I have a project to design a 60 Hz, two winding transformer with a 120 V primary voltage, 14 V secondary voltage and 54VA power rating.

I'm supposed to use :

1. Standard E-I laminations
2. M-36 magnetic steel for the core
3. Standard wire gauges for the windings.

 

I pulled those links out of your post, as you are the designer. What progress have you been able to make?

 

well I have found a lot of information but I have not made much progress because I don't know exactly how to calculate and determine which of the E-I laminations and wire gauge I'm supposed use...based 120/14 voltage transformation and 54 VA power rating

 

Do i just pick any lamination and wire gauge? or do i have to pick one based on the power rating, primary and secondary voltages...

 

Do i just pick any lamination and wire gauge? or do i have to pick one based on the power rating, primary and secondary voltages...

As for the wire gauge there are rules set down by the NFPA (National Fire Protection Agency). They have several books out there that are usually adopted as law throughout the states. One of these is the NEC, sort of an electrician's bible. The NEC gives tables of wire sizes vs. amps and insulation. But I'm not sure if these tables apply in your case. There are books, such as the White Book, that deals with the concerns of manufacturers.

Then again, it may be that the tables in the NEC apply. If so, you can consult it at your local library or buy it for about 30 bucks or I'll give you the data from my 1990 copy of the NEC.

 

do you know how to put E and I laminations??? before putting the laminations you have to first make primary and secondary windings on the core. The formula to calculate the turns is N1/N2=V1/V2=I2/I1

 

do you know how to put E and I laminations??? before putting the laminations you have to first make primary and secondary windings on the core. The formula to calculate the turns is N1/N2=V1/V2=I2/I1

As I read the problem the laminations are standard. I'm guessing there are formulas I'm not aware of for this standard and also for the core material. The only thing left is wire size. But, yes, by all means your equations, salmanshaheen, are necessary for the calculation of turns of wire.

 

Thanks guys but is there any equation that directly relates any of the electrical parameters(voltage, power rating, current, turns ratio or resistance...) to the any dimensions(length, surface area, volume..) of the the core, lamination or wire to be used

do you know how to put E and I laminations??? before putting the laminations you have to first make primary and secondary windings on the core. The formula to calculate the turns is N1/N2=V1/V2=I2/I1

@salmanshaheen thanks but i've definitely used that equation...if i'm not mistaken its:

V1/V2 = I2/I1 = N1/N2 = n

where V1 and V2 are the primary and secondary voltages respectively; N1 and N2 are the number of turns on primary and secondary side respectively while n is turns ratio.
From calculation I've determined my turns ratio to be approximately 9 : 1 but how do i determine the exact (or an appropriate) number of turns for each side?

 

As for the wire gauge there are rules set down by the NFPA (National Fire Protection Agency). They have several books out there that are usually adopted as law throughout the states. One of these is the NEC, sort of an electrician's bible. The NEC gives tables of wire sizes vs. amps and insulation. But I'm not sure if these tables apply in your case. There are books, such as the White Book, that deals with the concerns of manufacturers.

Then again, it may be that the tables in the NEC apply. If so, you can consult it at your local library or buy it for about 30 bucks or I'll give you the data from my 1990 copy of the NEC.

Thanks PRS but i don't think the tables apply in this case (or at least not at the design stage I'm at now)...but if i happen to need it later on, I wouldn't mind your 1990 copy

 

The first thing to do is determine what size laminations you need for a transformer that can deliver 54 VA. You can find tables that will give you this information, but a quick and dirty way to do it is to look up some commercial transformers at a distributor or manufacturer such as Triad. Find one that can supply 54+VA and get the dimensions.

Check out a manufacturer's web site for lamination specifications:

http://www.tempelsteel.com/products.asp?cat=11

A lamination with a 1 inch center leg width in a square stack would probably do a little over 54VA:

http://www.tempelsteel.com/products.asp?id=161

M36 steel is quite lossy and probably wouldn't be used to make a commercial transformer that pretended to a normal level of quality. Your professor probably specified it so you would have to pay attention to core loss. The M36 designation means that the steel has a nominal loss of 3.6 watts/pound at 15,000 gauss peak. Normally, M6 would be used for good transformers; this steel has about .6 watts/pound at 15,000 gauss.

If you don't already have the formulas for peak flux density vs. core cross section and applied turns/volt, you can probably find it somewhere on the Tempel site.

You have to decide how much temperature rise you can tolerate. If the transformer is allowed to get hotter, it can supply more power. Standard commercial units are often designed for a maximum hot spot temperature of 100° C. With proper materials, transformers can be designed for hot spot temperatures of 180° C (class H), but that's more expensive.

You must decide whether your transformer will be convection cooled, or whether you will have forced air (a fan). You can find tables in various references that will tell you how many watts a given lamination stack can dissipate with convection.

Normally a good design choice is to allocate about half of the available winding space for the primary and half for the secondary. Determine how many turns you need to give a peak flux density of around 15000 gauss with your 120V primary voltage (remember that's RMS). Since you will be using round wire, there will be space not filled with copper. A 60% packing factor will be a reasonable starting point. Let's say you need N turns to get the desired flux density. Half the winding window is available for your primary, so divide that area of half the winding window by N and pick a wire gauge with .6 times that cross-sectional area. Then you should be able to get N turns of that size wire for the primary in half the winding window.

Determine the number of turns the secondary needs to give the desired output voltage, and pick a wire size that will allow you to wind that many turns in half the winding window, in the same manner as the primary wire size was determined.

Don't forget that the output voltage will drop when the transformer is loaded. If your professor requires the given secondary voltage when loaded, you will have to take the IR losses in the primary and secondary copper into account.

Calculate the current in the primary when delivering 54VA, and knowing the resistance per foot of the chosen wire size, and the length of wire in the full primary winding, you can calculate the DC resistance of the primary, and from that calculate the I^2*R primary copper losses. Calculate the weight of the core and knowing that the core has losses of around 3.6 watts/pound (for a flux density of 15,000 gauss), you can get the core loss.

Double the primary copper losses (to account for the secondary copper losses which should be about the same as the primary losses), add the core loss and see if the total losses are able to be dissipated with convection cooling with no more than a 100° hot spot temperature.

Assuming that secondary copper losses are the same as primary copper losses isn't quite exact. If your transformer is concentric wound, meaning the primary (or secondary) is wound on the bobbin first, and the secondary (or primary) is wound on next, the last one wound on occupies more volume because the Mean Length of Turn (MLT) is greater for the outer winding. You might want to take that into account eventually.

This will give you the loss when you first energize the transformer at room temperature, but don't forget that when the transformer gets hot, the resistance of the copper wire goes up. You have to take that into account when calculating the total transformer loss, and the consequential hot spot temperature.

All this is assuming your load is a resistive load. If this transformer is to supply a capacitor input rectifier, you will have to derate the power output, so you would have to use larger laminations to get the same 54 VA into a rectifier supply.

Your final design may require that you iterate some of these steps, but this should do for a first cut at the design.

Here are a couple of web sites that may be useful:

http://ludens.cl/Electron/trafos/trafos.html

http://www.cramerco.com/reference-tramyst.htm

 

The first thing to do is determine what size laminations you need for a transformer that can deliver 54 VA. You can find tables that will give you this information, but a quick and dirty way to do it is to look up some commercial transformers at a distributor or manufacturer such as Triad. Find one that can supply 54+VA and get the dimensions.

Check out a manufacturer's web site for lamination specifications:

http://www.tempelsteel.com/products.asp?cat=11

A lamination with a 1 inch center leg width in a square stack would probably do a little over 54VA:

http://www.tempelsteel.com/products.asp?id=161

M36 steel is quite lossy and probably wouldn't be used to make a commercial transformer that pretended to a normal level of quality. Your professor probably specified it so you would have to pay attention to core loss. The M36 designation means that the steel has a nominal loss of 3.6 watts/pound at 15,000 gauss peak. Normally, M6 would be used for good transformers; this steel has about .6 watts/pound at 15,000 gauss.

If you don't already have the formulas for peak flux density vs. core cross section and applied turns/volt, you can probably find it somewhere on the Tempel site.

You have to decide how much temperature rise you can tolerate. If the transformer is allowed to get hotter, it can supply more power. Standard commercial units are often designed for a maximum hot spot temperature of 100° C. With proper materials, transformers can be designed for hot spot temperatures of 180° C (class H), but that's more expensive.

You must decide whether your transformer will be convection cooled, or whether you will have forced air (a fan). You can find tables in various references that will tell you how many watts a given lamination stack can dissipate with convection.

Normally a good design choice is to allocate about half of the available winding space for the primary and half for the secondary. Determine how many turns you need to give a peak flux density of around 15000 gauss with your 120V primary voltage (remember that's RMS). Since you will be using round wire, there will be space not filled with copper. A 60% packing factor will be a reasonable starting point. Let's say you need N turns to get the desired flux density. Half the winding window is available for your primary, so divide that area of half the winding window by N and pick a wire gauge with .6 times that cross-sectional area. Then you should be able to get N turns of that size wire for the primary in half the winding window.

Determine the number of turns the secondary needs to give the desired output voltage, and pick a wire size that will allow you to wind that many turns in half the winding window, in the same manner as the primary wire size was determined.

Don't forget that the output voltage will drop when the transformer is loaded. If your professor requires the given secondary voltage when loaded, you will have to take the IR losses in the primary and secondary copper into account.

Calculate the current in the primary when delivering 54VA, and knowing the resistance per foot of the chosen wire size, and the length of wire in the full primary winding, you can calculate the DC resistance of the primary, and from that calculate the I^2*R primary copper losses. Calculate the weight of the core and knowing that the core has losses of around 3.6 watts/pound (for a flux density of 15,000 gauss), you can get the core loss.

Double the primary copper losses (to account for the secondary copper losses which should be about the same as the primary losses), add the core loss and see if the total losses are able to be dissipated with convection cooling with no more than a 100° hot spot temperature.

Assuming that secondary copper losses are the same as primary copper losses isn't quite exact. If your transformer is concentric wound, meaning the primary (or secondary) is wound on the bobbin first, and the secondary (or primary) is wound on next, the last one wound on occupies more volume because the Mean Length of Turn (MLT) is greater for the outer winding. You might want to take that into account eventually.

This will give you the loss when you first energize the transformer at room temperature, but don't forget that when the transformer gets hot, the resistance of the copper wire goes up. You have to take that into account when calculating the total transformer loss, and the consequential hot spot temperature.

All this is assuming your load is a resistive load. If this transformer is to supply a capacitor input rectifier, you will have to derate the power output, so you would have to use larger laminations to get the same 54 VA into a rectifier supply.

Your final design may require that you iterate some of these steps, but this should do for a first cut at the design.

Here are a couple of web sites that may be useful:

http://ludens.cl/Electron/trafos/trafos.html

http://www.cramerco.com/reference-tramyst.htm

@The Electrician Wow! thanks a lot I'm getting to work now

 

to calculate power, current or voltage (either primary or secondary) the formula is P=VI
remember that theoretically your input and output power should be same which is the beauty of transformer but in practical world due to losses your secondary power will be less than primary. Another important formula to be used for selecting the area of core is A=√w/5.6 Then calculating the area you can find turns per volt and the formula to be used is T/V= 5.6/A Keep in mind that this 5.6 used in T/V formula is constant which is to be used just for silicon sheet, if you are using iron sheet then the value should be changed I guess that is 7.6