LTSpice and transformer series resistance

[ssbothwell KJ6RSG]

modeling transformers in LTSpice. this is a simple question but i suppose it might spiral into something larger.

when modeling mosfet push-pull amplifiers in LTSpice the fet's drains, and drain power supplies (V3 and V5 in the screenshot), are tied together through the output transformer which re-combines the two out of phase signals.

unless i add a series resistance to inductor L8, LTSpice reports an error about a voltage loop between power supplies V3, V5, and L8. i usually just add a 1ohm series resistance to L8 but i have no idea if this is the correct way to resolve the problem.

how should i go about setting the transformer's series resistance? can i just measure the resistance of an actual transformer i intend on using in a physical circuit, or is there some calculation i can do based on core size and number of turns on the core?

i'm asking this because i noticed that setting higher L8 series resistance gives pretty dramatic differences in output power in ltspice.

[KF1Z]

Most wire charts give the resistance per 100 feet.

Easy math to determine the resistance between two points.

[W4NEQ]

When you right click on the inductor to set properties, you should assign a realistic series resistance - this is essentially determining the Q of the coil.  

Depending upon many factors, for non-air wound coils, typical Q's range from 50 to 200.   So if you're at 3.9 mHz, 50 uH has about 1000 ohms reactance.  If the Q is 100, that's 10 ohms effective series R.   Unless you're operating at a much lower frequency, 50 uH seems like a lot.

For a broadband transformer try Xl = 6-7 times stage output impedance.

The realistic series R (at RF) will usually be significantly greater than the DC resistance.  This is due to skin effect, parasitic capacitance, core loss, etc.

Chris

Edited for clarity.

[ssbothwell KJ6RSG]

thanks chris. what are the equations used to determine reactance and Q  for a given inductor?

[W4NEQ]

Inductive reactance is simply 2PI * F (in Hertz) * L in Henries

or:

http://www.66pacific.com/calculators/xl_calc.aspx

Inductor Q = reactance / resistance, but this doesn't help much.

Due to the wide range of variables when using ferrite cores, Q is really hard to calculate, and it is easy to overestimate.  There are Q meters and test gear to measure it, but these are generally complex or costly.  The easiest way is to get the data from the manufacturer, or in the case of winding your own torroidial coils, the core manufacturer.  Amidon used to have Q curves for their Iron-Powder cores published, but I don't recall seeing them for ferrite cores, which are what's typically used for broadband transformers.   

Ferrites #61 and #43 are commonly used for baluns and broadband transformers.  At least for #61, for 2-7 MHz, if you used a Q of 150 you wouldn't be too far off.  It might be better, maybe 200.   

A silver-plated, edge-wound 20 amp air-core coil can deliver a Q of 400, but most air-core coils would probably be in the 100-250 range.

Often, the manufacturers will simply note "best Q will be in the x to x frequency range. Fair-Rite does this.

Others on the forum may have more specific Q data for broadband transformers.

Chris

[ssbothwell KJ6RSG]

thanks chris. i use #43 ferrite, would a Q of 150 be a safe estimate for this material as well?

[W4NEQ]

I don't have much first-hand experience with 43, other than it having a much higher permeability - requiring less wire.  When I've measured coils wound on it, they have significantly lower Q than #61, as there seems to be more of a resistive component in the mix.  Having said that, it's very popular for broadband cores, especially the two-hole binocular cores.  Perhaps this is due to the increased coupling due to higher permeability ...  I dunno.

There are some magnetics specialists on this forum - I would be interested in their take on this ...

[Rob K2CU]

Did you try using one 12V power supply? That will probably solve the initial problem.

Unless set otherwise, the voltage sources as well as the inductor have zero ohms resistance. LTspice is probably gagging on that situation. if you must use two supplies, try setting the sources' internal resistance to 0.1 Ohm. Without some resistance, you run the risk of a divide by zero problem.

[W1RKW]

Solomon,
have you seen this article from Linear?
http://cds.linear.com/docs/LT%20Journal/LTMag-V16N3-23-LTspice_Transformers-MikeEngelhardt.pdf

I have NI Multisim here on one of my computers.  It works based on SPICE models and has a handful of air transformers. I don't know if the models would help here but one model is of a generic air core transformer.  Maybe you can glean some info from here.


##################  Model Data Report  ##################

============= SPICE Model =================
.SUBCKT nlt_pq_4_56 2 1 4 3
    R0 2 5 66
    L0 5 6 146m
    A0 (6 1) (7 0) nlt_pq_4_56_lcpl_0
    A1 (7 nlt_pq_4_56_mcore
    A2 (9 3) (8 0) nlt_pq_4_56_lcpl_1
    L1 9 10 34.66m
    R1 10 4 14.88
.MODEL nlt_pq_4_56_lcpl_0 lcouple(num_turns=1.622K)
.MODEL nlt_pq_4_56_mcore core(
+   H_array=[0 13.89 51.58 222.22]
+   B_array=[0 1 1.5 1.7]
+   area=130u length=76.15m
+   input_domain=10m fraction=TRUE mode=1)
.MODEL nlt_pq_4_56_lcpl_1 lcouple(num_turns=790)
.ENDS
============= Model template =================
x%p %t1 %t2 %t3 %t4 %m

[W4AMV]

I don't have much first-hand experience with 43, other than it having a much higher permeability - requiring less wire.  When I've measured coils wound on it, they have significantly lower Q than #61, as there seems to be more of a resistive component in the mix.  Having said that, it's very popular for broadband cores, especially the two-hole binocular cores.  Perhaps this is due to the increased coupling due to higher permeability ...  I dunno.

There are some magnetics specialists on this forum - I would be interested in their take on this ...

The Q of ferrites, particularly the lower permeability units is published directly from some manufacturers. Higher permeability is usually not as it is lower. However, the Q can still be found from the published values of the loss factor (reference the LOSS TANGENT). If you take the product of the loss factor and the initial permeability and find its reciprocal, the Q will result. For material 43 at 850 perm, we have Q values of 6-10 at 100 kHz-1 MHz. The material 61 at 125 perm, provides a Q factor of 130-350 at 1-10 MHz and is obviously higher. The equivalent series R values are obtained as mentioned earlier and 10 ohm to 1 ohm, higher perm to lower perm is a typical range. See below a chart of a range of materials with their associated perm and loss tangent as well as some scribble notes. This curve would provide you the info to find Q and therfore Rs for a range of materials. Chart from Fair-Rite. Hope this helps.

[W4NEQ]

The choice for #61 seems quite intuitive, but there are many optimized amplifier designs where the broadband transformers were #43, or other apparently low-Q core.  This would suggest that minimizing copper losses (with fewer windings) is more important than a Q of ten where one would always, at best, have a 10% resistive core loss.

Is it because at flux levels approaching saturation the cores behave differently? 

Chris

[W4AMV]

Chris, the approach to broadband transformer design using Ferrite is quite different than simply designing a high Q inductor using the same. Your point is well taken and simply looking at the core loss is not sufficient. There is the matter of the transformer coupled wires operating as such (an ideal transformer) at low frequency and then as a transmission line at higher frequency where the core permeability drops significantly. You mentioned earlier in a post the need to get the reactance of the transformer large compared to the resistive impedance level, say in a 4:1 case, 50 ohm to 200 ohm matching system. I need to achieve at low frequency a reactance above 200 ohms so as to not impact the transmission loss. At the same time at higher frequency I need to keep the transmission line length of the same transformer which now operates as a coupled transmission line, short in physical length (<< 1/4 wave) and Ruthroff in an old paper addresses that issue. Thus the higher perm material is engaged. Finally, controlling the capacitive coupling between windings enters into the picture and the higher frequency roll off of the transformer now operating as a transmission line complicates matters! This whole topic is treated well in several textbooks and deserves a dedicated forum discussion. 

Alan

[W4NEQ]

Well, alright.
Assume, we are operating the broadband transformer at the low end of its frequency range.  OR assume we are using a binocular core type conventional RF transformer - where transmission line coupling effects are minimal.

Does a Q of 10 not apply to the inductance of the windings because the resistive load on the output cancels all but the (minimal) stray inductance?

[W4AMV]

Good question and here attached the explanation. In addition, for FUN, measured a 850 perm 4:1 transformer, consisting of 1 Turn and 2 Turn only on a 2-hole binocular core. Prior to the measurement, I measured the Q of the self-inductance of the 1 T and 2 T primary and as expected it is low, in the neighborhood of 2. The equivalent series R of each side is in the range of 24-88 ohms. The XMFR is terminated into 12 ohms and the transform to the 2T side is indeed ~ 48 ohm despite this low Q! However, the key here is that the equivalent resistance due to the permeability losses is a PARALLEL component and the model of the XMFR is attached below. The SERIES losses are uniquely associated with the copper winding losses and are quite small. So, the principal components of the XMFR which account for its operation and this "paradox" are the PARALLEL R obtained from the Q measurement or the loss factor and the inductance due to the permeability of the core. An additional series L is due to leakage inductance. So long as the equivalent Rp obtained from Rp=Rs(Q^2+1) is large compared to the terminations, the LOW WORKING Q of the self inductance of each side of the XMFR is not a limitation. Again, see attached .jpg drawing. NOTE: Q in the prior equation is Xs/Rs where Xs is the self-inductance of the XMFR primary or secondary winding.

I might add, circulating back to the original post-question, there are a pair of 2-R values required to model the simple BB XMFR. One small value, 1-2 ohm in series and one in parallel with the XMFR winding to handle the so-called magnetization loss. It is the magnetization loss R value that may be measured via a Q measurement or calculated from the loss factor  of the self inductance of each side of the XMFR and then transformed to its parallel equivalent as highlighted in the drawing. Its series value maybe as high as 20-100 ohms, possibly higher. However, its parallel equivalent value will usually be much larger than the desired transformed R value (hopefully) and hence neglected.

[W4NEQ]

Thank you.   I've always been weak in motors and magnetics, and this helps.  It does seem far from intuitive.

Chris