Hello I'm interested in making a lab power supply with a lm317 or alm338k voltage regulator but I dont know how to calculate the current with my wires and I would like the voltage output of the power supply to be 1.25V+- to 30V+-

can any one show me how to calculate it?

I already have the E's and I's core from a 3V-12V step down power supply that had an output of 2A and it got burned out..

You can look up the current carrying capacity for different wire gauges, for example the American Wire Gauge: American wire gauge - Wikipedia

It is much easier to start with a 100 VA, 220 V transformer and just rewind the secondary, rather than attempting to wind both sections.

I agree with jremington.

Depending on the transformer, it should be possible to take it apart and rewinding the secondary coil. However it might be a bit tedious work.

I already have the E’s and I’s core

You’re saying you already have the laminations for the transformer?

I don’t think you can make a 34V supply without exceeding the 40V limit of the LM317.

To guarantee you have an output, you need a minimum of 2.5V across the LM317. So that takes you to 36.5V. Now consider your line voltage at the wall can vary 10%, even if you could get your secondary voltage to exactly 36.5V, when the line voltage goes up 10%, your voltage will be at 40.5V. If your line goes down 10%, you will be at 32.85V which means the 317 will stop working. So in essence you will have to make your LOW LINE 36.5V which puts you way over the 40V rating of the 317.

It is like rmetzner49 says. LM317 will burn out at 40+ volt.

You can use an LM317HVT which can handle up to 60V input.

But there is a limit on the current, 1.5 Amp is max.

You can do 2 lineair regulators in parallel when you give them a small series resistor (about 0,1 ohm) at each output to make sure the current is 'equally' devided between the two regulators.

(Putting regulators in parallel can be tricky and is not highly recommended because you must be sure currents are equally divided and cooling is adequate so that both regulators work at same temperature. )

What was the the transformer before you turned in into a small pile of scrap copper and electrical steel? For example, was it a mains rated part? Did you count the primary and secondary turns before you stripped them off? Did you measure the thickness of the windings(s)? Did it have plastic bobbins? I could keep going but I think it is most likely a lost cause...

If you don't know the original rating and the turns used to achieve that rating, your chances of repurposing the E & I's is about... well... zero. You'll do much better to re-purpose a known rating close to what you need. 220-240v, 50 hz primary, 100 va would be a good starting point. Or whatever is the correct combination in your country of residence.

220-240v, 50 hz primary, 100 va would be a good starting point.

Remember if you are using a full-wave diode bridge you will need about 1.5x your desired current in your transformer. A 3A transformer will be OK. Unless you are going full-wave center-tapped, in which case you need only 1X the current but 2x the voltage.

JohnLincoln:

Do you also have a coil former to put your windings on?

Do you have any data on the Es and Is that you have?You will probably need around 2000 turns for a 220V primary, but the actual number depends on various things, such as the magnetic. properties of the laminations, and the cross sectional area of former.

This will need to be a thin gauge wire.The secondary will a few hundred turns, depending on the actual voltage you decide you need. Once you know the number of turns in the primary, you can calculate the number of turns in the secondary, because the ratio of the input voltage to the output voltage is equal to the ratio of the number of turns in the primary to the number of turns in the secondary.

This will need to be a thicker wire.A correctly designed transformer will have the same amount of copper in the primary as it does in the secondary/secondaries.

Yes I made my own wooden coil former because I couldnt find one and I know how to calculate the voltage of the secondary coil but how do I know witch thickness and length I should use with considering the amount of heat it produces and, I am going to use the lm338k for that power supply and would like the out put of the system to be from 1.25V+- to 30V+- and the output of the transformer to be 34V because there will be a voltage drop after the diode bridge and all the other things .

Transformers are best calculated by using the transformer Equation:

Urms = 4,44*B*A*N*f

Where

Urms = the rms voltage in one winding

4,44 = constant (pi*sqrt(2))

B = max flux density in the core in Tesla. This is a material dependent value, normally you can use 1,2T for non oriented cores.

A = smallest section area of the core

N = number of turns in the winding

f = frequency used (50 or 60 Hz)

Solving the equation for number of turns gives you

N = Urms/(4,44*B*A*f)

When designing linear regulators for high current and a large adjustable voltage range one often encounters the following problem.

The unregulated voltage Uin must be somewhat higher than the largest output voltage Uout because of the voltage drops in the regulator. For Uout = 34 you will need Uin >= 36,5V. If you lower the output voltage the voltage drop in the regulator increases. For n output voltage of 1,5V you get a voltage drop of 35V. if you want 3,2 amps from your regulator it will have to dissipate 3,2*35 = 112w. This is quite a lot of heat (think of 2 60W incandescent light bulbs) requiring a thermal resistance rfom chip to ambient of 0,9 K/W which is a very large heat sink. Also a lm338 has a thermal resistance junction-case of 1,4 K/W which is to high.

There are a number of solutions to this problem:

- Build a regulator using discrete components
- Use pre-regulation with scr's on the AC side
- Use a switched regulator