Power Current Voltage Resistance: (Spot welder)

Other Hardware General Electronics
1

i've just finished building a microwave transformer based Spot welder. I had some issues with the welder at first which i managed to resolve;

-it wasnt welding! just heating the metal. i figured out that it was due to the current not being high enough. all resolved now, but i dont understand the reason.
-cant measure current since it i expect that it will blow my multimeter.

therefore tried to calculate as follows;

trasnformer rated 700W
Primary transformer coil: 240V , I=~3A

Secondary coil, initially with one turn of wire
assume 90% efficiency in transformer (dont know if this is reasonable??)
Power out is 630W
measure voltage of 1v
I= 600A ?? (p=i*v)

i understand that this current would be fine for welding, but it didnt, so i think actual current was much lower.

edit: increasing number of turns in secondary-> higher voltage, therefore lower current?! (i = p/v), assuming power out is the same


however, we also know ohms law (i=v/r)
if i use the same length of long wire for secondary coil (no change in resistance), if i increase number of turns from 1 to 3 or 4, output voltage increases from 1v to 3v and current is also increased linearly.

when i did this, the spot welder started to work and the more turns i added, the current visibly increased (i know since it burnt through metal much easier/quicker)

**So, finally, my question is, what is wrong with my logic / Power based calculations for current???**
**any other input/feedback/comments would be great**

next step for me is to build a high current shunt so that i can measure the current output directly
2

Primary transformer coil: 240V , I=~3A

Did you actually measure 3A or is that the "rating"?

assume 90% efficiency in transformer (dont know if this is reasonable??)

That's probably reasonable under normal/nominal load but not with a shorted output and probably not with nearly-shorted output. (It has zero efficiency at with an open or shorted output. :wink: )

however, we also know ohms law (i=v/r)
if i use the same length of long wire for secondary coil (no change in resistance), if i increase number of turns from 1 to 3 or 4, output voltage increases from 1v to 3v and current is also increased linearly.

Did you measure that voltage. More turns adds resistance so the voltage (with a load) may not increase as expected.

next step for me is to build a high current shunt so that i can measure the current output directly

There are other ways of measuring current. Since it's AC I'd try an "amp clamp" type meter.

3

Nothing wrong with your logic. What you are missing is that transformers are not a one-way power supplying device. The current in your secondary also induces current into the primary winding, although out of phase.

Don't measure the secondary current. Measure the primary current as you try to weld. You will see it drop as the secondary tries to increase.

The ideal solution is to have the primary winding with larger wire and fewer turns.

Paul

4

DVDdoug:
Did you actually measure 3A or is that the "rating"?
That's probably reasonable under normal/nominal load but not with a shorted output and probably not with nearly-shorted output. (It has zero efficiency at with an open or shorted output. :wink: )
Did you measure that voltage. More turns adds resistance so the voltage (with a load) may not increase as expected.
There are other ways of measuring current. Since it's AC I'd try an "amp clamp" type meter.

hi Doug - the 3A is calculated/estimated from power rating 700w and voltage of 240v, i=p/v?
on the secondary coil, indeed i measured the voltage of 1v with 1 turn. (3v with 3 turns etc..)
in order not to increase the resistance with more turns, i used the same length of cable for 1to 5 turns while i did my tests.

i still dont understand why the power trasnfer rule suggests higher current with lower voltage (i=p/v), whereas ohm's law suggests opposite of higher current with higher voltage (i=v/r)... (i remember never getting my head fully around this when i was studying at age 16!)

indeed for current, i saw those amp clamps, but i just invested in a DMM which doesnt have this feature and dont really want to get more toys just yet!

cheers

5

Paul_KD7HB:
Nothing wrong with your logic. What you are missing is that transformers are not a one-way power supplying device. The current in your secondary also induces current into the primary winding, although out of phase.

Don't measure the secondary current. Measure the primary current as you try to weld. You will see it drop as the secondary tries to increase.

The ideal solution is to have the primary winding with larger wire and fewer turns.

Paul

thanks Paul - i do intend to measure the current on the primary also while under load. my multimeter should manage with this current.. i will also measure secondary as and when i sort out a suitable shunt.

so if i undertand you correctly, the power laws (p=iv) cannot be applied so simplistically due to induction type effects? hence why i dont observe 600A?!

but what confuses me, is how power law gives an inverse relationship between voltage and current, and ohms gives a linear relationship..

6

You are computing based on DC values, while working with AC voltages and current, which, as you suggest means your "resistance" must include inductance values.

Paul

7

i still dont understand why the power trasnfer rule suggests higher current with lower voltage (i=p/v), whereas ohm's law suggests opposite of higher current with higher voltage (i=v/r)... (i remember never getting my head fully around this when i was studying at age 16!)

Ohm's Law is a law of nature (with man-made units of measure).

If you have 600 Amps at 1V you'll have 1800 Amps at 3V (assuming the same load, of course). That's 5400W.

And since conservation of energy is also a law of nature, that's more than 5400W (22.5 Amps) into the primary depending on efficiency, with the additional energy converted to heat. Possibly burning-out the transformer.

8

Your 700 watt transformer it likely to be a number you got from the microwave unit itself. That is a marketing number, not reality. Remember AC deals with RMS values. Marketing will take the peak current X peak voltage to give you 700 watts for their advertisements. That is a true value only when sealing with the peaks.

Paul

9

I still don't understand why the power transfer rule suggests higher current with lower voltage (i=p/v), whereas ohm's law suggests opposite of higher current with higher voltage (i=v/r).

Your confusion is this:
I = P/V, so if you lower V and keep P the same then then I must increase to compensate for lowering V. However, what you are missing is that for P to remanin the same when you lower V you must also lower the resistance of the circuit in order to get more current for the lower voltage. You can't just lower V and expect I to magicly increase, it won't. If you lower V and leave everything else the same P will decrease.

I = V/R means that if you increase V or reduce R (or both) then I will increase. This is a different equation to the above and is telling you something different, it does not contradict the above, it gives you different information. It does not tell you anything about the power, but if you combine the 2 then you get this:

I = V/R and
I = P/V
But we don't know what P is that this point so:
P = IV
So
Taking I = V/R if you increase V then I also increases, and if you put that into P=IV then P increases by V^2
I hope you now see that when you change V you also change P by the square of V.

I think your confusion comes from not realising that P varies with V. In your original question you said:

Power out is 630W
measure voltage of 1v
I= 600A ?? (p=i*v)

How do you know that's the power out? You don't. You're just assuming it because of the rating of the transformer, you have not taken R into account, which includes the resistance of the metal you are welding. I think you are assuming that the resistance of the metal is negligible, which cannot be the case because if the resistance of the metal were negligible it would not heat up (much) because power is also P=I^2*R, and if R id very low then the power dissipated in R, the metal, is also low. The metal must have significant resistance otherwise you would not be able to weld it.

Back to your problem, you need to know the resistance of the metal then you can work out what current you need to drive through it to dissipate 630W, and what voltage you need to achieve that current, then from that how many turns on your transformer.

I think you are making the common mistake of thinking that because a power supply of whatever type can supply some particular power that it must therefore supply that power. It doesn't, it supplies the power the load takes, which in this case depends on the resistance of the metal.

10

DVDdoug:
Ohm's Law is a law of nature (with man-made units of measure).

But not in the way you think. Its an experimental result about the behaviour of metallic conductors (with some exceptions).

V = IR

is not a statement of Ohm's law, its a definition of resistance and is tautologically true (nothing to do with nature,
but true by definition)

The actual Ohm's Law is that for metallic conductors resistance is constant (at a given temperature).
For other materials resistance can depend on voltage and current (often very markedly) - or put another
way current is proportional to voltage in metallic conductors.

It doesn't work for the super conducting state in metals, and it breaks down for very narrow films or wires and
for extremely high current densities.

11

sapre:
So, finally, my question is, what is wrong with my logic / Power based calculations for current???
any other input/feedback/comments would be great

You forgot that the workpiece you are wleding has significant resistance. To transmit power to a load of a certain resistance you need enough voltage and enough current. You want the secondary impedance to match the load.

Lets assume the transformer has about 1V per turn, and 600W. A single turn has a source impedance of 1/600 = 0.0017 ohms (for max power),
and can put 600W into a 0.0017 ohm load. With a 0.1 ohm load it can only put 10A through it before the voltage
runs out - that's 10W max.

With a 0.0001 ohm load it would be limited to 600A, which is only 36W (and in practice would blow the primary fuse - you do have a primary fuse?)

Say your workpiece is 0.1 ohm, then you want 7.7V at 77A to get 600W out of the secondary, ie 8 turns.

V = sqrt (RP)
I = sqrt (P/R)

Once you know the volts/turn for the transformer, calculate the secondary voltage you want from sqrt(RP) and
thus determine how many turns.

To measure the low resistance of typical welding setup you'll need to do a Kelvin measurement (4 wire) using
some power source capable of an amp or so of current.

You had problems because your load resistance was more than the single-turn impedance of the transformer.

12

thanks all for the detailed explanation - i understand the mistake i was making with the power rating of the power supply. seems quite obvious now i know the answer!

one thing i did struggle with was measuring the resistance with the multimeter. everything i measured (primary, secondary coils and a piece of thick wire) seemed to give the same reading of 0.5 ohm! i believe that DMMs are not very accurate at low resistance, so assume that this was the problem.

thanks for the tip MarkT on Kelvin measurement of resistance, i have started looking into this. still also on my Todo list is making/calibrating the high current shunts so i can take proper measurement of the (high) output current.

good news is that the spot welder is working well. other good news is that i learnt a thing or two along the way!

13

sapre:
Nne thing i did struggle with was measuring the resistance with the multimeter. Everything i measured (primary, secondary coils and a piece of thick wire) seemed to give the same reading of 0.5 ohm! i believe that DMMs are not very accurate at low resistance, so assume that this was the problem.

Correct, they are not good at really low resistances. Did you also measure the resistance of the work piece you are trying to weld?

Good news is that the spot welder is working well. other good news is that i learnt a thing or two along the way!

Good! :slight_smile: