AArgh! Power supplys!

[angie1199]

Ok, so I've got my 230/40vAC-18vAC transformer.

So I look up building a power supply and come up with a circuit based on one on the 'net. I upgrade the components listed to support more current. I build it according to the schematic.

It gives me 2.5vDC and not 5v DC!

v AC at transformer out - 18v AC
v DC after rectifier diode - 16v DC
v DC after smoother and regulation circuits - 2.5v DC :S

Any ideas please?

(Schematic and stripboard layout uploaded.)

 

[angie1199]

Doh! forget them!

 

[Nigel Goodwin]

Is this with, or without, a load?.

If it's without a load, then you've probably connected the 7805 the wrong way round?.

You should also be VERY aware that the 7805 is going to get EXTREMELY hot under any reasonable load, and will require a substantial sized heatsink. Main reason for this is that the transformer is far too high a voltage to feed a 5V regulator.

 

[Sig239]

Hello Angie1199
Check the polarity and rating of your electrolytic cap. Also check your solder joints and stripboard "breaks". How much current do you expect to draw? IMHO I would suggest using a 12V transformer, as this will cause the regulator to generate less heat. What current rating is your 18V transformer? Are any of the components getting hot?

 

[Sig239]

lol sorry Nigel Your post didn't show up until I Posted mine.

 

[angie1199]

The transformer was an earlier problem I had, i.e. VA and how I know how many I needed, so I used an old transformer I got here which worked in it's original case but was too bulky.

I needed 18V AC for a capacitor discharge unit, and a 5V DC for LEDs display.

The CDU requires 16-24V AC max 2.5A but only while charging it which apparently takes milliseconds

The LEDs require total max 1.5A

Thus the 5A fuse then the 2A fuse.

I've already got a heatsink (twisted vane) and nothing seems to get hot. I've looked at the datasheet and I copied the image but I am assuming the VR image on datasheet is from the lumpside facing. If its the lumpside away then it's back to front.

The capacitor is wired with the positive wire from the fuse to the positive terminal on the capacitor (and the capacitor hasn't exploded)

I'm not sure whether my multimeter is working properly so new one should be here tomorrow.

Re-checked the voltage after the rectifier and it says 25V DC!!! this is from 18V AC in. Is this right????? How does the rectifier make the voltage grow? I thought the voltage should drop after the rectifier by about 1V per diode....

Also re-checked the voltage out and it started at about 4.6V DC and gradually dropped to 1.6V DC. Once it had dropped to under 2V I wired in an LED and it didn't need a resistor! It was sooooo dim

I think I need help...... I'm going mad...... the world is against me......... <G>

 

[RODALCO]

18 x 1.41 = 25 Volts DC that is ok.
The 2200 µF Capacitor smoothens the ripple and bring it up to 25 Volts at no load.
Through the bridge rectifier 1.2 Volts is lost because of the diodes in series.
If you have 2 x 12 volts 5 watt car bulbs, wire those in series to check your power supply before the rectifier.

Check the regulator , perhaps wired up wrong as already said. Could be a dud too.

 

[RODALCO]

Just had another look at the schema.
The regulator is drawn in ok, you are sure it is on the board , right way around?

Also check that the negative from the regulator is continuous. otherwise it has no reference.

A good, reliable multimeter is a must for fault finding.

Check all solder joints and see if components are loose or tracks are broken.

 

[OutToLunch]

you might have the 7805 backwards. Recheck the datasheet carefully.

The 18Vac output is the RMS value. This means that the peak voltage of the sinusoidal voltage about 25V, so the output of the diode bridge will be right around there. If you want that to be 18VDC, then you might want to get another linear regulator and supply the 18V with that.

A 25V input to a 5V output linear regulator that will see a max load current of 1.5A is going to fry that 7805 real quick. The 7805 will be dissipating 30W! Even if you're input to the 7805 was 18V, that is still 19.5W. Way too much for any single pass element, even with a heatsink. There's no way to remove the heat quickly enough in a realistic scenario.

Solutions... a transformer with two secondaries would work. Then you just rectify and regulate for your 18V and 5V. Another option would be a switching regulator, but that's fairly complicated.

 

[Optikon]

If I read correctly, you need 5V supplying 1.5A to your LED. Well, you've got 25V input that means the 7805 needs to dissipate 30 Watts! Don't do that. Reduce input voltage to 7805 & use BIG heat sink. Or buck converter for the 5V.

 

[philba]

If this is to run only 1 led. I guess it won't matter much. If it is to run lots of LEDs (and nothing else), why not put them in series and use a variable regulator to produce a voltage of a couple of volts above the sum of the LEDs' forward voltage. why not dissipate the voltage in the LEDs and not the regulator? also, it will draw much less current.

 

[angie1199]

Just had another look at the schema.
The regulator is drawn in ok, you are sure it is on the board , right way around?

Yep, it's marked with the wiggly thing for ac and that's where the input is. The pos and neg are wired accordingly too.

Also check that the negative from the regulator is continuous. otherwise it has no reference.

Eh? Check it with what?

A good, reliable multimeter is a must for fault finding.

Yep. New one on order.

Check all solder joints and see if components are loose or tracks are broken.

I'd checked twice and they appeared ok. So now I've removed all components after the 2200 Cap and replaced the 2200 Cap with a new 1000uF so when the new multimeter arrives I can check from there.

 

[angie1199]

you might have the 7805 backwards. Recheck the datasheet carefully.

Did that, it doesn't show whether the image of the 7805 is lumpy side facing or away though.

The 18Vac output is the RMS value. This means that the peak voltage of the sinusoidal voltage about 25V, so the output of the diode bridge will be right around there. If you want that to be 18VDC, then you might want to get another linear regulator and supply the 18V with that.

A 25V input to a 5V output linear regulator that will see a max load current of 1.5A is going to fry that 7805 real quick. The 7805 will be dissipating 30W! Even if you're input to the 7805 was 18V.

Solutions... a transformer with two secondaries would work. Then you just rectify and regulate for your 18V and 5V. Another option would be a switching regulator, but that's fairly complicated.

Now here's where I get confused. How does 18vAC secondary turn into 25vDC after going through two diodes that according to the info I've seen uses approx 1V per diode?

I looked at the datasheet and, apart from not really understanding it, it says Io 20mA to 2A. Is that not good? Also the product I purchased stated it was a 2A 5V regulator.

I've just aquired a 12v AC transformer rated 4.5A. would that be ok to use for the 5v DC circuit?

 

[angie1199]

Reduce input voltage to 7805 & use BIG heat sink. Or buck converter for the 5V.

Ok, so if I built a 15v DC regulated supply, then after that a 12v DC regulator, then after that a 5v DC regulator that would be reducing the power gradually, would that work?

 

[angie1199]

If this is to run only 1 led. I guess it won't matter much. If it is to run lots of LEDs (and nothing else), why not put them in series and use a variable regulator to produce a voltage of a couple of volts above the sum of the LEDs' forward voltage. why not dissipate the voltage in the LEDs and not the regulator? also, it will draw much less current.

It's to power 50 LEDs, but each has to be wired seperately as they each must be switched independantly. If I could wire them in series I would, believe me. This project has turned into a real major one!

 

[kchriste]

Did that, it doesn't show whether the image of the 7805 is lumpy side facing or away though.

If you look at the "lumpy side" with the legs pointing down, then the input is at the left and the output is on the right and the center leg is ground.
If your transformer has a center tap you can rewire it like this and dissipate less heat in the 78S05:

 

[angie1199]

If you look at the "lumpy side" with the legs pointing down, then the input is at the left and the output is on the right and the center leg is ground.
If your transformer has a center tap you can rewire it like this and dissipate less heat in the 78S05:

Aha, ok, it was wired up ok.

I don't have a center tap on the transformer. It's hard-wired with two wires each end.

However, (just to cofuse the issue) I have just been given some 12V AC transformers that are used to power halogen 12V downlighters. They are rated :
20-60VA
230V, 50Hz in
11.6V-EFF out
4.9A
ta 50C tc 85C

Assuming that I'm not a total idiot to this, logically if these transformers can cope with 12V AC halogen bulbs then if I rectify the voltage it would then be lower than my current transformer, therefore less watts, and less heat?

 

[OutToLunch]

Now here's where I get confused. How does 18vAC secondary turn into 25vDC after going through two diodes that according to the info I've seen uses approx 1V per diode?

AC voltage is usually expressed in terms of it's RMS value. Since an AC voltage is continuously changing, it is easier to consider how power is dissipated on what you may consider to be more of an average basis. So, if you have a 230V AC voltage and look at it on an oscilloscope, what you would see is a sinusoid with a frequency of 60Hz (for the US) and a peak of +/- 325V. Now place a 1 Ohm resistor across that AC voltage. The instantaneous power dissipated in the resistor will vary as the voltage swings up/down. The RMS power will be 230W. Similarly, if you had a 230V DC source and placed a 1 Ohm resistor across it you would get a continuous power dissipation of 230W. I probably did not explain that very well...

Anyway, to get the peak voltage of a sinusoidal signal that is given in RMS, just multiply by sqrt(2) - note that this is only true for sinusoidal signals.

OK, so your 230V-12V AC transformer will be putting out 12V RMS, which yields a peak voltage of 17V. When you rectify the secondary, you should get about 15V DC (at light load it will likely be higher). The transformer should be capable of handling the required 1.5A of load current if you are using linear regulation to get 5V.

I looked at the datasheet and, apart from not really understanding it, it says Io 20mA to 2A. Is that not good? Also the product I purchased stated it was a 2A 5V regulator.

The linear regulator can indeed pass 2A through it - the silicon was designed to handle 2A as were the bond wires. The problem with passing current through a linear regulator is that there is always an inherent voltage drop across the input/output. This means power dissipation (P=V*I) which means heat is being generated. If your input voltage is only two or three volts higher than you output voltage, then your still going to dissipate 3-4.5W. Not absolutely horrible if there is a heatsink and good airflow.

You can step down the voltage gradually with linear regulators. This will spread the total power loss across more devices. The circuit will still be dissipating a lot of power, though.

You may want to seriously consider buying an AC/DC converter off the shelf. Astrodyne makes a nice little AC/DC regulator ready to go...
https://www.electro-tech-online.com/custompdfs/2006/08/PWB.pdf
The 10W model will more than fit your needs.

 

[angie1199]

Woohoo!!!

One 5V DC power supply

Nothing gets hot, nothing blew up.

Not even an electric shock!

Hmmmm, what next? Ah yes, an infrared light detector....

Thanks all who helped.

 

[angie1199]

Anyway, to get the peak voltage of a sinusoidal signal that is given in RMS, just multiply by sqrt(2) - note that this is only true for sinusoidal signals.

OK, so your 230V-12V AC transformer will be putting out 12V RMS, which yields a peak voltage of 17V. When you rectify the secondary, you should get about 15V DC (at light load it will likely be higher).

If your input voltage is only two or three volts higher than you output voltage, then your still going to dissipate 3-4.5W. Not absolutely horrible if there is a heatsink and good airflow.

You can step down the voltage gradually with linear regulators. This will spread the total power loss across more devices. The circuit will still be dissipating a lot of power, though.

You may want to seriously consider buying an AC/DC converter off the shelf.

Thanks OutToLunch, Understood that (sort of). I'll see how this thing goes with the LEDs. I alredy blew up one regulator the first time I put it together. This time I think I sorted out any shorts.

I checked the voltage after the rectifier and it says 12v (ish) and after the capacitor it started at 11.5 and gradually grew I suppose because it's gathering the charge to smooth it but there was nowhere for it to go as I hadn't finished adding the rest of the bits.

I saw an AC/DC convertor on sale here in the UK. It was £25 !!!!!! Think I'll give this a go before spending all that. Alternately if this blows I may use a 12VDC regulator and a bigger resistor to power the LEDs.

Thanks again.