The electrical length is unstated so it's impossible to say how well it will isolate.
There's a graph including "phase imbalance"
http://www.coilcraft.com/misc/ja4220imb.html but what does "imbalance" mean with regard to a single ended output? It's nonsense. And it's clearly not being used against a differential source or load, because an inductance of 15uH should be pulling significant phase shift down at 0.1MHz.
The bandwidth of a transformer is maximized when it is used at its characteristic impedance. The low frequency equivalent model uses leakage inductance and parasitic capacitance to represent the high frequency cutoff (which is really the lowest (1/4 lambda) mode of a transmission line transformer). The ratio sqrt(LL / Cp) has units of impedance, which is the impedance that should be matched to (for a given winding, using the respective winding-referred values, of course). The cutoff frequency of course is 1 / (2*pi*sqrt(LL * Cp)).
A proper TLT has bandwidth above this cutoff frequency, but how much depends on the design. A Guanella design is an autoformer, and has unlimited bandwidth, or rather, bandwidth not limited by its electrical length (LF cutoff is given by how much inductance the TLs have, and HF cutoff is limited by how precise the matching is at junction points, and the TEM cutoff of the TLs used).
Unfortunately, you can't get DC isolation with that type of transformer (only high common mode AC impedance).
So, isolating transformers must incur at least one TL delay, which sets the cutoff frequency.
If you have some of these transformers, you can take apart and measure the winding length. This is essentially the transmission line length, give or take speed of light and velocity factor.
Mind the common mode response. At low frequencies, common mode simply looks like a capacitance (the interwinding capacitance), but comparable to the cutoff frequency, it becomes complex, and mismatched to the ports. Consider an equal (say, positive-going) current step, applied to each terminal of the ("floating") primary. This current goes into each end of the transmission line, dropping a voltage I * Zo across each TL port (so the primary terminals both rise in voltage), and carrying the same current into the output terminals. Thus, the output voltage (unbalanced) will rise by Zo * I, while the other (ground) terminal remains constant (or rather, is our reference, so by definition, well, yeah). The primary nodes are thus at 2*Zo*I (on the 'top' end of the transformer) and Zo*I (on the grounded side).
After one TL delay, the voltage from each end of the winding has propagated to the other, and the primary CM voltages jump up by Zo*I. Superposition continues, and after a delay, the primary terminals will be at (N+1)*Zo*I and N*Zo*I, respectively.
As time goes on, the ratio (N+1) / N approaches 1, which means the primary terminal voltages are getting closer together, in a relative sense. Of course if we aren't applying a constant current step, but rather use a controlled impedance source, the current will gradually taper off, and CMRR at low frequencies will be inversely proportional to frequency, so, arbitrarily high for arbitrarily low frequencies. But we're concerned with CMRR at high frequencies, so this isn't much help, except to say we need a very short TL indeed, in fact len(TL) / lambda(Fmax) ~= CMRR.
Following the isolation with a CMC does improve the above situation, because the voltage imbalance was induced by current flow, and if we prevent current from flowing into the output node, we improve CMRR by exactly that much.
Indeed, it would seem necessary that, on applying a step input, the voltage should first be dropped entirely across the CMC, then as time goes on, it "sloshes" into the isolation barrier instead. Thus we have an LC resonant circuit, formed by isolation capacitance and CMC inductance, and we should want this to be at a suitable cutoff frequency (so that the attenuation near Fmax gives us, basically, the actual CMRR we should expect), and well damped (so we don't get a valley in CMRR at the cutoff frequency).
If the signals being sensed are nice and low impedance (like a high side gate driver, which I'm guessing is the underlying motivation here?), then the CM impedance will be dominated by the series resistors (which act in parallel, so ~112.5 ohms). (Oh, and obviously, any mismatch in source impedances will directly translate to loss of CMRR.) For, say, 40dB CMRR into 50 ohms, we need 5kohms at Fmax. Which is kind of unbelievable at >> 100MHz, but if it's a distributed CM filtering action (in which case the 5kohm figure is a trans- one, i.e. Vout / Iin), that works out. If Fmax = 1GHz (it's probably higher for the transformer in question), that implies L = 0.8uH. And then, whatever Cp is, let's say 10pF [1], comes to a resonant impedance of 282 ohms and a resonant (common mode cutoff) frequency of 56MHz (so that, with a slope of -40dB/dec, we expect < -40dB CM attenuation by > 500MHz, as intended).
The resonant impedance being higher, means we have a series resonant circuit, with ESR (the series 112.5 ohms), L (0.8uH) and C (10pF), with a Q factor of (282/112.5) = 2.5. This will lead to a modest peak near 56MHz, however, most CMCs will be lossy in that frequency range, so additional damping probably isn't required. (For EMI or power supply filtering applications, where you might not get that loss effect, you should add an R+C in parallel with the C, or an R || L in series with the L.)
So, with a suitable value common mode choke (one which has a quite high impedance at high frequencies, or an equivalent filtering value), maintaining high CMRR at and beyond 200MHz should be achievable.
[1] If it's a transmission line that exists (known Zo, delay, velocity factor), then you know its LF equivalent (L and C) from fundamental constants. The impedance of free space is 377 ohms; real TLs typically have an impedance lower than this, due to geometry shaping the fields, increasing the capacitivity and reducing the inductivity. Additionally, any loading in the TL's volume (dielectric permittivity or magnetic permeability) changes Zo by the ratio of sqrt(mu_r / e_r). Most TLs are partially or fully filled with a dielectric around e_r = 2 to 5, and few use any magnetic material. Since velocity factor is c_o = c / sqrt(e_r * mu_r), and making the assumption that mu_r = 1, we can turn c_o into e_r, then into the ratio on Zo.
Bringing all these factors together: traditional 50 ohm coax (c_0 = 67% of c, so e_r = sqrt(c / c_0) = 2.23) with 1/2 wavelength corresponding to 1GHz (t = 0.5ns) has a physical length of 10cm. The un-dielectric-filled impedance is 50 / 0.67 = 75 ohms, so the capacitivity is:
C = (377/50) * (c / c_0)^2 * e_0
and inductivity
L = mu_0 / (377/50)
Or, C = 98.6pF/m, and L = 0.25 uH/m. And therefore, for a 0.1m length, L_eq = 25nH and C_eq = 9.9pF. (Note that 50 ohm = sqrt(25n/9.9p) is still true, so the math checks out.)
Tim
Home
Help
Search
Login
Register
