power transformer current

20va power transformer electronic formulas

hi folks
i do have a question about transformer
for example the primary of my power supply is 220
and the secondary is 9-0-9
and it says 20VA
How do i know the output current of this transformer?
is it 20/ 9 or 20/(9*1.414) or 20/(9*1.414-0.7 )?
0.7 is the voltage drop of a diode
i will use half wave rectification
Thanks in advance

current transformer full-wave rectifier

The apparent power is 20VA.
That means the output current is 20/(9*2)=1.1A in each secondary

hi pal
thanks for ur fast reply
but what do u mean "it is 1.1A in each secondary"
?
Do you mean 9-0 , 0-9 and 9-9 these 3 pairs are 1.1A?
Thanks

You basically have two secondaries, each producing 9V at 1.1A.
You can draw 1.1 A from 9-0 and 1.1A from 0-9.

hi vvv,
thanks so much for replying
I asked a guy about this , but he gave me this answer
which made me so confused

he said "your 20VA 9-0-9Vac transformer has a maximum continuous output of 20/[9+9]=1.1Aac.
If you use this to feed a capacitor input filter to get a +-12Vdc supply the maximum continuous output is approximately half the AC output i.e. ~550mAdc.
If you run the transformer at it's maximum rating of 550mA it will run hot, right up at the maximum that the manufacturer designed it for. This internal temperature will be about 130degC.
I suggest the maximum continuous demand be limited to about 50% of rating i.e. 280mAdc continuous."

he said something diff from your statements
now i am so confused
hmmmmmmm
do i misunderstand something here?
Please help
regards

Added after 26 minutes:

hi VVV
sorry i forgot to ask you what if i connect 9 and 9 2 9's

then how much current would that be?
Thanks

As often happen the true is somewhere in between so both statements are correct, depending of transformer working condition .
I do not want to analyze problem in deep. I just want to show you a direction you can follow and do a precise analyze.
Lets us simlify analyse by taking into account only the transformer resistances.

The most comfort condition for transformer is when its load is resistive.
The current is sinusoidal and its momentary changes lays between 1.4 * I nominal - through zero to + 1.4 I nom. (I nom. = 1.1 A)

When you load transformer with rectifier and capacitor the (standard for almost all electronic equipment) the secondary current have form of pulses. In case of 50 Hz mains the half cycle last 10 msec. The capacitor is charged with 2 msec. pulses with peak value about +/- 6A.

Since the heat generated by wounding is equal I * I * R
(I,R = secondary/primary, current/resistance)
then you can calculate the power that worm up transformer for both types of load.

As conclusion you should get that short pulses with high value generate the higher power than sinusoidal current during resistive load.

The worst case is when the capacitor is charged not by rectifier diode, but through phase controlled thyristor without the inductance in series.

hi

hmmmmmmmm this is too hard for me to understand SOS!lol
In case of 50 Hz mains the half cycle last 10 msec. The capacitor is charged with 2 msec. pulses with peak value about +/- 6A.

how come 50 hz half cucle = 1/100 sec?
why is the cap charged in 2/1000? how to calculate this?

pulses with peak value about +/- 6A. ???????? why is it +/-6A?
hmmmmmm
gosh i am totally lost now


thanks
regards

The difference is basically between DC output current of a rectifier circuit and the RMS value causing the transformer losses. The term RMS contains the complete definition of the measurement. Form the square of current momentary value, average over a meaningful interval, e. g. a cycle of the waveform and calculate the square root. As a result, the said short, high capacitor current causes thermal effects much higher than it's respective DC value (simple mean).

A half wave rectifier, that uses two windings alternatingly, has higher transformer losses than a full wave rectifier. It may have advantages for a low voltage supply due to lower rectifier volatge drop anyway.

To simonwai999,

As I understood you have a transformer with two 9V windings connected seriously. In this case you will have about 20/(9x2)=20/18=1.1A available continuous current. For half-Wave (single diode) rectifier theoretically output DC voltage should be about 18x0.45=8.1V RMS. RMS (Root Mean Square) is equal to hypothetical DC value that will produce the same amount of heat as discussed value of AC. The actual power what your transformer will consume from the power line is about 15% more for this range of transformer’s power, so it will be about 20*1.15=23W. The difference between consuming from power line energy and actual output will be the power that dissipated inside your transformer and heat it up (23-20=3W). Actual calculations for output voltage, current, power and heat are not easy and empirical. If overheat is concern try to find bigger transformer. Output voltage and ripple will depends heavily on rectifier circuit, filter capacitor and regulator. You also need to consider diode’s and regulator’s heating loss.

hi pal
thanks a loti understand most of it ,but
about this part

"For half-Wave (single diode) rectifier theoretically output

DC voltage should be about 18x0.45=8.1V RMS. RMS (Root Mean Square) is equal to

hypothetical DC value that will produce the same amount of heat as discussed

value of AC."

my doubt is what is this figure 0.45?
how do we get it?
is it 1.414/ 3.14?
please tell me more
regards

The factor 0.45 output voltage would be seen only without a filter capacitor. I understood so far, that your intention is to get a filtered DC output, but I may be wrong.

hi
i wanna connect 9-0 with a bridge rectifier to get around 12vdc then
filter the input with a cap
for the input of 7805
then a filter cap and a bipassing cap to get 5vdc output

so the 0.45 factor is used for half wave rectifier right?
if so i will not need this factor then
regars

If you want to use your existing 9-0-9 transformer, the discussed helf wave rectifier would allow a higher output current than a full wave rectifier. With two isolated 9 V windings, paralleling the windings and connecting a brifge rectifier would be better.

With the 20 VA transformer, about 1A at 12V would be the full load with a suitable capacitor (e.g. 2200 uF). Idle voltage can be expected higher, e. g. 15 V.

hi
with one A at 12 VDC is enough for me
i will only connect a 7805 in the circuit
thanks

One drawing is worthy more that 1000 words.
See attachment.

hi pal


very clear drawing
thanks a lot
regards

I contradict the not less than 4700uf in this case. Please consider the available voltage margin. Making the ripple much lower than necessary only increases transformer losses due to a higher crest factor.

hi
in the drawing
?700 uf
i actually cannot see the first digit of this cap value
4700uf for this power supply hmmmmmmmm
hope the guy who drew this will reply us
regards

The drawing suggest 10000 uF and the text says not less than 4700 uF. That's generally a good suggestion, if the unregulated voltage is required to have a lower ripple. 10000 uF means about 0.8 Vss ripple at 1 A DC load with a 9V transformer, 2200 uF means somewhat below 2.5 Vss, leaving minimal difference voltage of > 4 V at the 7805. A cheap design would possibly use 1000 uF with around 4V ripple and a capacitor AC current near to the maximum rating, resulting in a reduced lifetime.

Most DIY electronicans (and some professionals as well) would simply assemble a circuit and watch the results - that's also a way to learn.

hi fvm thanks

but would you kindly tell me how u got all these figures?
how did u calculate it?

if the unregulated voltage is required to have a lower ripple. 10000 uF means about 0.8 Vss ripple at 1 A DC load with a 9V transformer, 2200 uF means somewhat below 2.5 Vss, leaving minimal difference voltage of > 4 V at the 7805. A cheap design would possibly use 1000 uF with around 4V ripple and a capacitor AC current near to the maximum rating, resulting in a reduced lifetime.
10000uf = 0.8vss ripple at 1A DC?
2200uf = somewhat below 2.5 vss?
leaving minimal diff voltage of 4v at 7805?
please teach me how to calculate all these
thanks a lot
simon

Added after 4 hours 54 minutes:

hi fvm
is the equation on this site for us to get the ripple voltage?
thanks
http://en.wikipedia.org/wiki/Ripple_(electrical)

Added after 8 minutes:

yes i tried to use the equation from that site to calculate
2200uf


ripple voltage (Vr) = 1 / 2(50)(2200/1000000)
Vr= 1 / 2(5)(22/1000) = 1/ 220/1000 = 100/22 = 4.54( it is not 2.5 as you said)
I wonder why
Thanks