# Soldering iron, need higher voltage

1. Dec 16, 2015

hey folks, so heres the deal, I have a isolation transformer that I use for my self wound irons.one of them is working good , heats up nicely and i solder with it all the time, the other one is an old iron , its bigger i want to use it were bigger parts need to be soldered.
heres the problem, my transformer outputs about 42 to45 volts AC on secondary, I wound the iron but I guess i made the coil a bit too long , it heats up but not as hot as it would need to be , now i basically have three options,

option nr.1 take the iron open and adjust the coil, now i would not like this idea because i already tightened the iron and packed it really good and the mica thermoelectric insulation is very very fragile.

option nr.2, take the secondary winding and put some extra turns around the core to step up voltage.

option nr.3 and this is the question , if i would attached a bridge rectifier and some smoothing capacitors it would make the voltage from AC to DC and around 63 volts DC, since 45vAC x1.414= 63.63,
now would rectifying the AC into DC heat up the iron more than the 45 volt AC?
I think it should because the power that goes into a given resistance is determined by the resistance and the voltage across that resistance which determines the current, so a 63v DC supply should push more current than a 45v AC supply correct?

2. Dec 16, 2015

### Staff: Mentor

I would expect that option 3 actually reduces the heating. While the peak voltage without load is 63 V, with a load you won't have a constant 63 V. You get some messy fluctuating voltage, probably from a bit above 45 V (not necessarily) to significantly below that, depending on details of the rectification.

Option 2 should work.

3. Dec 17, 2015

### jim hardy

Your suggestion 3 would work only if the transformer and capacitor are preposterously oversized relative to the iron. Here's why.
Between line voltage peaks your capacitor will discharge ,
lowering the voltage,
as mfb stated.
Can you put a number on rate of discharge? Δv/Δt =i/C, volts amps and farads

Now, when next peak comes along and capacitor charges back up
instantaneous current is really high because you are moving charge only during the brief interval where incoming powerline sinewave is above capacitor voltage. ΔQ = iΔt .
but your supply only replenishes charge during the brief interval surrounding sinewave peaks when Vincoming exceeds Vcapacitor.
So incoming current is not sine shaped but flows in short , large gulps .
those gulps of high current create more voltage drop in transformer windings than one would expect
so your cap won't get charged all the way to 1.414 Vincoming..
A drastically oversized transformer can however supply those high current peaks.
And an oversized capacitor can absorb the charge.

Here's an article explaining the phenomenon, just one of many from a google search
http://www.zen22142.zen.co.uk/Design/dcpsu.htm
observe short current bursts (I call them 'gulps' ). and compare their instantaneous to average value.
The poor little transformer is overloaded during those brief intervals and will drop quite a bit of voltage..

If you measure the DC resistance of your transformer windings you can estimate expected voltage at whatever current your iron draws.

old jim

4. Dec 17, 2015

thanks for the input, well yes I am aware of this problem , also the reason why old equipment which had a transformer as its main power supply had really big capacitors especially amplifiers, were solid steady voltage under varying load is necessary.

well my transformer that I use is not a small one, I just got one old transformer and since it buzzes it cant be used anywhere else so i use it as an isolation transformer but its bulky, rated for 630 watts but probably can do more with some voltage sag , so i guess it comes down to putting a decent size capacitor and just a bit of experimenting , watching what happens, measuring voltage under the iron load.

5. Dec 17, 2015

### BvU

Let me point out that there is no point whatsoever in using a capacitor or a rectifier for this application. The power delivered to the soldering iron is not influenced by adding a capacitor or a rectifier. In old amplifiers you needed the capacitors because you needed DC with a minimum of AC on top.

If you can't or won't shorten the coil in the iron, then adding to the secondary coil in the transformer is your only option. Since it's primary coil is rated a whopping 630 W there's plenty of room there. Hope the secondary coil wire is thick enough, though:
That's really good advice. Alternatively you can measure the AC voltage without load and with the soldering iron connected.

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6. Dec 17, 2015

### jim hardy

that's not quite so.
Here's a picture from another Google search , and it's worth a thousand words.....
http://www.geofex.com/Article_Folders/Power-supplies/powersup.htm

Heating value is RMS
With an oversize transformer, a rectifier and enough capacitance you can drive Vripple to near zero.
If ripple is small enough that voltage Vpeak-Vripple between peaks does not fall below .707 of Vsecondary 's peak, you have most certainly raised the RMS value of that odd voltage waveform. Draw a horizontal line at .707Vpeak and see...

But as always, one experiment is worth a thousand expert opinions. Try it and see.

old jim

7. Dec 18, 2015

### BvU

Thanks for the lecture, Jimbo...
What if the internal resistance of the secondary coil is about the same as the load resistance ?
I sure second that !

And would very much like to hear what DC resistances and what AC voltages with and without load Salvador has measured !

8. Dec 18, 2015

well I too agree with Jim , when the voltage is rectified it rises to a higher value , a higher voltage can push more current through a given resistance than a lower voltage, thta's for sure, ofcourse as long as the supply can stay at that higher voltage.
a light bulb for example glows brighter when connected to rectified 230v mains, than when simply connected to 230v mains , because when the mains is rectified with sufficient smoothing capacitors the voltage rises to 325v DC.

Ok I will go and try it ut and see what happens.

9. Dec 18, 2015

### BvU

Is this a thought experiment or a genuine observation ?

10. Dec 18, 2015

### Staff: Mentor

In this special case, rectification won't help. Your previous post was way more general (too general).

11. Dec 18, 2015

that my friend is the truth, a higher voltage makes the filament hotter pushing more current through it.
I can report my experiment which went well, I took a diode bridge, soldered an additional 470uF capacitor to it and the output voltage went from 45v AC to 55v DC (under load) without load it should be about 63 but since im too lazy to solder the connections off again i'm just relying on maths.
clearly the capacitance is a bit too small for this load so the voltage stays only at 55v DC but I tried the iron and it works just good, the iron is now hot and i can melt large chunks of tin with it , so actually the less capacitance is fine , if it would be at the full rectified 63 volts under load it actually might get too hot after some time.

so yes increased voltage does matter for higher power , even if that increased voltage came from mains peaks rectified and stored into a capacitor.
I added a few pictures , the diode bridge is small and connected right to the transformer togethr with a capacitor.

well but i can't agree to what BvU said earlier that this doesnt do anything , if the rectified dc voltage can be kept at a steady level (enough capacitance) and if the supply (transformer in this case) itself is strong enough then this works.

also when i said that rectified mains makes a bulb filament brighter than simply mains , here is a video were this can be seen.

this video basically talks about different stuff but around 2:50 you can see him attaching bulbs to both 120v AC and then to 170v DC and the 170v DC is brighter.

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12. Dec 18, 2015

### BvU

Well, it was fun watching the video. And now I know how a capacitor can conduct AC while blocking DC. The AC zigzags through. Of course.

What I meant is your "example" is not an example. It was something you invented. A 220V bulb would go poof (but I admit I never tried that experiment). You can buy bulbs that can stand 325 V over there ?

The lunatic in the video probably uses 220 V bulbs ?

I agree. My picture was of a situation with a high internal resitance and a low load resistance. Unlikely with a 630 W transformer. My bad.

--

13. Dec 18, 2015

### sophiecentaur

Yes. Maximum Power theorem rules - but you may need to be cooling your power supply.

14. Dec 18, 2015

### sophiecentaur

Quoting the way a light bulb will behave can sometimes be misleading because their resistance changes so much over their possible brightness range. I agree that a light bulb can be an useful substitute for a voltmeter when the source resistance is nice and low. Otherwise it can give funny answers.

15. Dec 18, 2015

### jim hardy

This thread brings me great joy -

we saw a lively discussion and and experiment that clarified the actual situation.

A question well stated.is half answered
this one was too skimpy on details when first posed
and y'all interacted , arriving at a better understanding.

That transformer in the photo looks preposterously oversized for a 40 watt iron
indeed one sized for the iron wouldn't be up to the job Sal asked of it.

Well Done, guys !

old jim

16. Dec 18, 2015

### sophiecentaur

It strikes me that you can either have a low impedance transformer (giving a voltage source) that would stay cool or run a much cheaper transformer with a heat sink and mounted in a vented box. That would probably be a lot cheaper - if a bit non-standard. It would need to have a high unloaded output voltage and dissipate 40W.

17. Dec 19, 2015