Hello people,

I've built a linear type dual power supply for the purposes of driving low current op-amp audio circuits. The design is relatively simple; a 220v - 12v transformer which i'm actually running on UK mains (which reads (247-ish)... this subsequently boosts the output voltage to around 18v or so.

The waveform was little "bent-over" on the scope but once I smoothed it off with a few 4700uF electrolytics I had nice steady voltages. I've also used postive and negative vregs and a few 1uF ceramics for good measure. Finally i added an op-amp feeback loop setup on both sides with potentiometers so that i can control the voltages.

The circuit appears to working nicely although I haven't really applied any load yet apart from my scope and my multimeter.

The transformer I bought is a cheap chinese thing which i got off amazon for about 4 quid and it doesn't say the maximum current rating. All it says is 220V primary 12V secondary and 2 Watts of power. So I know that in order to calculate my current I can do P/V but should I use the primary V or the secondary. To make things more confusing I'm not actually getting the exact Voltage due to the tranformer being designed for EU mains.

If I measure the output of the Vregs (at maximum V) I get about 14V (+ and -) give or take a few millvolts which I’m guessing is just down to vreg’s tolerances. So If I do 2 Watts/ 14 Volts I get 0.142A. if this is correct then that’s not too bad seeing as though I only intent to use this power supply to drive a few op amps at a time and I can keep the gains low and leave the main workload to the power amplifier…. but am I calculating this correctly??

And also, do you think using this transformer may cause me problems? Seeing as firstly, it’s blatantly pretty poor quality (I can’t even find a datasheet for it) and secondly it’s not desgined for UK mains… that being said it does appear to be working okay.

Please see schematic attached.

Thanks in advance

 

Oh yeah... and here is a link to the amazon listing for the transfomer: -

https://www.amazon.co.uk/Power-Tran...373&sprefix=ac+to+ac+transfomr,aps,189&sr=8-4

thanks again

 

18vac = 25vdc after rectifier and smoothing.
Isn't there a source of LV transformer's like that at local HW stores similar to here in N.A.?
You are only getting 1/2 wave rectification WRT common (GND)

 

If I measure the output of the Vregs (at maximum V) I get about 14V (+ and -) give or take a few millvolts which I’m guessing is just down to vreg’s tolerances. So If I do 2 Watts/ 14 Volts I get 0.142A.

Transformers are normally rated while driving a resistive load like a heating element. When you use a diode and capacitor to make DC the transformer will output a little less power. The diodes only conduct during the peak of the wave. ("flat top")

The "2 watts" is power from the transformer not power to the load. So some of the power is going into heat in the 7815. With out doing the math maybe 1/3 of the power is heat and not making it to the load.

 

So I know that in order to calculate my current I can do P/V but should I use the primary V or the secondary.

If you want th primary current you use the primary voltage. If you want the secondary current you use the secondary voltage.

And I hope you realize that your opamp based regulator can only output about 20 mA.

Bob

 

So If I do 2 Watts/ 14 Volts I get 0.142A

As ronsimpson alluded to, due to the high peak current pulses of a rectified sinewave, which generates high RMS transformer currents, the maximum full-wave rectified DC current is about 0.6 times the rated transformer RMS current or 0.6 * 2W / 12V = 100mA.

 

AC voltages are stated using RMS (root mean squared) values.
12VAC of a sine wave has peak value of 12 x 1.414 = 17V.
After one diode drop you will get about 16.3VDC on a reservoir capacitor with no load.
The average voltage into a load resistor is another story altogether.

Average current x average voltage = average power

Derate your transformer to 80% stated power to take into account power losses.

 

Derate your transformer to 80% stated power to take into account power losses.

From 5c007.pdf (hammondmfg.com):


As you can see, the derating factor for DC output current is 62% of the transformer rated RMS output current for a bridge rectifier with a capacitor filter.
The 0.6 value I used was not arbitrary.

 

I was only derating for power loss in the transformer, not for circuit configuration.

If I follow this correctly, you have to derate the power by 0.9 x 0.62 = 0.56 for full wave bridge with capacitor filter.

This is for Hammond transformers. Mileage will vary for other manufacturers.

 

European mains is 230V ± 10%. Anything specified for use in EU + Britain must work from 208V to 253V.
(British mains is 230V +10%-6% and continental Europe mains is 230V +6%-10%.)
I've used a lot of Chinese wire-ended transformers and generally they have been good quality - if it looks like the one in the picture (and I know I can't tell much from a picture) I wouldn't expect it to cause problems. It's now £10 and that seems expensive.
As @MaxHeadRoom says, as you only have a single secondary so you are getting half-wave rectification.
See
https://www.rapidonline.com/vigortr...hassis-mains-transformer-6va-2x-0-12v-88-3712
https://cpc.farnell.com/unbranded/835-469/transformer-6va-230v-12v-12v-cha/dp/TF01571?st=transformer
for something more appropriate to what you need.

 

Thanks very much for all the help guys. I'm spending a bit of time looking at transformer calculations videos on youtube this morning. I didn't realise that transformers are generally rated as RMS but now that you lot have pointed it out that not only makes sense but also explains quite a few things regarding the readings i was getting on my scope.

And I hope you realize that your opamp based regulator can only output about 20 mA.
Bob

And thanks for pointing this out BobTPH. Perhaps I should just settle for a non-adjustable supply straight from the regulators or possibly even get hold of some adjustable regulators.

Really appreciate everyones input

 

If just use a rectifier and a couple of smoothing caps, you'll quickly learn how to make op-amp circuits with good power-supply rejection, immunity from ripple, and tolerant of variations in supply voltages!
However, I'd recommend a 7812/7912 setup. Good enough for most circumstances, and at least you know that if there's something funny happening on the op-amp output, it's not caused by the power supply.

 

If I follow this correctly, you have to derate the power by 0.9 x 0.62 = 0.56 for full wave bridge with capacitor filter.

Why do you include the 0.9 value for the voltage?
I'm referring to the current derating value only.

This is for Hammond transformers. Mileage will vary for other manufacturers.

The derating is from the RMS output current rating of the transformer, which mainly comes from the IR losses in the windings' resistance.
Don't see how the manufacturer of the transformer would affect that mileage. (?)

 

1) I am derating the power, not the current alone.

2) IR losses in the windings' resistance varies from one manufacturer to another.

 

1) I am derating the power, not the current alone.

Why?
Derating the current already derates the power.

2) IR losses in the windings' resistance varies from one manufacturer to another.

Of course, but the transformer current rating is mostly based upon the particular transformer's winding resistances, so the same derating should be valid for all power transformers.

 

In my experience building power supplies, performance of transformers did not meet the manufacturer's ratings.
The one exception has been transformers made by Hammond.

 

In my experience building power supplies, performance of transformers did not meet the manufacturer's ratings.
The one exception has been transformers made by Hammond.

Fair enough.
I would have assumed that most transformers from a reputable manufacturer would meet specs, but I suppose many these days are from Asian suppliers with dubious quality.