# Confused with transformer VA/aperage rating

#### Simas Chomentauskas

Joined Nov 3, 2017
11
Hello, this might be answered several times already, but I can seem to pick correct keywords to find an answer :]

1) I need a linearly-regulated PSU of 9V@2A. Using three 7809 in parallel with load resistors to regulate.
2) Picked a transformer with VA of 40 for the job (2x12V@1.67A, in parallel = 12V@3.34A), expected to be more than OK.
3) Able to to get only ~1A of clean supply, at 1.5A it already is way overloaded (regulation of, traffo gets really warm, voltage before regulator drops to 9.5V).

Seems the transformer, although seems more than adequate for the job, fails to deliver required current (9Vx2A = 18W required, 40W rated. Fails even at 13W). My question is - did I missed something in selecting the required transformer or should I suspect it's quality and stated ratings?

My calculations:
- I need 9V@2A and will use 2x4700=9400 uF smoothing caps.
- This will give me 2.13V ripple voltage (2A / (2*50Hz*0.0094F) )
- And all-time required peak voltage of 15.63V ( 9V (output) + 2.5V(regulator dropout) + 2.13V (ripple) + 2V (diode drop) )
- This translates into 11.08V@2A requirements for transformer and I stepped it up to 12V@3.34A

Just to check it with the other calculation method:
- At 9V@2A draw, voltage before regulator should be 13.75V, more than enough for 7809 to work correctly ( 12*1.41V (secondary rms) - 2V (diode drop) - 1.05V (Vripple/2) ).

Last edited:

Joined Jul 18, 2013
19,157
That would only be a 40va transformer, with regulators and the capacitors used, this most likely pushes it over the limit.
The larger the smoothing capacitors, the higher VA needed.
Take a current reading of the AC might show more, but a 60va or > may be needed.
Max.

#### The Electrician

Joined Oct 9, 2007
2,750
My calculations:
- I need 9V@2A and will use 2x4700=9400 uF smoothing caps.
- This will give me 2.13V ripple voltage (2A / (2*50Hz*0.0094F) )
- And all-time required peak voltage of 15.63V ( 9V (output) + 2.5V(regulator dropout) + 2.13V (ripple) + 2V (diode drop) )
- This translates into 11.08V@2A requirements for transformer and I stepped it up to 12V@3.34A

Just to check it with the other calculation method:
- At 9V@2A draw, voltage before regulator should be 13.75V, more than enough for 7809 to work correctly ( 12*1.41V (secondary rms) - 2V (diode drop) - 1.05V (Vripple/2) ).
You're not taking into account the winding resistances. What is the primary voltage rating? What are the DC resistances of the primary and the two secondaries? What was the RMS current in each of the secondaries under your stated load?

#### Simas Chomentauskas

Joined Nov 3, 2017
11
Took some more digging but still can't the answer :/

Primary is 220V rated, measured 78ohms, each secondary is 2ohms (so 1 ohm with 2 in parallel). With output of 1.5A current in joined secondaries was 1.8A, seems well within transformers limits.

I guess this has something to do with this (found in another similar thread): ,,Note that you need to derate a transformer's current rating about 50-60% when it is used in a rectifier-capacitor DC output circuit, i.e. the maximum DC output current should be no more than about 50-60% of the transformer RMS rating!!!"

It seems I just have to take that into account while choosing the transformer, but I can't (and would like to) understand the nature of this. Taking into account the number in later quote - I should expect only 50-60% of the rated transformer current output in real life. That makes roughly 1.6A out of 3.34A rated which is exactly what it's giving now.

At 1.6A in secondary I get roughly 1.3A after regulators and it's exactly the point where everything starts to fail and drop.

Last edited:

Joined Jul 18, 2013
19,157
): ,,Note that you need to derate a transformer's current rating about 50-60% when it is used in a rectifier-capacitor DC output circuit, i.e. the maximum DC output current should be no more than about 50-60% of the transformer RMS rating!!!"

.
As I said in #2!
I may still have the formulae for this somewhere.
It relates to the charge current of the smoothing capacitor and is dependent on the load and size of capacitor, essentially due to the rate of charge after each half cycle peak caused by the resultant ripple current.
This results in increased sizing of the transformer due to this effect.
Max.

#### nsaspook

Joined Aug 27, 2009
6,473
The rectifier-capacitor DC output circuit creates a non-sinusoidal current waveform in the transformer. Analogy: Instead of the nice smooth field rotation of energy flow of a V8 engine you have the thumper strokes of a Norton motorcycle. The non-sinusoidal current increases total ohmic (IR winding real resistance/power becomes a larger component of total impedance) losses because the current pulse peaks are usually much higher than the peak levels of the sinusoidal current for the same power level during one complete cycle of the waveform.

https://www.dataforth.com/rms-revisited.aspx

Some non-sinusoidal current effects on power transformers.
Ohmic losses: windings
Stray losses: eddy current, core

#### crutschow

Joined Mar 14, 2008
23,487
the maximum DC output current should be no more than about 50-60% of the transformer RMS rating!!!"
It seems I just have to take that into account while choosing the transformer, but I can't (and would like to) understand the nature of this.
It's due to the high RMS transformer current caused by the rectifier capacitor filter.
Such a circuit draws high current at the peak of the AC waveform to generate the average DC output current.
Since the power lost in the transformer winding resistance is proportional to the square of the current, this high peak current translates into a high RMS current value.

This is shown in the LTspice simulation below:
With a 1 ohm winding resistance and for a DC load current of 1.09A, the peak transformer current, I(R1), is about 3.61A and the RMS current, is about 1.77A.
The ratio is thus 1.09/1.77 = .615 (where the 50-60% derating comes from).
The actual ratio depends upon the transformer winding resistance and the value of filter capacitance, as those both affect the peak current for a given DC output current.

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Joined Jul 18, 2013
19,157
@Simas Chomentauskas
The paper I have on it was the result of a empirical test by the Univ of Mich.
Max.

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#### Simas Chomentauskas

Joined Nov 3, 2017
11
Wow guys, that was very educating, especially the paper with clear guidelines for target ripple.

All seems to be clear now!

#### The Electrician

Joined Oct 9, 2007
2,750
Wow guys, that was very educating, especially the paper with clear guidelines for target ripple.

All seems to be clear now!
What meter did you use to measure the secondary current under load?

#### Simas Chomentauskas

Joined Nov 3, 2017
11
Freshly aquired Mastech MS8268

#### crutschow

Joined Mar 14, 2008
23,487
regulation of, traffo gets really warm, voltage before regulator drops to 9.5V
You do understand that at a 2A output, the power dissipated by the regulator is about ((12*1.4-2) -9V)) * 2A = 11.6W(?).
This means the regulator will require a good heatsink (e.g. ≤ 5°C/W).

#### The Electrician

Joined Oct 9, 2007
2,750
Freshly aquired Mastech MS8268
Looking into the user manual for the MS8268 I find this:

The MS8268 does not read the true RMS value, so your measurement of the secondary current is too low. The transformer runs hotter than one would think based on your measured secondary current.