Transformer Currents

[Hero999]

Yes, that's fine or use two 4700:mu:F capacitors as I just said.

 

[neon]

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,

kva or va as far transformers are concerned means that you can get the max rating as specified as kva. ei a 1kva means 1kv/amp at 100v you can get 10a.but at the same time if the transformer is rated as 50v max output then the rating changes to 1kva will be 1kv/50 or 20 amps max. KVA is realy a way to express total power out put out of a tranformer it can have one secondary or many the kva does not change it is total power. you may get 30va out of one output none out at the other. get it?

 

[neon]

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


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


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**

the rating of the diode is 1 amp but the regulator is not i ruther blow diodes then regulators if i must. besides a 1n4000 series can conduct 10x amps before the heats blow it up. bad sugestion i think. And the drop out voltage realy mean the minimun voltage differential to keep the regulator working properly.

 

[neon]

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

bad idea if you put that much capacitance then include a diode across the regulator to prevent reverse voltage damage. 100mfd is not neccessary you have a regulator remenber?

 

[Hero999]

In this situation I don't think that's a problem because I can't see how the output voltage could ever exceed the input voltage. I suppose there's a very low chance that there could be a short across the filter capacitor but in that case it would overheat anyway.

 

[neon]

regulator

with 2500 mfd and 8.5 volts the cap should go to peak voltage with no load 12v dc. with a load of 2amps you haven't reached the 3.3 a the transformer is capable of delivery. at the same time look at the spec of lm7805 more the 2 amp it will shut down cannot deliver the current. so now you need a by pass device to pass the 2 amps. there are T05 cans that can deliver it but not T120. REMOVE THE 100MFD OUT CAP not neccessary use a 1mfd

 

[adrianbodor]

Could u please draw me a 3A 5V supply with 78T05

I'm very confused about the filtering
What I want is this circuit to be able to deliver 3A (almost) at constant voltage of 5

I have seen circuits with high loads attached and voltage dropped significantly to about 4 and a half, maybe they were bad designed, i don't know

 

[Hero999]

Could u please draw me a 3A 5V supply with 78T05

No, because it's the same circuit as the one I posted but with the LM78T05 instead of the LM7805 and a much larger filter capacitor.

I'm very confused about the filtering
What I want is this circuit to be able to deliver 3A (almost) at constant voltage of 5

Why are you confused?

I've posted the formula, just increase the size of the capacitor in the circuit I posted to >9000:mu:F. It doesn't matter whether you use two 4700:mu:F capacitors, three 3300:mu: capacitors or 9 1000:mu:F capacitors.

I have seen circuits with high loads attached and voltage dropped significantly to about 4 and a half, maybe they were bad designed, i don't know

Yes the were badly disigned.

They probably used too smaller filtering capacitor or a 6V transformer.

 

[adrianbodor]

Ok I decided to make a schematic for this high current type of supply

**broken link removed**

T1 = 9V AC 3A - 18VA
B1 = 4A 100V bridge rectifier
C1 = 10000uF 16V
C2 = 10 or 100uF 16V
C3, C4 = 100nF 104 cer
LED and resistor

Where is better to put a fuse? On AC line, let's say a 3A 250V fuse
I saw fuses also on the regulator input

 

[Hero999]

The regulator has current limiting so you don't need a fuse to protect it.

However putting a slow blow 100mA fuse on the primary side is advisable.

 

[adrianbodor]

I searched for that on a french site, I discovered exactly what i need but the fuse was 200mA, however I don't get it how the fuse can face it if current drawn is 3A

Does in it blow the fuse that way?

 

[Hero999]

The current on the primary is much lower than the secondary.

Assuming the 18VA transformer is 80% efficient and the supply is 230V, at full load the input current will be 18/(.8*230) = 98mA.

Use 200mA it's better than no fuse at all and make sure it's slow blow or it will blow when the filter capacitor charges.

 

[adrianbodor]

... O saw my previous post and I made an error, It's a 27-30VA transformer not 18VA
I already learned what slow blow or quick blow means so I don't ask you about this

a "slow-blow" rating, means that it can handle moderately high currents for a short time without damage. Such a fuse might be used in an appliance containing a motor. If the motor is rated to consume 2 Amps during continuous operation, it could be protected by a slow-blow fuse. When the motor is first turned on, it will draw much more than 2 Amps for a few seconds, and the slow-blow rating of the fuse ensures that the fuse doesn't blow every time the motor starts. A quick blow fuse is actually just a standard fuse designed to blow instantly

Could u tell me the formula to use when selecting a fuse on primary for an aleatory circuit

 

[Hero999]

I've just told you that, see the calculation for an 18VA transformer in my previous post.

 

[adrianbodor]

OK let's see i was wrong about the transformer I corrected above
It's 30VA for 3A not 18 so 80% efficiency means 30/(0.8*230V)=163mA so a 200mA slow blow fuse will do the job

 

[neon]

In this situation I don't think that's a problem because I can't see how the output voltage could ever exceed the input voltage. I suppose there's a very low chance that there could be a short across the filter capacitor but in that case it would overheat anyway.

DISCONNECT THE LOAD AND TELL ME ABOUT IT.

 

[Hero999]

What's your point?

Disconnect the load and the input voltage will just rise more and the regulator will adjust the output votlage accordingly. There may be a brief spike on the output but the capacitor will abosorb most of it.

I don't see any need for a diode from the output to input because there's no chance that the input voltage could drop below the output. I might agree if the regulator was being powered from a solar cell or dynamo but it isn't. When the mains is disconnected the huge filter capacitor across the rectifier will stay charged for much longer than the little capacitor on the output of the regulator. When the filter capacitor slowly discharges the little capacitor on the output of the regulator will discharge back through it so slowly that it won't damage it.

 

[neon]

What's your point?

Disconnect the load and the input voltage will just rise more and the regulator will adjust the output votlage accordingly. There may be a brief spike on the output but the capacitor will abosorb most of it.

I don't see any need for a diode from the output to input because there's no chance that the input voltage could drop below the output. I might agree if the regulator was being powered from a solar cell or dynamo but it isn't. When the mains is disconnected the huge filter capacitor across the rectifier will stay charged for much longer than the little capacitor on the output of the regulator. When the filter capacitor slowly discharges the little capacitor on the output of the regulator will discharge back through it so slowly that it won't damage it.

you actualy don't get it if you disconnect the load with a big cap there the voltage remains now disconnect the input power.Now you understand? why the diode with a big cap on the output REVERSED VOLTAGE ACROSS THE CHIP blow it up. Don't believe me try it.

 

[Hero999]

How can it reverse voltage across the chip when the voltage on the massive 4700:mu:F capacitor is greater than the voltage on the tiny 100:mu:F capacitor which will rapidly discharge through any load connected to it?