Electronics > Projects, Designs, and Technical Stuff
Transformer Impedance and DC Resistance
joelrobertson:
Hi all! This is my first post here, so please let me know if there's anything I could do better :)
I'm working on a commercial project in my first job out of uni. It involves a transformer-isolated amplitude-shift-keyed 1 kHz signal output (7.5 Vpp unloaded). The existing circuitry uses a through-hole output transformer for isolation transformer (this one). I'd like to replace the transformer with a surface-mount part for manufacturing cost reasons, but I'm having trouble finding a replacement.
The current transformer specifies a 150 ohm primary impedance and a 150 ohm secondary impedance. I understand this is the result of a 1:1 winding ratio. However, I'm unsure if there are any other considerations about what this impedance really means.
The only transformers I can find that seem to be suitable replacements in surface-mount varieties have a primary/secondary impedance of 600 ohms, which I understand is common for telephony applications. This, of course, corresponds to a 1:1 winding ratio also. My concern is that the DC resistances are very different: 5.8 ohms (max) for the existing part, and more like 100 ohms for the surface mount part. The source impedance on the primary side of the transformer is currently 100 ohms, and the recommended load gives a total load impedance on the secondary of the transformer of about 115 ohms. This gives an output voltage of 4 V pp when loaded.
So I'm wondering the following things:
* What does the impedance of a transformer really mean, and is it fine to substitute a 150 ohm for a 600 ohm part?
* What impact will a change in DC resistance have on the operation of the circuit?
* Can anyone see I'm looking at this the wrong way, and help me with a path forward?
No one else at my workplace seems to know much more about all of this than I do, so I'm hoping someone here will have done something similar before! Thanks in advance.
T3sl4co1l:
Transformer impedance can be somewhat arbitrary. It may simply mean what impedance it was designed for, and tested at.
Transformers do have a characteristic impedance. This is the system impedance for which the transformer offers maximum bandwidth. The design impedance is usually close to this, for signal purposes at least (but often not for power applications, for a number of reasons).
So, system impedance?
RF systems are modeled as a signal chain. A signal is carried on a transmission line, with some characteristic impedance. Sources and loads have impedance as well, and we expect maximum power transfer when these impedances match (namely, they will have equal resistances).
This isn't too different from introductory circuit theory. At this point, it's more semantic, grouping pairs of wires into transmission lines (or assuming a global ground and treating a single wire as such).
Transmission lines connect to ports. A port is an arbitrary element with two terminals, such that the currents are always equal and opposite (i.e., no common mode current leaves the element) and the voltage difference across those terminals is all the port can "see".
Ideally, a two-winding transformer is a 2-port element.
An element with multiple ports, can share some amount of voltage or current between its ports. In this way, we can have an arbitrarily complex network, and analyze it in terms of signals entering and exiting it -- we can understand it empirically, without having to know its internal circuitry and solve for a transfer function.
In terms of introductory circuit analysis, we're breaking up what is potentially a very complex circuit, into few very simple circuits, isolated from each other, coupled through this arbitrary element, which we can describe with a few parameters. Now we can solve for input, reflected and transmitted power very easily!
These parameters, are usually given as s (scattering) parameters, but any orthogonal combination of impedances, conductances or ratios will do -- there are seven in common use, and formulas to convert between all of them. Note that you need one set of parameters for each frequency, which is why tables of values are usually provided.
So -- we can set up a system, with sources and loads and transmission lines of a given system impedance, and send signals back and forth and observe the transmission and reflection characteristics of a DUT plopped down in the middle.
If we use a transformer as the DUT, and vary the system impedance on both sides (in general, for a transformer of some ratio), we maximize bandwidth when the impedances match. Great, right?
So that's where that comes from.
Alright; what's an equivalent circuit for a transformer, anyway?
This is the 2nd order approximation. At the heart, an ideal Np:Ns transformer, which has self inductance which acts in parallel with the windings. Winding DC resistances Rp and Rs add in series. This sets the low frequency cutoff, F_L = DCR / (2*pi*Lp), where regardless of system impedance, the transformer is just too lossy to be worthwhile. (There's also a loss component in parallel with the inductance, to represent core loss if applicable.)
At high frequencies, leakage inductance becomes relevant. Leakage inductance is also added in series. This introduces a cutoff at Zs / (2*pi*LL), where Zs is system impedance.
When self-inductance is much greater than leakage, the leakages can be lumped to one side or the other, the other part being "pulled through" the transformer ratio, i.e., LLp(tot) = LLp + LLs * (Np/Ns)^2. Alternately, we can use the nonideal transformer model, with some Lp, Ls and M (mutual inductance), or k (coupling factor).
The capacitances complete the model: the winding self-capacitances represent the proximity of the windings to themselves, and the isolation capacitances, the windings to each other. These similarly introduce an HF cutoff, but combine with the inductance as a split pole, which is over- or under-damped depending on impedance. When Zs = sqrt(LLp / Cp), it's critically damped, and the cutoff is sharp, flat and as high as it gets.
The isolation capacitance acts in parallel with the self-capacitance, but also acts in the common mode, between windings -- worsening the isolation. (This is not included in the 2-port RF model; you'd typically use a partial 4-port model to measure this.) This may or may not be helpful in modeling a given application, but it is often important.
So, all that said; what do you need in a transformer? Guessing your bitrate is pretty low (less than 1kbps), so you don't need all that much bandwidth. Since we know bandwidth is determined by the ratios of resistance to inductance or capacitance, we know that, if a 600 ohm transformer is used at 150 ohm, it will have a 4x lower LF cutoff (nice, but maybe not helpful here?), 4x worse losses (due to DCR), 4x higher capacitive cutoff (yay?), and 4x lower leakage cutoff (boo). The LF cutoff is limited by the DCR-Lp cutoff of course.
You can guess, a transformer that's ~100 ohms DCR, isn't going to be very useful at 150 ohms. :) The HF cutoff may not be important (it tends to be in the multiple kHz for little audio transformers). It may be important for wave shape, because phase shift.
Another factor which isn't covered here -- saturation. This is the nonlinear effect of core inductance, namely, that it drops as peak current rises above the saturation current level. At a given frequency, this limits the terminal voltage on the transformer; or say for a square pulse, it limits the area (height * width). Saturation scales with part size, so it could well be that the SMT transformer can't handle the flux you need to put into it (i.e., 7.5Vpp / (8*pi*1kHz) = ~600 µVs).
Have you considered a custom part? Worth quoting at least, see if it can end up cheaper than your current solution.
Alternately, a vastly more complicated solution may end up cheaper and smaller, despite itself! Transformers can be emulated with op-amps, within their common-mode voltage range that is. If more range (or less power consumption) is desired, an isolated amplifier can be made with optos (although, making one with consistent and stable gain is harder). Perhaps the signal itself doesn't need to be isolated, only power and data -- use a DC-DC converter and digital isolator, and run the modulator directly on the line. (Industrial RS-485 links are normally made this way.)
Tim
iMo:
FYI: my 600:600ohm trough-hole transformer is 56:66ohm wiring resistance. It works fine at 1kHz.
I would take the smd part and adjust the other components around.
The Electrician:
joelrobertson, you say that you have 4 V pp at the load. Do you have a specification as to the minimum acceptable output level?
Do you have a low frequency impedance analyzer available?
You should measure the insertion loss of your existing transformer. You probably should try to find a replacement transformer with an insertion loss not much worse than what you have.
What are the dimensions of your existing transformer, and how much smaller do you need your replacement to be?
The Electrician:
Pico Electronics has a broad line of surface mount transformers.
Here's an example page showing some 100 milliwatt units: https://picoelectronics.com/node/13352
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