I've got a Miller Thunderbolt 225 (your basic AC transformer welder), it's
been a very reliable unit for general welding but I sure would like to be
able to get DC out of it.
Does anyone sell a diode bridge and choke assembly that goes on the output
Any plans for such an add on accessory?
Will a bridge of 4 300amp diodes (2+ and 2 -base on a heatsink ground) work
without a choke?
Since a welder is regularly shorted ( striking arc, stuck rod etc.) on the
output, is a special type of diode or bridge needed to handle the overload
Spud: ve7ifd and that would be at telus dot net
I have an old Miller Thunderbolt 225 and the voltage/current curves show a
maximum current of about 280A when the voltage is 0 volts. This would be the
voltage when the rod is stuck. In normal use say running at 150A instead of
full power the stuck down current would be less because of the action of the
constant current shunts in the transformer. So, a bridge of 4 300A diodes is
quite adequate. Also, each diode only conducts half the time so you have
even more of a safety factor.
The choke will add stability to your arc. One way of looking at it is that
the field collapses and provides some current when the voltage goes through
zero to keep the arc alive. I did some internet searching on this and
recommendations range from "there is enough inductance in a set of long
leads to achieve this" to "try and find an old one somewhere". Another
poster recommended wrapping a few turns around a large bolt.
I built a rectifier for mine and I opted for another solution: Get an old
microwave oven transformer (biggest you can find). Remove the windings by
grinding the weld holding the E laminations and the I laminations together.
Microwave transformers are assembled cheaply so the Es and Is are not
interleaved thankfully. I wrapped as many turns of welding cable as I could
around the center of the Es; only about 5 turns I think. You don't want to
put the I laminations back on as they were. You must have an air gap or the
high welding current will saturate the core. I put a piece of panelling
between the Es and the Is to ensure a space.
I actually used 2 old transformers in series giving me about 9-10T total or
thereabouts. No scientific measurements done on it but it works fine. The
choke can be in either lead. My diodes are 12 35A 600V bridge rectifiers in
parallel with equal lengths of #18 supply wire on each leg of the bridge to
give some load balancing resistance. Your 4 diode method is better.
Your diodes should be rated at 150V or better. Put a 0.01-0.1ufd capacitor
with a voltage rating of 200V or more across each diode. This will protect
the diode against voltage spikes caused when the arc is broken and the
fields collapse. You can also put a 0.1ufd across the + and - output leads
Make sure you mount your diodes on a heat sink and try and get it in the
fan's air flow. I built mine in an old desktop computer box and put a couple
of muffin fans in to blow air on the bridges. I don't think they get too hot
given the intermittent nature of welding but heat is the enemy of
semiconductors so good cooling is cheap insurance.
Hope this helps,
Hello Bill and TMT,
A chap on another group gave me a few clues on how to
set the gap.
Connect the primary of the stick welder to a variac.
mains supply ----variac---stick welder primary
Connect say a 100 amp meter shunt and the choke under
test across the secondary of the stick welder like this.
welder terminal------meter shunt--- choke-----welder terminal
Across the shunt, connect a digital voltmeter and an oscilloscope.
Turn up the variac from zero volts and look at the DMM and
interpret the amps from the DMM (eg 50mV may equal 100 A)
Look at the oscilloscope waveform as you turn up the current
flowing in the shorted out secondary of the stick welder.
If you see the waveform changing from a nice sine wave to a
slightly distorted sinewave at say 40 amps you will know that
the choke core gap isn't big enough.
Reset the gap with thicker packing and test again.
Like you two guys, I found very little information
on the web on how to build and test chokes. Well,
information that wasn't written in mathematical
gobbledegook for hobbyists like me.
Wow sounds simple & complex. CAUTION on the variac - make sure it can handle
the current the stick welder needs. I have half a dozen around the
house and only the one I will get from Dad in a few months can handle that!
What is the Mains current - Mains voltage - simple VI = Power
lets say a 220 at 25 amp - 5.5KW That is a big Variac. Maybe your mains
current isn't that high.
by your comments and CAUTION
I can tell that you have not read
what I said.
Turn up the output of variac slightly, just above zero
volts, is maybe what I should have said and then you
would not have issued cautions. I apologise for not
making myself crystal clear.
We are only using the stick welder as a piece of
high current test equipment NOT as a welding
We are testing chokes for the saturation point
The variac output voltage is adjusted by hand from
zero to low volts while watching the digital mutimeter.
The primary voltage to the testing transformer (which
just happens do be a stick welder) is low. OK?
The secondary resistive load is the shunt.
Power is volts times amps as you pointed out.
Voltage across the shunt is 50 milli Volts ( 0.05 V )
0.05Volts times 100 amps equals NOT MUCH.
We are NOT welding!
Just using the welder as a high current source at
near zero volts output and with low voltage input
to the transformer (which just happens to be a stick
welder because it is convenient to use).
A small variac used as above will do nicely for this
choke gap setting excercise.
The transformer has copper loss and core loss so the unloaded transformer
would be drawing some current and has power loss. When you shunt out the output
except through a choke (aka coil with an iron core...) the secondary delivers
current. This makes the primary deliver power through the core and draws
current through the variac. The current is proportional to the turns ratio.
Without the transformer in the normal operational magnetic curve position,
the reactance will be different.
I would fuse the variac lead to protect it. Inductive inrush currents and
reactive currents can be hard to deal with.
I have designed and made transformers and have a background in Tube designs.
Caution means to be 'cautious' not to 'stop doing this stupid'. Science
lost hundreds of highly educated professionals to the unknown inrush
of current from this or that flowing in their body.
John Cright> >
Now, how about giving the original poster and hobbyists like me a few
clues on how he should build the choke assemby for his AC to DC
In case you missed the original request, here it is,
"I've got a Miller Thunderbolt 225 (your basic AC transformer welder),
it's been a very reliable unit for general welding but I sure would
like to be able to get DC out of it."
Hi. How much is it worth to you? :-)
Seriously, I am planning to write up a document on how to make a
micro-welder out of a defunct microwave oven.
The design of the reactor is straightforward if it is taken step by
step. The key reference is by Eric Lowdon.
Estimate the energy storage required by drawing the plot of sine
squared on graph paper.
1. Use Hanna curves for the required core size as a function of
energy per cycle.
2. Design around this core size according to the book. This is an
iterative process with the wire size and the gap thickness as
variables. Start with a fixed wire size, calculate filling factor and
turns. Select a gap to prevent saturation. Compute inductance and
heat dissipation. Adjust wire size to obtain desired result.
1. Use a junk 60 Hz transformer from an old battery charger or TV or
flea market. I saw somebody selling them for 10 cents per pound.
2. Disassemble the core and measure the path length and cross section.
3. Bifilar wind the thick wire and quadfilar wind the thin wire, or
more, and fit as many turns as possible.
4. Compute a gap size to avoid saturation.
5. Compute the resulting inductance.
6. If insufficient, repeat process and place inductors in series.
7. If multifilar winding is too much of a drag, single wind the
primary, and bi or trifilar wind the secondary and place it in series.
Then, parallel the inductors to obtain sufficient current capacity.
Warning: this procedure takes a long time. Make sure it is what you
want to do.
Also, the above (very brief) discussion above makes it quite clear that
the inductor design given in the aircraft alternator welder article
will not work, since its inductance will be too low. The inductance is
lower because these alternators typically produce much higher
frequency. Also, if you draw it on graph paper, the output is 3-phase,
so you are adding sine squared + sine squared alpha + sine squared
beta, which has much less area (graph paper squares) in the dips.
Any interest in a booklet?
In reference to the pdf file from hobartwelder, I believe there is a gross
On page 21 (the plans) figure 7-1 is not right! The two upper Diodes are
connected together at their
bottom end, which shorts out the transformer, as well as the fuse/circuit
breaker. You get the point.
What I would do first is to measure the output open circuit voltage.
Closed is close to zero. - close enough anyway.
I'd find out how it was measured - likely RMS and then generate Peak and
if I had a Drantz power line sensor/logger I'd hook it up to the output
and do some welding. What is the maximum peak hits. Baring that, some
meters have peak holding or maximum holding. Ratio up from that for protection.
Find the Current the Thunderbolt can generate on a short - one considers 250
down rated to 225 for specs. So the current would be 400 amp rectifiers.
The 400 amp would handle the peak inrush current from 0 to max stick current.
(e.g. when the stick sticks and the transformer hums...)
The voltage is rated by the peak inverse voltage that would be seen - this is
tough to guess on my half - what is the Thunderbolt open voltage ? Does the
surge or oscillate when the process is started ? or does it have a controlled
signature that drops from open to short in a linear or near linear method and
when the arc is cut. Transformer regulation and core design determines how
'wild' it is.
A good amp probe attached to a scope could show so much. Mine is packed now for
Now for waveform shape. What a single or ganged (parallel) rectifier (power
provide is a half sine wave. The Positive or Negative pulse would be seen while
alternation is almost flat. Duty cycle is about 50%.
During the other 16ms of 'off' time the rectifier, mounted on a large copper
large Al plate - best with fins and a blowing fan to cool - perhaps a demand
type based on temp.
This would be a DC without a choke - used to smooth the waveform. Not much good
with one diode.
When four diodes (Rectifiers) - this is expensive as 400 amp rectifiers are
one would build a Full Wave Bridge and this would take the typical sine voltage
and convert it
into a double positive or double negative waveform and the duty cycle is about
If one wanted a constant current output (or something like it if we keep a blind
eye closed :-) )
we would add a choke in series or between the bridge and the output lead. A +/-
exactly prior to the bridge would be best - then both - the line out of the
then proceeds to the choke. What this is trying to do is simply keep the
current the same value
by building an electric field on the higher currents and clasping the field and
current to make up for the loss. Since the currents we are talking about are so
to electronic power supplies, the regulation has to be tremendous. It is used
to aid and not
smooth the waveform. In other words, it is a smoothing device. But it will not
like a battery or a near pure constant current.
My guess is this : The coils used are to generate a high voltage oscillation
due to being out
of regulation. I believe these are just to improve stick arc by making a
alternation when the stick current drops (as the hand moves...).
A regulation coil or choke to control a stick welder is a dream. Consider
running a stick
at 90 amps and another at 150. What coil will be useful on both ? Now kick it
up a bit to 200.
You can consider the core power of a transformer or choke to equal 100 watts in
a core the size
of a large mans fist. Not much power considering a welder. What is the size of
the transformer core ?
The Miller OCV on low current range is about 80V which is a safety maximum
for welders. High range it is about 55V.
Max stuck down current on the high current range is about 280A according to
the Miller VI curves.
Assuming a 300A max current limited by the constant current action of the
shunted transformer a 150A diode in a bridge (50% conduction) will just make
Any transient is likely to be produced only when the arc is broken and the
transformer and any reactor in the DC circuits fields collapse. I think
commercial stick welders only use 200 PRV diodes.
be attached to a scope could show so much. Mine is packed now for a move.
The shunted transformer design provides for constant current for stick and
tig welding. The regulation for stick welding is overly important and the
transformer provides sufficient CC regulation. The inductor is not really
there for smoothing in the "nice DC" sense; it is there so the current does
not go to zero when the voltage goes to zero thus keeping the arc lit. You
can look at this in the sense that the collapsing field generates current to
keep the arc going when the voltage is zero or you can consider the inductor
provides a few degrees of phase shift for the V and I. Note that once the
arc is struck the voltage needed to keep going is only a few volts, nothing
like the OCV.
The ideal reactor would be variable to match the current.
Well if you want to quote the formula relating the average DC current for a
full-wave rectifier as a function of the peak current per cycle in a bare
rectifier circuit. The formula for the peak current in a half-wave rectifier
is the same but without the amazing factor of 2, ie, Imax= 3.14 I output. So
the full-wave diode only has to work half as hard.
It shows the maximum (peak of the sinusoid) current in the secondary as a
function of the rectified average DC current. The principal current rating
for a rectifier diode (the one typically listed) is average current, not
peak. My original point is that the average current through each diode in a
bridge is 1/2 the average output current.
Most folks are worried about calculations for these diodes because,
the higher the amperage, the more expensive the diode. I have a
simpler solution. If you live near a large city, go and visit the
local scrap metal yards. Usually they will let you look around for
stuff on your own. Find the biggest diodes that you can and make your
own bridge rectifier. The reason that I am mentioning this is that
most metal scrap yards only have prices for metal. i.e. aluminum
might be a buck a pound, etc. When you walk up to a metal recycling
yard, they don't have a price for electronic components. They see you
with this tiny piece of metal in your hand and usually, you walk away
with a small fortune in semi-conductor components for almost free.
You can test them with a multimeter right on the premises, so you
won't be taking any scrap home with you. I have built many things
like this. My plasma cutter came from the metal scrap yard. It cost
me $30 for the unit and only $6 for the components that needed
replacing. It pays to dig around a bit. Big business is very
wasteful and most items in the scrap yard really don't need to be
there. Motors usually fail because of bad bearings or brushes.
Bearings and brushes are cheap. Therefore, if you need that 5 hp
motor for your shop air compressor, you can have it for usually less
than 20 bucks.