# Coil Design software?

1. Feb 27, 2014

### hobbs125

Hi everyone,

I have been designing wideband transformers for a while. My biggest problem is the time it takes.

So, I am looking for good software for coil design. I need something that can calculate from inputs the inductance, total number of turns, number of turns per layer, resistance, and leakage inductance of the transformer.

Anyone know of any affordable software that can do this.

There are a lot of good multi layer coil calculators online but I can't find any that do all of the above or any that can be downloaded.

2. Feb 28, 2014

### Baluncore

Wide band transformers come in many topologies and with many different core materials.
Can you be more specific about your geometry and material? How wide is wide band?

3. Feb 28, 2014

### hobbs125

Yeah,

I am using a ferrite E core material 3C90 Ferroxcube 55/28/21. I am trying to design a wideband xmfr which will give me a 1:3 step up ratio to provide 90V across a 100 ohm load at 5kHz.

4. Feb 28, 2014

5. Mar 1, 2014

### hobbs125

Wow, exactly what I was looking for.

Thank you!

6. Mar 1, 2014

### hobbs125

Ok, So now I have come across another problem.

I am trying to drive a 100 ohm load with 90V which requires the secondary coil to provide 900mA.

The problem I am having:
The ferrite core has an Ae of 353mm^2 and a Bsat of 300mT.
When I calculate the min turns to avoid saturation at 5kHz and 30V applied I get 56 turns.
The core has an AL of 6300, which gives me a primary L of 19.7mH.
19.7mH @ 5kHz has an impedance of 618 ohms, so with 30V applied Primary Imax will only be 48mA.

So, I already have a problem. The primary coil impedance is too high.

When I input what I need into the coil design software it states something like "Required core size not available".

I need the secondary coil impedance to match the load (100 ohms). Also, since the pulses are unidirectional the coil cannot have too much leakage.

What is my next step?
I thought about looking into gapped cores but them I'm afraid the gap will make the leakage inductance too high. Then my regulation will be bad and waveform distorted. Can anyone here give me any suggestions?

7. Mar 1, 2014

### Baluncore

That 48 mA is the magnetising current required when the secondary is unloaded, it is not real power as it is 90° phase shifted from the voltage. The in-phase, real current, will be greater.

“Required core size not available” suggests too much current for the coil design is causing saturation, so reduce the number of turns.

It would help if we knew your application.
Is this an audio power transformer ?
Is this a pulse or a sine wave ?
What is the pulse repetition rate ?
What is the frequency range, Fmin to Fmax ?

A 1:3 turns ratio = voltage step up ratio.
Output voltage is 90 V into a 100 ohm load. Output current is 900 mA.
Input voltage is 30 V, real input current is 2700 mA.

8. Mar 1, 2014

### hobbs125

The coil is switched using a single FET using a 5kHz 50% duty cycle unidirectional pulse. Frequency range is 1-5kHz.
I just need a coil with a 1:3 step up ratio that will give me enough current to get 90V pulses across the load that are the same length as the primary pulses.

I have been trying to design the coils as shown in my previous post. However, I must be doing something wrong. I get a very low voltage across the load (100 ohm resistor). I thought it was due to the secondary inductance being too high, limiting the max secondary current. Am I incorrect in my thinking? Does the secondary inductive reactance limit current? I thought it was considered the source impedance along with it's resistance?

Last edited: Mar 1, 2014
9. Mar 1, 2014

### Baluncore

One problem with square waves is that they need high order harmonics to keep them square. Transformers with magnetic core material do not like flat topped waveforms, they prefer sine waves. There is an alternative.

As you are dealing only with square waves, a more efficient solution would be to obtain a switching supply producing 90VDC, with a 1A capability. The load can then be driven using a single mosfet. An alternative might be to use a mosfet H-bridge to differentially drive the 100 ohm load from a 45VDC or a more common 48VDC supply.

If you are having trouble finding a transformer solution then I would suggest you first design a 1 watt transformer. When you have done that you can scale it up to a bigger core.

The inductance of the driven winding on a transformer decides the no-load magnetising current.
There is an in-phase magnetic coupling between the two windings that is in parallel with the inductance of the windings. Inductance does not limit the current. Current is limited by load and saturation.

When you specify a “1:3 step up ratio” you should also include the units since there are several different interpretations. Ratios do not usually need units, but here they are needed to prevent ambiguity. You know what you mean but others don't. Consider;
Is the Voltage stepped up and the Current stepped down ?
Is the Current stepped up and the Voltage stepped down ?
Or is it Impedance that is being stepped up by the square of the turns ratio ?

10. Mar 1, 2014

### hobbs125

Ok,

So it is a 30V primary 90V secondary.

I have been able to design coils which gave me good square waves. The problem has been getting good regulation. On my last design I minimized the turns ratio as well as the turns on each coil in an effort to minimize leakage inductance. However, after testing the coil I still had horrible regulation. I'm not sure but I think I can attribute it to the core material I was using, powdered iron. After reading more about it I realized it has a distributed air gap. So the leakage inductance turned out to be very high because of it, and the voltage across the secondary load was 500mV, instead of the 8V it should have been.

Anyways, I thought the inductance limited the current in some way. Thank you for clearing that up.

One other question-How does impedance matching work in this type of circuit?
I assumed you matched the secondary coil XL to the load.

Last edited: Mar 1, 2014
11. Mar 1, 2014

### Baluncore

An example;

Output voltage is 90 V, (into a 100 ohm load). Output current is 900 mA.
Therefore the output impedance is 90V / 0.9A = 100 ohm, obviously.
Output power is 90V * 0.9A = 81 watt.

The turns ratio is 1 : 3 so;
Input voltage is 30V, input current is 2.7A.
Therefore the input impedance is 30V / 2.7A = 11.11 ohm.
Input power is 30V * 2.7A = 81W (The conservation of energy is reassuring).

The impedance has been transformed by a factor of 100 / 11.11 = 9.00
Which is because the impedance ratio is the turns ratio squared.

Looking out of the transformer's secondary is seen a 100 ohm load.
Looking into the transformer's primary is seen an 11.11 ohm load.
When viewed through the “transformer” the output load impedance is “transformed” to appear as a different impedance.
An ideal transformer does not have an impedance. It has an impedance ratio.

12. Mar 2, 2014

### hobbs125

Ok, now I understand.

Thank you very much for your help. I appreciate it:)