How to choose the transformer rating for linear power supply voltage regulator

[Ehsan]

I am designing a lab power supply. I intended to have a range of 0-30v up to 2A.

So I went out and bought a 24v-0-24v transformer with rating of 2A.

I chose 24v because that gave me 33v peak DC voltage after rectification, I I kept about 3v for voltage drop out so I expected a raw 30v available, but I think something is wrong here with my choosing the secondary voltage also because I can't go up to 30v with 2A current.

Now after the rectifier bridge and regulator circuit, if I try to get a DC current of more than 1.2A the transformer secondary voltage drops significantly. So I guessed I am overloading it. I measured the AC currents at full load as follow :

It is a center tapped transformer, when I connect the ammeter in series with the center wire the AC current reads 1A. When I put it in series with each windings it reads 0.8A. This reading are when the load consumes 1.2A of DC current.

Now my question is that to get 2A of DC current at max 30v what rating of transformer should I buy?

(Right now with my 24v-0-24v transformer and a full bridge rectifier I only can have a reliable regulated voltage at rating of 15v, 1A. Increasing either the voltage or current impacts the regulation and output voltage drops considerably. )

 

[(*steve*)]

the short answer is that you should have a transformer with a secondary current rating 1.414 times the dc current

 

[Ehsan]

(2A DC) * 1.414 = (2.83A AC)

So I need to get a 3A transformer, that's why my current 2A transformer doesn't work!

Thank you Steve.

 

[Ehsan]

ok I searched for that 1.414 number to see where it comes from and I found these two articles that I would like to reference here for future users. (I hope I knew this information before purchasing my transformer!)

http://www.hammondmfg.com/pdf/5c007.pdf

http://www.powervolt.com/techan01.html

 

[(*steve*)]

Yes, and you'll note that often they specify something higher than 1.414. The reason for this is that with a capacitor holding up the output voltage, all the current is drawn near the peak of the voltage from the transformer.

Given the 1.6 to 1.8 multiple, you might even go for a 3.5A transformer.

Interestingly, "green" switchmode power supplies circumvent a similar issue by employing a boost converter after the rectifier so they can draw current through the full mains cycle. For a linear power supply you could do something similar, but it wouldn't really be worth it.

 

[hexreader]

I am a little confused.... ... but I guess my understanding is poor.

If the 24-0-24 transformer is being half-wave rectified using two diodes (don't know if that is the case here), then won't the average current in each winding be half of the AC output value?

... or to put it another way.... If each winding can provide 2A AC 100% of the time, then can't it provide 4A AC half of the time?

I would have thought that the 2A transformer should be able to provide a 3A DC final output and remain with specification for average current.

I can well imagine that 24V AC might be insufficient to give 30V DC final output under load though.....

My maths is terrible, and I have not used any, but by estimation and a limited understanding, I reckon the problem is of insufficient voltage, not insufficient current.

I am happy to be corrected and proved wrong.....

EDIT: maybe I had better start reading and try to understand the links above - I might be able to answer my own question...

 

[Ehsan]

Hexreader:

A half-wave rectifier does not have two diodes. It only has one diode. Those with two diodes are called full-wave rectifiers.

I am using "Dual Complementary Rectifier Circuit" to get both positive and negative voltages which is something similar to this :



A 2A transformer can only provide (2A / 1.414) = 1.414A DC current, and Steve is right: In practice you should go for 1.6 or 1.8 multiplier.

24v AC is the rms value so Vp = 24 * sqrt(2) = 33.9v.

34v is the DC unregulated peak value coming out of secondary winding.

But one should not go for the peak value but the average value !

Average DC voltage can be calculated by this formula: V_avg = (2/pi) * V_peak .

So in my case average voltage is : V_avg = (2/pi) * 33.9v = 21.58v

So basically with 24v-024v transformer I can only get 21.58v max DC voltage.

To get 30v DC one need to buy a 33v-0-33v transformer at least.

 

[Ehsan]

Can someone teach me how to write math formulas here ?

 

[Harald Kapp]

You're doing a good job the way you write them.
If I need more complicated equations, I write them in a math-capable editor (e.g. MS-word) and paste them as graphics.

By the way: Where did you get this equation from?

Average DC voltage can be calculated by this formula: V_avg = (2/pi) * V_peak

I guess you took it from a website like this ?
In that case the equation is valid for an unfiltered DC only. Since unfiltered rectified AC is not useful for a power supply, you typically add a filter capacitor (as you've donde). Ripple voltage is now a function of the current and the equation changes to

(image copied from above website).

 

[Ehsan]

I took it from here, scroll way down the page:

http://www.rfcafe.com/references/electrical/sinewave-voltage-conversion.htm

 

[Harald Kapp]

That's for unfiltered rectified AC, as I guessed.

For the design of a regulated power supply you take Vpeak = 1.414*Vrms-2*Vdiode
where Vrms is the AC output of the transformer and Vdiode is the voltage drop across one rectifier diode.
With Vripple in mind you get Vmin = Vpeak-Vripple.
Vmin needs to be at least Vout-Vdrop
where Vdrop is the dropout voltage of the regulator (voltage drop across the regulator from in to out). Add a suitable safety margin for stable operation under worst case conditions and after long time wear of e.g. the capacitor.

 

[Ehsan]

I was talking about the output average DC voltage NOT the mere ripple voltage. But you explained it very clearly. Thank you.

Oh I forgot the voltage drops in diodes.

So to choose the secondary voltage of transformer we don't consider the 'average voltage' at all ?

 

[Harald Kapp]

The average voltage is useless.
You need the peak voltage at the storage capacitor. This will define the max. voltages your components have to withstand (specially the capacitor itself).
You need the ripple voltage to know which min. voltage will be available from the storage capacitor to power the regulator. Or vice versa from the min. voltage needed for the regulator you determine the max. allowed ripple voltage and from that and the output current you can calculate the size of the storgae capacitor.

 

[(*steve*)]

No.

However Harald's method hides a devil in the details.

Following that method you might decide to use a very large value of capacitance (for low ripple) and note that your DC voltage is relatively high.

The devil in the detail is that current only flows from the transformer when the instantaneous voltage is higher than the capacitor voltage (ignoring diode voltage drops).

since the power delivered to the capacitor is equal to the power being drawn from it, the fact that current can only flow for a brief period near the peak of each half cycle means that the peak current can be very high.

Lets assume your ripple is 10% of the peak voltage. This means that for about 30% of the time the transformer voltage exceeds the minimum voltage for your 10% ripple. However, the voltage is rising during the first half of each peak, this is when the capacitor charges. So all the current is drawn during 15% of the waveform.

Lets assume that these figures were calculated on the assumption of a 1A DC current. This will produce a peak current from the transformer of about 6.7A.

A significant amount of the losses in the transformer are resistive. If the resistance of the transformer secondary is R ohms, and the 1A is drawn continually through the full cycle, losses are 1R watts. If you draw 6.7A for 15% of the time, the average losses are 6.7R watts.

Apart from the increased dissipation, there are also IR losses that will depress the peak voltage from the transformer. In extreme cases the transformer can saturate.

Fortunately for all of us, transformers are pretty rugged and tend to put up with all sorts of abuse without much complaint.

 

[(*steve*)]

The "No" in my last post is in response to the question about Average voltage being used.

 

[hexreader]

Sorry to break the flow of the thread again - just wanted to say thanks - I understand a little better now