Current transformer check-over
Thomas W, Thu Oct 30 2014, 02:29PMHello 4hv,
Im working on an X-Ray transformer driver and i need to sort out primary current sensing.
However i don't know to0 much about this this subject and need a bit of help.
Okey, here is my math:
Primary max current: 100A peak (don't want to really go above this)
100A to 250mA:
--1:400 ratio
0-2.5V output:
2.5/0.25 = 10
--10R resistor
2.5*0.25 = 0.625W reisistor, i shall use a 2W one.
That makes a resolution of:
1A / 400 = 0.0025A
10*0.0025 = 0.025mV
25mV across the load resistor per A on the primary.
The main thing is i have to wind 400 turns. The frequency will be 20kHz to 50kHz
I am thinking of doing 2 CTs on this, one on the actual main circuit watching the current through the transformer itself in the fullbridge, and another one on the input from the mains.
I plan to be able to set a current limit on the mains in and the transformer, allowing me to keep myself from blowing the fuse should i decide to run from a 13A socket, or if i use a smaller transformer and don't wish to break that.
Does anyone have any input on this?
Thanks!
Re: Current transformer check-over
dexter, Thu Oct 30 2014, 04:47PM
read Uzzors2k guide for his inverter
winding 400 turns on a toroid can be dreadful... so why not cascade CT's like ppl are doing for DRSSTC's, that would mean two toroids each with 20 turns
dexter, Thu Oct 30 2014, 04:47PM
read Uzzors2k guide for his inverter
winding 400 turns on a toroid can be dreadful... so why not cascade CT's like ppl are doing for DRSSTC's, that would mean two toroids each with 20 turns
Re: Current transformer check-over
Newton Brawn, Fri Oct 31 2014, 03:28AM
How about each transformer 400:1 be replaiced by 2 tranformers 200:1 ? the primary (100A) in parallel and secondary (250 mA) in series. The cores to be identical
Or just wind the 400 turns, it is more reliable,
Newton Brawn, Fri Oct 31 2014, 03:28AM
How about each transformer 400:1 be replaiced by 2 tranformers 200:1 ? the primary (100A) in parallel and secondary (250 mA) in series. The cores to be identical
Or just wind the 400 turns, it is more reliable,
Re: Current transformer check-over
radiotech, Fri Oct 31 2014, 04:42AM
Current transformers can be used with multiple turns through the eye. This is
done commonly in electrical work where small diameter wires are involved,
to measure, for example the current in a motor feeder.
The flux in the core is ampere turns.
This may help you with your 400: 1 ratio.
100 amp* 4 turns (in eye ) = 400 ampere turns.
radiotech, Fri Oct 31 2014, 04:42AM
Current transformers can be used with multiple turns through the eye. This is
done commonly in electrical work where small diameter wires are involved,
to measure, for example the current in a motor feeder.
The flux in the core is ampere turns.
This may help you with your 400: 1 ratio.
100 amp* 4 turns (in eye ) = 400 ampere turns.
Re: Current transformer check-over
Thomas W, Fri Oct 31 2014, 11:42AM
I think i might look into cascading them, more updates as they come. Im currently working on a few schematics.
Thomas W, Fri Oct 31 2014, 11:42AM
I think i might look into cascading them, more updates as they come. Im currently working on a few schematics.
Re: Current transformer check-over
Newton Brawn, Sun Nov 02 2014, 10:31PM
Thomas,
Classical approach, industry equipment
I Think you can define a priori that the primariy (100A) may have just one tun.
Then the secondary (250mA, 2.5V) has to have 400 turns.
The resistence of the secondary wire to be 20 times LESS less than the 10R you gave, so the secondary wire resistence about 0.5R.
Supose that the mediun lenght of one sec turn 50 mm, total lenght secondary about 400X0.05 = 20meters .
THIS Means 0.5R/20m = 0.025R/m (25 homs per kilometer ...) (EDITED)
See wire table,
The wire selected to be #19AWG.
See wire table again,
#19AWG >> 100 turns per sq cm, so the core window has to have 4.0sq cm minimun area to accomodate the secondary. (or more)
Try to find a core that have a "eye" or widow of 4sqcmm area, look the junk box.
Then check the maxi induction in the core,
DEspite a ferrite can work up to 0.2 to 0.3 tesla, keep the max induction +- 0.01 tesla or less to increase the magnetizing inductance.
Max inductace = Bmax = Vmax/( 4xFXNxA)
were
Bmax = tesla = T
Vmax = peak voltage, = V, = 2.5V
F= frequency, hertz, H = 20000Hz
N = # turns = 400,
A= area, sq METERS (yes, meters) = core cross section
assuming that you can find a core with 1 sq cm = 0.000100 sqm,
The Bmax = 2.5/(4x400x20000x0.001) = 0.00078T
That is acceptable.
The 1sqcm core trafo can work at 1.5Khz... Check!
Chek the numbers , again
Cheers
Newton Brawn, Sun Nov 02 2014, 10:31PM
Thomas,
Classical approach, industry equipment
I Think you can define a priori that the primariy (100A) may have just one tun.
Then the secondary (250mA, 2.5V) has to have 400 turns.
The resistence of the secondary wire to be 20 times LESS less than the 10R you gave, so the secondary wire resistence about 0.5R.
Supose that the mediun lenght of one sec turn 50 mm, total lenght secondary about 400X0.05 = 20meters .
THIS Means 0.5R/20m = 0.025R/m (25 homs per kilometer ...) (EDITED)
See wire table,
The wire selected to be #19AWG.
See wire table again,
#19AWG >> 100 turns per sq cm, so the core window has to have 4.0sq cm minimun area to accomodate the secondary. (or more)
Try to find a core that have a "eye" or widow of 4sqcmm area, look the junk box.
Then check the maxi induction in the core,
DEspite a ferrite can work up to 0.2 to 0.3 tesla, keep the max induction +- 0.01 tesla or less to increase the magnetizing inductance.
Max inductace = Bmax = Vmax/( 4xFXNxA)
were
Bmax = tesla = T
Vmax = peak voltage, = V, = 2.5V
F= frequency, hertz, H = 20000Hz
N = # turns = 400,
A= area, sq METERS (yes, meters) = core cross section
assuming that you can find a core with 1 sq cm = 0.000100 sqm,
The Bmax = 2.5/(4x400x20000x0.001) = 0.00078T
That is acceptable.
The 1sqcm core trafo can work at 1.5Khz... Check!
Chek the numbers , again
Cheers
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