[Rob Strand]

So the question is: shall I use the transformer with current ratio as in proposed in those articles, or I shall go for more, especially if I consider the fact that transformer current is peaking up to 800mA when circuit starts?

No need to worry about peaks.  The transformer heating depends on RMS current.  You need to add some series resistance to the transformer to represent winding resistances (which determine transformer regulation).

For DIY, over-specing is cheaper than brain-pain. Go at least 24VA and if you see a 40VCT part use that. (48VCT makes 35V DC which may be your regulators' Max.)

12VA will easily do it from a load perspective.  However, I'm not 100% on the how much current is going to be drawn from the raw DC lines (and if current will be drawn on those at the same time as the regulated lines.)

There's a couple of things which keep the voltage up:
- The open circuit voltage of a transformer is higher than the rating
- If the load is less than the rating then the voltage drops less and the voltage is higher.

One thing is the open circuit transformer voltage is higher than the rated voltage.  A 12VA transformer might have about 15% regulation so under no load the voltage of an 18V rms transformer will be 18Vrms * 1.15 = 20.7V rms.   For a 6VA load on *each* 18Vrms that mean 0.33A full load.  The transformer resistance in *each 18V arm *is (20.7-18)/0.33 = 8.2 ohm.

So for the spice model use a voltage source with 20.7V * 1.414 = 29.3V pk  and 8.2 ohms in *each* arm.   PRR's 10 ohms are equally good estimates.

You probably should re-do the calculations for 10% regulation and see if it still works.  Unless you already have have the transformer and know the regulation.

Once you get that far try dropping the input voltage by say 10% to see how you go with mains regulation - as PRR already mentioned.

[PRR]

> You need to add some series resistance to the transformer to represent winding resistances

It's in his sim. Unlike my coal-fired sim, his has a field in the "battery" for "Rser" which I ass-ume is a series resistance.


The other fields must be checked against the sim's docs. I would ass-ume 0 is a DC offset, 25.38 is a *peak* of the SINE, and 50 is the Hz. (Since he has stacked two 25.38Vpk sources, the "18V AC CT" note is wrong, it is 18-0-18 or 36V CT, which explains some of my confusion.)

Note that 36V 0.8A implies the winding will drive a 45 Ohm resistor with AC. Assuming regulation is 10%-20%, that is 5 to 11 Ohms total winding resistance, or 3 to 6 Ohms each side; "30" may be wrong. (30 twice means more Ohms in the transformer than its rated load, which is awful hot and huge un-sag.)

[Rob Strand]

It's in his sim. Unlike my coal-fired sim, his has a field in the "battery" for "Rser" which I ass-ume is a series resistance.

Good eyes!  I forgot some sims have that (the sim I use is ancient).

Note that 36V 0.8A implies the winding will drive a 45 Ohm resistor with AC. Assuming regulation is 10%-20%, that is 5 to 11 Ohms total winding resistance, or 3 to 6 Ohms each side; "30" may be wrong. (30 twice means more Ohms in the transformer than its rated load, which is awful hot and huge un-sag.)

Agreed, 30 ohms is far too high.

[rankot]

PRR and Rob are really great and helpful, as always! You guys really opened my eyes on this! <3

Checking current RMS, it shows some 143mA, so I'm pretty safe with even 2x300mA rated transformer. I can locally buy only 2x444mA, so that's why've asked, and I will go with that one. https://www.mgelectronic.rs/trafo-print-myrra-220v-16va-2x18v

Cheers!

[Rob Strand]

Checking current RMS, it shows some 143mA, so I'm pretty safe with even 2x300mA rated transformer. I can locally buy only 2x444mA, so that's why've asked, and I will go with that one. https://www.mgelectronic.rs/trafo-print-myrra-220v-16va-2x18v

On the datasheet you can see "no-load voltage"  of 22V.  So that transformer has about 22% regulation.
About  9 ohms equivalent resistance in each arm in each 18V winding.