what determines the max current that can be drawn from a transformer's secondary coil?
how could i figure out what the max current i can get out of a transformers secondary coil? example: a welding transformer
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V2 = V1*N2/N1, where Nx is the number of windings, N1 primary, N2 secondary. (in absense of losses), and
I2 = I1*N1/N2
Yes, roughly speaking the input side current can be calculated as
I1 = V1/R(primary)
technically these are inductors, so R is actually frequency and inductance dependant and the numbers change a little, but for the sake of brevity...
That will give you a ballpark idea. The efficiency of the transformer is unknown, so the actual output will be somewhat lower than the calculation.
I'm not such a fan of calculating unless someone else has already put a device thru the empirical hoops and would prefer measuring the input side current, but whatever...
regarding the ga of the output wiring, that pretty much determines the maximum current that can be delivered no matter what, as it will heat up and melt if it goes beyond its capacity. (and it's why I included it in the note above, due to the nature of your original question title.)
I2 = I1*N1/N2
Yes, roughly speaking the input side current can be calculated as
I1 = V1/R(primary)
technically these are inductors, so R is actually frequency and inductance dependant and the numbers change a little, but for the sake of brevity...
That will give you a ballpark idea. The efficiency of the transformer is unknown, so the actual output will be somewhat lower than the calculation.
I'm not such a fan of calculating unless someone else has already put a device thru the empirical hoops and would prefer measuring the input side current, but whatever...
regarding the ga of the output wiring, that pretty much determines the maximum current that can be delivered no matter what, as it will heat up and melt if it goes beyond its capacity. (and it's why I included it in the note above, due to the nature of your original question title.)
Nov 21, 2009. 10:39 PMtwenglish1 (author)
says:
here are my calculations:
winding ratio:
winding ratio = voltage secondary / voltage primary
voltage primary = 120v
voltage secondary = about 20 or less(i can rewind the transformer if i need a lower voltage)
ratio = 20 / 120
winding ratio = about 0.17
input current:
when i measured the resistance of the primary coil it came out as .5 ohms could this be right?
I = V / R
I = 120 / 0.5
I = 240 - this can't be right, there is no way this transformer is drawing 240 amps from the main. it has to be less than 20 amps cause its only on a 20 amp breaker
winding ratio:
winding ratio = voltage secondary / voltage primary
voltage primary = 120v
voltage secondary = about 20 or less(i can rewind the transformer if i need a lower voltage)
ratio = 20 / 120
winding ratio = about 0.17
input current:
when i measured the resistance of the primary coil it came out as .5 ohms could this be right?
I = V / R
I = 120 / 0.5
I = 240 - this can't be right, there is no way this transformer is drawing 240 amps from the main. it has to be less than 20 amps cause its only on a 20 amp breaker
Nov 21, 2009. 11:23 PMtwenglish1 (author)
says:
i am going to guess it is maybe 5 ohms because
I = V / R
I = 120 / 5
I = 24 which is alot more reasonable
so about 24 on the primary of the transformer
secondary current:
current secondary = current primary / winding ratio
current secondary = 24 / 0.17
current secondary = about 141 amps
the most the 12 gauge windings can handle is about 40 amps
is 40 amps all it will be able to draw? or will it try and draw 140 and overheat melting the secondary windings?
I = V / R
I = 120 / 5
I = 24 which is alot more reasonable
so about 24 on the primary of the transformer
secondary current:
current secondary = current primary / winding ratio
current secondary = 24 / 0.17
current secondary = about 141 amps
the most the 12 gauge windings can handle is about 40 amps
is 40 amps all it will be able to draw? or will it try and draw 140 and overheat melting the secondary windings?
These are inductors - you can't get meaningful values of inductance / impedence by measuring resistance.
If you want heavy-current you'll need a big heavy transformer (in the foot-crushing range of "heavy")
L
If you want heavy-current you'll need a big heavy transformer (in the foot-crushing range of "heavy")
L
Yeah, sorry, wasn't thinking re: input side. contact the transformer mfg for the specs on it's performance. Guessing won't do.
You'd still need quite a large transformer to handle 30amps, your reference to overheating in the secondary is appropriate.
L
L
Nov 22, 2009. 9:10 AMtwenglish1 (author)
says:
how about a microwave transformer? i have 2 and i can wind about 20 turns on each one, with about 20 volts each, the current has to be about 40 amps cause i was able to weld with both of them hooked together. both of them together was about 40 volts
. Figure out what gauge/gage wire you have. Strip the (enamel) insulation and measure with a dial caliper or similar. Compare measurement to an AWG table.
. Look up ampacity for that gauge wire. May get lucky and find that figure on same page as above.
. This will give you a ballpark figure.
. Look up ampacity for that gauge wire. May get lucky and find that figure on same page as above.
. This will give you a ballpark figure.
The primary side voltage and current, the efficiency of the transformer, and the secondary winding ratio and wire gauge
Nov 21, 2009. 9:14 PMtwenglish1 (author)
says:
ok the primary voltage is 110/120v. how would i calculate the primary current? would it be current = volts / resistance of primary coil? the winding ration is secondary voltage / primary voltage right? if so that is about 0.17 and the secondary uses 12 gauge wire
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