Transformer Currents

[adrianbodor]

Hy,

I want to buy a toroidal from my local dealer but i have few doubts

I need a 9V transformer at 30VA My circuit draws over 2 amps
How can i know how much amps deliver this transformer?

Let me say my version and correct me please if i'm wrong
Following the formula 30VA/9V result 3.3 amps

If dual secondary of 4.5V each at same 30VA means each winding is about 15VA at 1.65 amps

My question is if I wire those windings in series (center taped) i get 6.4 volts after rectify at how much amps? 1.65 or 3.3?

This I don't know, in parallel or series what is the value for max current output if dual secondary is used rated at 4.5 V each and 30 VA for both.

Thanks for any help

Regards,

 

[adrianbodor]

I think in get it now

I followed elliot site i learned a bit from there.

There is no 1.65 amps, if my toroidal is 2x4.5 at 3.3 (30VA)

Then each winding is 4.5 V at 3.3 (15VA). I was wrong at this step
In parallel i have 4.5V at 6.6A (30VA)
I series i have 9V at 3.3A (30VA)

 

[Leftyretro]

I think in get it now

I followed elliot site i learned a bit from there.

There is no 1.65 amps, if my toroidal is 2x4.5 at 3.3 (30VA)

Then each winding is 4.5 V at 3.3 (15VA). I was wrong at this step
In parallel i have 4.5V at 6.6A (30VA)
I series i have 9V at 3.3A (30VA)

Yep you got it. Elliot site is a great one, lots of basic and practical circuits and explanations, to say nothing of all the great audio projects

Lefty

 

[adrianbodor]

A question thought about regulators

I know 7805 regulator can deliver max 1.5A.
My circuit draws about 1.3 amps. Can this regulator really deliver this max load with o good heatsink?

I intend to use a 6V 18VA 3A toroidal, after rectify i get about 8.4V @3A

Is there any losses of current during rectify (that's why i choose a 3A)? if not, i don't see the problem to use a 1.5A transformer exactly the output of 7805

Regards,

 

[mneary]

The size of the 7805 heatsink is determined by the power it must dissipate. If you have 8.4V in, and 5.0V out @ 1.3A, the 7805 needs to get rid of more than 4.4 watts. This will require a modest heat sink.

 

[rezer]

At 1.3A your regulator will get very hot. I would use a heatsink.

 

[kchriste]

I know 7805 regulator can deliver max 1.5A.

Be careful with that. Some 7805's can deliver 1.5A and others only 1.0A. Check the datasheet to be sure.
It is better to leave some headroom in all designs, so I would find a regualtor that is rated to run at 2A or more. The 7805 will not always be able to deliver it's max current rating under all conditions:

Example datasheet. (Yours will be different):
https://www.st.com/stonline/products/literature/ds/2143/l7805.pdf

 

[Hero999]

I intend to use a 6V 18VA 3A toroidal, after rectify i get about 8.4V @3A

That isn't enough, you've forgotten to acound for the voltage drop in the rectifiers (between 1.4V to 2.2V depending on the load) and the ripple. This will give 7.4V to 6.2V, minus say 1V ripple which isn't enough to ensure the regulator regulates properly.

Us a 9V transformer and a 16V 4700µF filter capacitor across the bridge rectifier and it should work perfectly.

 

[adrianbodor]

I don't get it one say is enough a 6V AC one say is not

With 6V AC i get 8.4V DC after rectify, how could i get 6.2 volts?

So to know, it's not a problem that my trafo have a current of 3A even if this regulator can max 1.5A, right?

 

[Hero999]

As I've just said, you haven't taken into account for the loss in the rectifier. Each diode will drop a minimum of 0.6V and a maximum of 1.1V and in a bridge rectifier the current passes through two diodes per cycle giving a total loss or 1.2V to 2.2V.

There will also be some voltage ripple on the capacitor which will reduce the useful voltage further as the voltage needs to always be above 7.5V for the regulator to work properly.

LTSpice says the peak is 6.8V and the troff is 5.2V which is far too low for the LM7805 to regulate properly - see attachment.

If the voltage is increased to 9VAC then the ripple will be between 9.4V and 11V which is more than enough, in fact with a 9V transformer, you could reduce the capacitor to 3300µF with no problems.

 

[adrianbodor]

Ok i'm a bit confused right now

Let's say i have a 12V AC trafo and a 2x12V AC trafo (wired in series)

What's the exact formula to find out final DC voltage after rectifier bridge for a dual in series or a single trafo

 

[Hero999]

If you connect them in series you get the sum of the secondary voltages, 24V.

Diodes drop a certain voltage (typically 0.6V to 1.1V for silicon) depending on the current flow. To calculate the peak output voltage we just subtract the diode loss from the AC peak voltage.

Why are you even thinking about using a 24V transformer?

There's no point in increasing the voltage to more than 9V because all the extra voltage will just be converted to heat.

 

[adrianbodor]

Of course i will not use a 24V transformer, it was just an example

I understand this so far

I need 7V on the input of the 7805, right.. now i will add 10% ripple voltage meaning 0.7V and aprox 1.4V diode drop resulting 9.1 meaning a 9V AC 3A is good for the project

Now for the capacitor i found this C=5 x i / Vp f

C: Capacitor value.
Vp: Peak voltage. ("Bridge output max voltage")
f: Frequency of the AC supply. 60Hz
i: Load current.


meaning C= 5 x 3 / (9 -1.4) x 60 = 3.300 uF
but I guess I will use a 4.700uF at 16V

by the way i calculated 3A because i will use LM78S05 witch is rated at 3A max.
So a solution for this power supply will be a 9V AC at 3A, 4 diodes 1N4004 (or 4A 50V bridge)
a capacitor elko 4.700uF at 16V and one elko 100uF 16V

To this I will add 2 MKT or ceramics capacitor on the input and output of the regulator.

Any comments are welcome

regards,

 

[Hero999]

Of course i will not use a 24V transformer, it was just an example

I understand this so far

I need 7V on the input of the 7805, right..

I don't know about he LM78S05 but the LM7805 requires 7.5V.

now i will add 10% ripple voltage meaning 0.7V and aprox 1.4V diode drop resulting 9.1 meaning a 9V AC 3A is good for the project

No, work out the peak voltage, then subtract, the diode losses and the minimum voltage required for the regulator to get the maximum allowable ripple.

Also the diode loss is normally 1.1V per diode at full load, not just 0.6V (look at the datasheet) which makes 2.2V.

[latex]V{peak} = 9 sqrt{2} = 12.73V\\
Ripple = 12.73-7.5-2.2 =3.03V
[/latex]

Now for the capacitor i found this C=5 x i / Vp f

C: Capacitor value.
Vp: Peak voltage. ("Bridge output max voltage")
f: Frequency of the AC supply. 60Hz
i: Load current.


meaning C= 5 x 3 / (9 -1.4) x 60 = 3.300 uF
but I guess I will use a 4.700uF at 16V

by the way i calculated 3A because i will use LM78S05 witch is rated at 3A max.
So a solution for this power supply will be a 9V AC at 3A, 4 diodes 1N4004 (or 4A 50V bridge)
a capacitor elko 4.700uF at 16V and one elko 100uF 16V

To this I will add 2 MKT or ceramics capacitor on the input and output of the regulator.

Any comments are welcome

regards,

That formula is rubbish, iot will grossly undersize the capacitor to size the capacitor use the following approximate formula:
[latex] C = \frac{I}{2f V_{Ripple}}\\
C= \text{Capacitance in Farads}\\
I=\text{Current drawn, in this case 3A}\\
f = \text{Mains frequency, in this case 60Hz}\\
V_{Ripple} =\text{The maximum ripple, in this case 3.03V}\\
C= \frac{3}{2\times 60 \times 3.03}=8.251 \times 10^{-3} = 82501 \mu F [/latex]
So for 3A, I'd use a 10,000µF capacitor to on the safe side.

There is a more accurate formula but it's complicated and this one is fine since it slightly oversizes the capacitor.

EDIT:
Your location says Romainia, I could be wrong but I think you're on 50Hz not 60Hz but you'll still be fine with 10,000µF

 

[adrianbodor]

Wow... i was close

What about the ripple, why is 3.03?
You wrote 12.72-2.5-2.2=3.03
Now I understand that 2.2 is diode loss, 12.72 is Vpeak but 2.5 what is it?

I saw that 78T05 has a voltage drop of 2.2V, can you explain what is this

Also could you point me to some good 5V 3A circuits with or without 78xx series
or maybe switching power supply that can deliver 3A to 5V

 

[adrianbodor]

I see that 2.5 is drop-out voltage for the regulator 78T05

Minimum voltage to operate is 7.2V and for bypass cap i will use a 330nF cap
What type is better for regulators ceramic or MKT polyester

 

[Hero999]

Wow... i was close

What about the ripple, why is 3.03?
You wrote 12.72-2.5-2.2=3.03
Now I understand that 2.2 is diode loss, 12.72 is Vpeak but 2.5 what is it?

Sorry, I meant 7.5V, 2.5V is just the dropout voltage.

I saw that 78T05 has a voltage drop of 2.2V, can you explain what is this

That's the dropout voltage of the LM78T05.

Therefore the minimum input voltage required for it to regulate properly will be 7.2V so you can increase the ripple to 3.33V

Also could you point me to some good 5V 3A circuits with or without 78xx series
or maybe switching power supply that can deliver 3A to 5V

The circuit is fairly standard and can be found on Google.

Use this one if you like but with the following modifications:

• Increase 2,500µF capacitor to 10,000µF.
• Increase the 0.1µF capacitor to 0.33µF
• I've never seen an 8.5V transformer, use 9V instead.
• Use the LM78S05 instead of the LM7805.
• Use some diodes with a 3A rating instead of the 1N004 which is only rated to 1A.

**broken link removed**
**broken link removed**

 

[adrianbodor]

Yeah, you're right i missed the diode max curret. Instead I will use a 4A 100V bridge and a 4700uF 25V elko, i already have one.
Maybe i will change this later

What about a filter ceramic cap of 100nF at the input of the regulator and a small 100uF 25V at the output of the regulator?


...LM78S05 is rated at 2A... LM78T05 is 3A

 

[Hero999]

4700:mu:F is too small for 3A but is alright for 2A.

You could use the LM78T05 if you like but there'd be no point with a 4700:mu:F capacitor becuase the ripple would be too high.

For 3A it needs to be a 10000:mu:F capacitor or two 4700:mu:F capacitors in parallel.

 

[adrianbodor]

What if I'll use more capacitor in parallel to reach 10000uF. I'm asking because i already have 30 pieces 1000uF 16V Low ESR 105 degrees Rubycon

ESR would rise that away, right? It is OK to use 10 in parallel?
It's a pretty dumb solution I know