Can you wire different taps from a transformer in series?

[Bosk]

Hi Folks,

I'm getting started in foam core cutting and I've got an AC transformer that has 6, 7, 8, 9 & 10 VAC taps on it selectable via sockets in the top. I'm using this to power the hot wire and for a 40cm bow this works brilliantly with some 0.3mm cutting wire. At 6v I get a .5mm kerf and at 10v it zipps through with a 1mm kerf.

As you have guessed the problem I've got is that it doesn't give enough heat to a 1.2m bow which is what I need to cut my wings with and it seems I need more voltage. I was therefore wondering if I could wire two of the taps in series to get more than 10v from it eg 6v + 8v to get 14V total? Now if it were a battery the answer would be yes without doubt. What I'm not sure about is the fact that these are different taps from the same transformer.

I am thinking I could build two simple 4-diodes bridge wave rectifiers, connect one to the 6vAC tap and another to the 8vAC tap (or the higher ones if needed) then connect the DC output in series and end up with a 14VDC output?

So here are my questions:

1. Would this cause the transformer problems or is a transformer happy to provide voltage from all it's taps simultaneously?

2. Does the wiring of the to DC outputs in series cause problems for the transformer?

3. Does this plan of mine have any merit in it at all.

4. Can this be done with AC voltage and remove the need for the diode rectifiers?

Any help would be greatly appreciated.

Regards,

Bosk

[vintage1]

No, these are taps, not separate windings.

HOWEVER if you full wave bridge the 10vAC and use a power electrolytic capacitor to smooth, you will get about 14v DC.

If you use a pair of half wavers you can get 28V DC.

[Chippie]

If you use a pair of half wavers you can get 28V DC.

Which is the same as a voltage doubler...btw....

[dmccormick001]

Will the DC current heat the wire the same as the AC?

[Bosk]

No, these are taps, not separate windings.

HOWEVER if you full wave bridge the 10vAC and use a power electrolytic capacitor to smooth, you will get about 14v DC.

If you use a pair of half wavers you can get 28V DC.
Vintage, do you mean something like this?

[jeffs555]

The voltage rating on a transformer is usually RMS (root mean squared). That means that when heating a resistor, your 10v transformer will produce the same heating as a 10 DC source. If you rectify and filter it, you will get more than 10 volts DC so it will heat the wire more.

My best guess however is that it would be cheaper and easier to just add a fixed 6 volt transformer in series.

[Bosk]

No, these are taps, not separate windings.

HOWEVER if you full wave bridge the 10vAC and use a power electrolytic capacitor to smooth, you will get about 14v DC.

If you use a pair of half wavers you can get 28V DC.

If the taps have physically separate connections, can it be done? They look like the first image where you need to physically change the connection between the black and the red sockets to select the different voltages.

I figured that because they had different physical connections, and the taps should look like the second picture, I should be able to connect more than one tap to the line terminal without them shorting out. They wouldn't get connected together until after the rectifiers have converted them to DC.

Am I just mis understanding this whole thing (very likely I must admit).

Regards,

Bosk

[Bosk]

The voltage rating on a transformer is usually RMS (root mean squared). That means that when heating a resistor, your 10v transformer will produce the same heating as a 10 DC source. If you rectify and filter it, you will get more than 10 volts DC so it will heat the wire more.

My best guess however is that it would be cheaper and easier to just add a fixed 6 volt transformer in series.
I hadn't thought of that... I really regret throwing out the "old" 19v laptop power supply a few weeks ago... I'd hung onto it for 3 years since the laptop died and decided I'd never need it last month... :mad:

[Chippie]

Your second picture shows an auto transformer with low voltage tappings...If that is the case I would NOT recommend you use it for foam cutting or any thing else...it doesnt provide isolation from the mains...You could suffer electric shock!


If you want to power a foam cutting bow, use a car battery......If you need ot vary the power you could build a small pwm controller for powering the bow via the battery....much safer.

[Chippie]

Vintage, do you mean something like this?
Yup thats a v. doubler cct

[Greywing]

I doubt your old laptop supply would have enough current capability to drive your cutting wire. That wire is a very low value resistor - hence high current. To maximize the energy delivered to the bow, be sure you aren't loosing significant voltage between the transformer and the cutting bow by comparing the voltage at the transformer with that directly at the bow. You want the largest wire (lowest resistance) possible between the transformer and the bow to deliver all the energy to the bow.

Regarding two rectifier circuits, I think if you draw out separate rectifier circuits you'll see that you are indeed shorting out some of the windings on your transformer (as shown in your second drawing) which wont work.

It seems your choices are limited to those presented earlier in this thread.

GW

Edit: Second what Chippie says on isolation. Don't cut foam while standing in water!

[Bosk]

I doubt your old laptop supply would have enough current capability to drive your cutting wire. That wire is a very low value resistor - hence high current. To maximize the energy delivered to the bow, be sure you aren't loosing significant voltage between the transformer and the cutting bow by comparing the voltage at the transformer with that directly at the bow. You want the largest wire (lowest resistance) possible between the transformer and the bow to deliver all the energy to the bow.

Regarding two rectifier circuits, I think if you draw out separate rectifier circuits you'll see that you are indeed shorting out some of the windings on your transformer (as shown in your second drawing) which wont work.

It seems your choices are limited to those presented earlier in this thread.

GW

Edit: Second what Chippie says on isolation. Don't cut foam while standing in water!

I posted late last night before I went to bed and realised while I was half asleep that I would be shorting out the winding, so it was a dumb question.

The transformer itself is designed to be used as a low voltage supply for powering soldering irons and they were used by the technicians at Philips in the assembly of TVs and radios so I believe that there is isolation from the 240V mains. To be honest I've used it like that for the last 30 years and never had a problem so I think I'm safe from that point.

I've also tried to make sure I've used the heaviest gauge wire I can from the transformer to the bow (using some Monster speaker wire which has a 2-3mm dia core, but maybe that isn't as good a conductor as I thought it was either.

[vintage1]

Yup thats a v. doubler cct
Not what I had in mind though.

If you use a pair of half wave rectifiers/caps you get plus and minus 14v wrt to one end of the transformer

[Bosk]

Not what I had in mind though.

If you use a pair of half wave rectifiers/caps you get plus and minus 14v wrt to one end of the transformer
That's what I thought you meant but I found this circuit in wikipedia under half wave rectifier/voltage doubler.

BTW what size type of cap are we talking here? It's been 20 years since I did electronics at school and I'm not sure what I'll need.I've got an old box full of 30yr old electrolytic caps that I got from Philips when they swept up the floor in the assembly line, but I'm not sure what I'm looking for now and how to check if they'd be able to handle the voltage and currents.

[Chippie]

Not what I had in mind though.

If you use a pair of half wave rectifiers/caps you get plus and minus 14v wrt to one end of the transformer

Ahh, thats not what I thought you meant...I now know tho' :p

[Bosk]

hi Folks,

I tried making this voltage doubler and found that it would happily double the voltage, but the current disapeared (well you know what I mean)

I found that while I could get up to 28V DC at the end of the bow connectors, as soon as I connected them to the bow wire the voltage fell to 0V. It didn't matter what length bow wire I had.

Any idea how I can manage to double the voltage AND still keep enough current flowing to heat the wire? I know that the supply is powerful enough to provide 30A at 10VAC so its not the supply... In my experimenting, I accidentally shorted the 6V supply when I forgot I was connecting one of the diodes across the terminals (left out the capacitor) and in about 3 seconds the 6A diode blew up and the connecting wire was glowing white hot and the insulation vapourised!


So... any help on what is going wrong would be appreciated. Also if you can suggest a suitable electrolytic that would smooth the DC and still pass through sufficient current I would appreciate it.

Bosk

[Greywing]

hi Folks,

So... any help on what is going wrong would be appreciated. Also if you can suggest a suitable electrolytic that would smooth the DC and still pass through sufficient current I would appreciate it.

Bosk
[FONT=Comic Sans MS]
I don't know what the resistance of the cutting wire is in your bow but I'm sure it is small - probably significantly less than an ohm, so the greater any series impedance in the circuit driving the bow, the smaller the voltage you will have at the bow terminals.

In the voltage doubler circuit you show in post #3, the capacitor connected to the top terminal of the transformer winding is in series with the load you want to drive and therefore needs to be as large a value as possible to minimize the voltage drop across this element. To get some feeling for what cap value is necessary, if we calculate the reactance of a 1000 µFd cap (using Xc = 1/[2*pi*f*C]), we get a little over 3 ohms which is probably at least 10 times as much as the cutting wire resistance. Even if the capacitive reactance was equal to the bow resistance, you would end up with a voltage divider getting you back to where you were without a doubler.

So you need a VERY large capacitor to minimize the series resistance of this circuit. One other complication is that this capacitor sees both polarities of voltage. Because of capacitor value requirements, one must use electrolytic capacitors and the necessity to be bipolar dictates 2 caps connected back to back in series - each at twice the required capacitance value for equivalance. Voltage rating should be at some margin above the pp AC voltage of the transformer outputs being used.

Without trying to compare the performance of a half wave doubler circuit as suggested by vintage1 with what you are attempting, first glance suggests it may be a better approach if for no other reason than it eliminates the series capacitor. However because your load is heavy, the filter caps would need still to be hefty to minimize voltage sag between the wave peaks and maximize peak voltage value and energy delivered to the load.

GW

[jeffs555]

GW is right about needing very large capacitors, but they don't need to be bipolar. The first capacitor charges up on one half cycle and then dumps its charge into the output capacitor(and also to the load) on the next half cycle. If selected correctly the first capacitor will not discharge much on the second half cycle so will never see a reverse voltage. To guarantee that, the first capacitor should be much larger than the output capacitor.

This would probably be a better way to wire the doubler as it uses equal value capacitors and also doesn't have one capacitor charging another. http://en.wikipedia.org/wiki/File:Bridge_voltage_doubler.svg

PS looked back and think this is what Vintage1 was refering to when he said 'two half wavers".

[Bosk]

Thanks gents.

I've been experimenting over the weekend with limited success. Don't have time right now to post, but in my lunch break I'll give you a break down of what I did and how it did/didn't work and with your help I might be able to solve this.

[Bosk]

I don't know what the resistance of the cutting wire is in your bow but I'm sure it is small - probably significantly less than an ohm, so the greater any series impedance in the circuit driving the bow, the smaller the voltage you will have at the bow terminals.I'm not sure either, my multimeter is 30+ years old and needs an old weird-style battery to do the resistance measurements and I haven't been able to find a replacement/substitute for it yet.

In the voltage doubler circuit you show in post #3, the capacitor connected to the top terminal of the transformer winding is in series with the load you want to drive and therefore needs to be as large a value as possible to minimize the voltage drop across this element. To get some feeling for what cap value is necessary, if we calculate the reactance of a 1000 µFd cap (using Xc = 1/[2*pi*f*C]), we get a little over 3 ohms which is probably at least 10 times as much as the cutting wire resistance. Even if the capacitive reactance was equal to the bow resistance, you would end up with a voltage divider getting you back to where you were without a doubler.

So you need a VERY large capacitor to minimize the series resistance of this circuit. One other complication is that this capacitor sees both polarities of voltage. Because of capacitor value requirements, one must use electrolytic capacitors and the necessity to be bipolar dictates 2 caps connected back to back in series - each at twice the required capacitance value for equivalance. Voltage rating should be at some margin above the pp AC voltage of the transformer outputs being used.

Without trying to compare the performance of a half wave doubler circuit as suggested by vintage1 with what you are attempting, first glance suggests it may be a better approach if for no other reason than it eliminates the series capacitor. However because your load is heavy, the filter caps would need still to be hefty to minimize voltage sag between the wave peaks and maximize peak voltage value and energy delivered to the load.

GWWhat I tried first was two 2200pf 25V electrolytics in parallel as that capacitor attached to the top terminal and a 6A rectifier diode as the first diode connected to the lower terminal. This gave me 15VDC and just enough heat to slowly cut through EPS. The diode did warm up a bit, so I ended up putting two in parallel. The diodes stayed cool and so did the electrolytics. So far so good.

I then tried to double the voltage by putting in the second diode and last capacitor from the second diode to the lower terminal. I only had one 2200pf electrolytic left (same as the first two) and tried the input as 6VAC. What happened was that the first two electrolytics in parallel stayed cool but the last electrolytic got VERY HOT, and the rubber end cap started to bulge after about 5 seconds. I disconnected it and it stayed hot for about 5 min then started making buzzing sounds!!! At this stage I put it far away from anything else in the cold garage and eventually it stopped making noises and got cold again.

My next experiments was to try using two 1500pf 40V electrolytics and a 1000pf 40V electrolytic in parallel to replace the single 2200pf one that got hot. These also got warm but nothing like the first one but though I had higher voltage it didn't heat the wire and testing the voltage across the ends of the bow wire had 0V, but without the bow wire it was giving me 28-30VDC.

Swapping the electrolytics around made no difference, though the two 2200pf ones would warm if they were the last in the circuit.

I even tried to make a full wave rectifier with 4 diodes and the two 2200pf caps hoping that I'd just wire that in series with another power source, but even that didn't give me any current when attached to the bow.

So at this stage I'm left with the only thing that will work in any way is the original 2x2200pf 25V electrolytics and 2x6A rectifier diodes in the left half of that circuit I posted. This will give me 14VDC on the 10VAC terminals and only just heat the wire I'm trying to use.

Without trying to compare the performance of a half wave doubler circuit as suggested by vintage1 with what you are attempting, first glance suggests it may be a better approach if for no other reason than it eliminates the series capacitor. However because your load is heavy, the filter caps would need still to be hefty to minimize voltage sag between the wave peaks and maximize peak voltage value and energy delivered to the load.I'm not sure what Vintage means by the 2x half wavers. I've tried looking on the web for what he means but I don't think I've searched correctly as the only thing I've come up with is the circuit I posted in #5 above. I've tried using the 6A diodes on each of the AC terminals but no matter how I connect them, I get 1/2 the output voltage as DC. The only way I get more than half the voltage is by using the capacitors as outlined above. I'm hoping that there is a way to get the voltage and the current without the capacitors, but I don't know how it goes together. Are you able to explain what he means?

Does it matter that the caps I've used have been wired in parallel and not single large ones? I'm trying to do this cheaply as I've got a box of old capacitors, mostly 1000pf and 40v or .47pf at 250V and bigger new ones seem to run at around $18-$20AUD at the local electronic component place so I'm trying to avoid having to spend $40-$50 on electrolytics.

Thanks again for any help you can provide.

[jeffs555]

I hope you mean uF and not pF. Are you sure you got the polarity right on the output capacitor?

Wire the circuit like this. http://en.wikipedia.org/wiki/File:Bridge_voltage_doubler.svg
I am sure this is what Vintage1 meant, and would be much better in this application.

[Bosk]

Is this the way the capacitors need to go in and what size do you think they should be? would the 2200uf be enough or do they need to be bigger, if so can I gang them up in parallel?

[jeffs555]

No, you have both caps backwards. The upper diode has the cathode connected to the output so that will be the positive terminal. The lower diode has the anode connected to the output so that will be the negative terminal.

PS
2200uF sounds low, but you can't calculate the required capacitor value without knowing the current draw of the wire. You can't just measure the resistance of the wire because the cold resistance will be significantly lower than the hot resistance.

[Brandano]

I still wonder if it's worth going through the hassle of doubling the voltage for an application that really requires a constant current source instead. Ultimately if the transformer can't deliver the proper wattage there's no circuitry that will cure it.

[Bosk]

I still wonder if it's worth going through the hassle of doubling the voltage for an application that really requires a constant current source instead. Ultimately if the transformer can't deliver the proper wattage there's no circuitry that will cure it.
There are two reasons.

1. It's an interesting exercise and it's been fun to pull out my box 'o' bits from 30 years ago.

2. It's a cheap exercise, if I've got suitable components. If what I've got will work then great. In Melbourne Aus, a 5A constant current power source of sufficient voltage is around $300 to buy and would probably cost at least half that in components to build even if I knew how.


The problem is that the transformer can deliver the wattage, it just doesn't deliver the voltage potential to cover the length of the bow wire. For instance, I accidentally shorted the 6VAC terminals in my experiment and my 14 inches of speaker wire that I used to connect the transformer to the the diodes glowed bright red and the insulation melted of it in around 3 seconds, just before the 6A diode exploded!!! It is able to heat a 300W soldering iron to a bright red on 10V so it's got the wattage, it's just a matter of keeping the current while driving up the voltage.

What I've got so far is almost actually usable it's so close. Hence the excitement and frustration.

[Bosk]

No, you have both caps backwards. The upper diode has the cathode connected to the output so that will be the positive terminal. The lower diode has the anode connected to the output so that will be the negative terminal.

PS
2200uF sounds low, but you can't calculate the required capacitor value without knowing the current draw of the wire. You can't just measure the resistance of the wire because the cold resistance will be significantly lower than the hot resistance.

Measuring the current will be interesting as the ammeter only works in mA. Maybe I can connect my eMeter which will work up to 100A with the shunt that came with it. So the circuit should be like this:

[Greywing]

The cap polarity is correct. Your capacitors can be used in parallel to increase the capacitance value. You would like the same capacity value in the positive and negative legs. Also be sure that your caps have a voltage rating at least as large as the peak voltage across them.

I don't know what current the e-meter measures so I can't comment on using that to measure your load current. Since your diodes are rated at 6A, if they don't burn up when (and if) you adequately light up your bow you are presumably near or below there rated current. I don't recommend this as a method to measure or confirm load current; however I think you did say you got a slow cut with the first circuit so if the diodes survived that it's worth trying in the new circuit.

GW

[earlwb]

Well, if the windings have been kept in phase, it may work, hard to say though.
Now there was some electronics classes that had lab tests with what happens when windings are in series or parallel. Here is a PDF of a lab class here:
http://www.google.com/url?sa=t&source=web&ct=res&cd=9&url=http%3A%2F%2Fwww.hirstbrook.com%2Fenergylab%2F Lab%25204.pdf&ei=DYw3Sre5N5ixtgeQvqHZDA&rct=j&q=connecting+transformer+windings+in+series&usg=AFQjCNFU6dtbs_2nMQCT9H-FeeM4P3j8bw

[jeffs555]

I believe his transformer has a single primary and a single tapped secondary so no problem with phasing. Getting the phases correct is only necessary when you have multiple transformers or multiple primaries or secondaries.

[Bosk]

I had a chance to try this today and found that I was able to double the voltage using 2500mf capacitors and 6A diodes without either getting hot. It still didn't push as much current as I'd like, but at least now it will heat the 1.2m bow sufficiently to cut well.

Thanks for all your help with this.

Last question on this topic... Do you know if increasing the size of the capacitors will increase the current flow? I could wire in some more capacitors in parallel and could get it up from 2500mf to 5700mf each, would this improve the current flow and provide better heating?

[jeffs555]

Yes.