RF Transformer - Page 1
Forum section: electronics / beginners

author: NivagSwerdna
author_rank: Super Contributor
timestamp: May 26, 2016, 10:19:56 pm
content: I bought some of these...eBay auction: #251401324712Minicircuits MCL T1-1T RF Transformer 200MHz 250mW 1:1 Centre-tappedI have never used anything like this before... I assumed it washttps://www.minicircuits.com/pdfs/T1-1T-KK81.pdfNow the datasheet says the impedance is 50 Ohms...But when I measure between Pins 1 and 3 or 4 and 6 I get 0.5 Ohm, almost a dead short....Am I missing the difference between resistance to impedance? Or have I actually got something that isn't 50 Ohms.Thanks in advance

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 27, 2016, 04:20:24 am
content: A couple definitions:1. System ImpedanceThe source generator can be modeled as an ideal voltage source in series with 50 ohms (Thevenin equivalent). There is also a 50 ohm termination or load, connected to the other side of the transformer.Almost exclusively, RF measurements are done with this configuration. It's easiest if everything is measured by the same standard; 50 ohms happens to be the accepted standard for pretty much all electronic test gear.2. Insertion LossIf you measure a 50 ohm voltage source, with no load (open circuit), you will read 2*Vo, and whatever frequency or waveform it's producing (say if you measure with a high impedance oscilloscope). When the source is connected to a termination resistor, you get a voltage divider. Two 50 ohm resistors are in series, current flows, and the voltage on the middle is Vo. If the load were a short circuit, I = 2*Vo / (50 ohm) would flow instead, and no voltage would be on the load (V = I*R, but R = 0).If you break the circuit between source and termination, and connect the transformer across the open terminals, you have inserted it in the system.Now repeat the voltage measurement. If the load has exactly Vo on it, there's no insertion loss (0dB). If it has, say, Vo/2, that's half voltage, or quarter power: 6dB insertion loss.This is shown in the insertion loss figure, as a function of frequency. Clearly, the transformer is not simply a hunk of wire, or a resistor. Something's going on there!3. What is a transformer?If it's wired up in the straightforward way (it's noteworthy they don't actually label what pins are used in the test), then one can assume the transformer presents a parallel inductance, which acts to short out the source at low frequencies.The source and load act in parallel, so the resistance is 25 ohms as seen by the transformer. (Has to do with Thevenin again.) Now, to figure out how much the inductance shorts out the source, we have to calculate the impedance as a complex number. The current through the transformer's inductance is not in phase with the voltage, so we can't ignore this. (...) Having figured this out, we can use a quick, simple rule: when X_L = R, the attenuation is 3dB. Looking at the table/figure, 3dB insertion loss falls right around 50kHz (= 0.05MHz, I'm just mentally shifting the decimal places). Now we can apply the formula,L = (Rs / 2) / (2*pi*F)Rs being the system impedance, 50 ohms, and F, the frequency. Apparently it's around 80uH.When you measure 0.5 ohms, you're making the same sort of measurement, except:- The meter is solving for resistance, for you.- Also, it most likely doesn't measure things as if it's a 50 ohm system in the first place...- You're measuring at DC, a frequency of zero. We should expect the insertion loss to be very high indeed at DC!- Of course, you're also not measuring the insertion loss (a transmission characteristic: i.e., signal goes in one side and comes out the other) -- you're measuring the return loss, in a sense (how much power is reflected back towards the source).If the transformer were an ideal inductor, we expect its reactance to go to zero at DC, because the reactance is X_L = 2*pi*F*L.But a real transformer is constructed with wire; judging by the 30mA DC current limit on this part, it's probably quite fine wire (~0.1mm dia?). So we should expect it to have a modest resistance as well, maybe 0.5 ohms?You need to account for contact and probe resistance, when measuring such small values (often, a meter reads ~0.3 ohms with the probes touching each other), but something around an ohm is quite reasonable.Indeed, since this resistance is connected in series with the source (it's *in the wire itself*!), we wouldn't expect much change in the insertion loss until it's comparable to the source resistance: more than 2.5 ohms, perhaps (which would give a 10% reduction in output voltage, or about 1dB insertion loss).In fact, insertion loss is minimum 0.36dB at 1MHz, where:X_L = 503 ohms is >20x the Thevenin system resistance (which suggests 0.44dB insertion loss, actually...)If the loss is only due to resistance, then the equivalent resistance is about 1.015 ohms. (Which sounds pretty reasonable; we also expect the resistance to be higher than at DC, because of complicated effects which increase the wire resistance. This is probably why the insertion loss rises slightly up to 100MHz.)We can even model the high frequency cutoff (insertion loss rising sharply above 100MHz), but this gets more difficult, because there are more complicated factors at work: leakage inductance, parasitic and interwinding capacitance, winding order (where the wires are placed, relative to each other, in the transformer) and more. The data isn't complete enough to get a good idea about these parameters, anyway.4. SummaryThe impedance of a transformer is not something you measure; it's a suggested, or "best at" impedance, for everything *around* the transformer! Over the frequency range where the transformer is intended to, well, transform, its impedance should be quite large, so as to avoid loading down the source. This loading effect is basically the inductance of the winding, and causes the lower frequency limit.Tim

author: NivagSwerdna
author_rank: Super Contributor
timestamp: May 27, 2016, 06:54:24 pm
content: Tim, Thank you so much!That exactly explains what I am seeing. I can proceed with my design now.

author: G0HZU
author_rank: Super Contributor
timestamp: May 27, 2016, 10:25:00 pm
content: One way to model a transformer across a wide bandwidth is to use s parameters taken with an n port VNA. If you have access to an RF simulator that can cope with 4 port data you can use the MCcts 4 port s parameter data to simulate the transformer quite well. Note that the centre tap on the secondary (5th port?) is grounded in this case. So you have to be careful how you use their model because it expects a grounded centre tap on the secondary. Otherwise you would need a 5 port model to model the transformer more completely.