# Increase the bandwidth of an RF transformer

I'm building an RF transformer for a custom application (~500KHz to ~5MHz) and my band edges are bumping into the roll off region on the transformer. It's not the end of the world, but it would be nice to increase the bandwidth to avoid this. Is there any practical way to increase the bandwidth of a transformer?

Here's another image showing Vin and Vout.

Edit: There is a very dumb mistake here - I was looking at 500Hz on the plot instead of 500KHz. The overall question regarding transformer bandwidth remains the same though.

• Transformers can be used from sub-audio up to many tens and hundreds of MHz. What is causing the problem with yours? Commented Mar 17, 2023 at 13:36
• You haven't yet modelled the self capacitance or leakage inductance, so don't expect a real one to be as good as your simulation. Commented Mar 17, 2023 at 13:37
• @Andyaka, Yes, but not with a single transformer design though, right? My understanding is that the inductance of the windings that drives the frequency limitations. Commented Mar 17, 2023 at 13:50
• Yes, higher inductance and worse coupling can create problems but, your picture doesn't give a clear representation of what is going on with your transformer. For instance, where is Cp, Rp and Rs? What are they meant to do? Why is Lt 57 uH? What makes you sure you can get a coupling of 0.99 (ambitious)? And, your AC response does not match your circuit at low frequencies. Commented Mar 17, 2023 at 14:00
• Please provide a winding schedule or diagram for your transformer. Commented Mar 17, 2023 at 14:20

Start with a 2nd-order equivalent circuit of a transformer:

T1 is a pair of coupled inductors with $$\k = 1\$$, equal to the magnetizing inductance of each winding, and giving the desired turns ratio. LLp and LLs (for $$\k \rightarrow 1\$$, we might pull both together into a single-side-referred equivalent) are the leakage inductance, the flux that isn't shared by both windings. (These are implemented by the $$\k\$$ factor in SPICE.) Rp and Rs are the winding DC resistances. Together, these give the 1st order transformer model. We can add capacitance Cp, Cs, Ciso1 and Ciso2 to approximate the winding self- and coupled (isolation) capacitances, which manifests because a winding is made of wire, and any two conductors (or portions of one, should it happen to approach itself as in a coil) have capacitance between them.

We can transform this into a common-ground equivalent, ignoring the isolation characteristics for now. In that case, we have:

simulate this circuit – Schematic created using CircuitLab

We can further transform this into LF and HF equivalent circuits, assuming the cutoff points will be far apart (which will be the case in a wideband transformer):

simulate this circuit

Note that these are all primary-referred values; the load resistance and secondary capacitance have been "pulled through" the transformer as their reflected equivalents, and the output voltage is simply assumed to be N2/N1 times the modeled value.

Note also I've replaced Rs and Rp (winding resistances) with source and load resistances, in the assumption that they dominate over the winding resistance (which is a safe assumption as we generally want to design transformers with low insertion loss).

Now we can analyze the two cases.

## Low frequency cutoff

Simply the single-pole L/R highpass filter formed by the Thevenin equivalent source and load resistances, and magnetizing inductance. (We can increase accuracy slightly by adding LL with LM, but by the above definitions, LL ≪ LM, so we don't care very much.)

To solve for the -3dB point, we enumerate three convenient conditions:

1. Source and load termination

Take the Thevenin equivalent, RS || RL. (I'm not writing the turns ratio here; replace RL with the reflected value.) This combines with the magnetizing inductance to give the cutoff: $$\F_c = \frac{R_{eq}}{2 \pi L_M}\$$. For the special case RS = RL, the cutoff is half what we would expect given the magnitude impedance alone (i.e., we get 25 ohms in a doubly-terminated 50-ohm system).

2. Source termination (load open-circuit)

Take just RS as the equivalent resistance above. This condition sometimes manifests in amplifier coupling applications, where the input impedance might be high (as in a MOSFET gate or vacuum tube grid). The source resistance defines the bandwidth in this case.

3. Load termination (source short-circuit)

RS has a minimum value equal to primary winding DCR; of course we cannot achieve the ideal case of zero, but we can approach it using low impedances (such as an op-amp, or with a switching circuit). This does maximize bandwidth, but the low impedance also quickly loses track of magnetizing current and therefore easily saturates the transformer. (Saturation is a nonlinear characteristic which can be covered by a separate question.)

We might also consider the current-mode case (load shorted), in which case the reciprocal of case 3 applies (swap load and source). Likewise for a CCS ("open circuit") source, such as an open-collector (or drain, or pentode plate) amplifier, case 2 applies (swap load and source).

## High Frequency Cutoff

We can use the same rubric as above, but notice the network is much more complicated: 3rd order as shown. It might be that one or the other capacitor is negligible (i.e. few turns). It might also be that we do not have three full degrees of freedom, so cannot make an optimal (e.g. named characteristic, e.g. Butterworth etc.) 3rd-order filter here, or that higher-order (inter-layer, inter-turn, or transmission line) effects begin to dominate and our analysis falls apart.

If nothing else, we can still calculate a lower cutoff in terms of the minimum given by any one element (Cp, LL or Cs) with respect to the system impedance, or respective (source or load) impedance if one or the other is short or open. That is, up to this frequency, we can expect flat frequency response; beyond, there may be peaks or dips that are not easy to predict from these (measured or assumed) values, and there may be more (flat within some tolerance) bandwidth available but we cannot predict it so easily beyond this point.

We might also work backwards from some Cp, LL and Cs, and find the bandwidth and characteristic impedance. Again, three parameters do not define these two in a unique manner, but in the event one parameter can be excluded, or all three work together to give a useful filter prototype, we will have a meaningful relation.

The relations are:

$$\Z_0 = \sqrt{\frac{L}{C}}\$$

$$\F_c = \frac{1}{2 \pi \sqrt{L C}}\$$

or $$\F_c = \frac{R}{2 \pi L}\$$ or $$\F_c = \frac{1}{2 \pi R C}\$$ if an L or C dominates, respectively.

Applying the above termination conditions, we have:

1. Source and load termination

Potentially all three elements participate. I'm not going to write out a cubic transfer function, nor the analysis of solving for an analytical type -- but suffice it to say, try all the above cases, and you should at least get something close.

LL dominant: Cp and Cs are small enough to ignore. Note that RS and RL act in series, so $$\F_c = \frac{R_S + R_L}{2 \pi L_L}\$$.

Cp or Cs dominant: $$\F_c = \frac{1}{2 \pi R_S C_P}\$$, or RL and CS.

We have maximum bandwidth when system and characteristic impedances match, i.e. $$\R_S = R_L = Z_0\$$, which will have

2. Source termination (load open-circuit)

LL-Cs acts as a resonant stub upon the input node. This causes a complementary dip and then peak in the voltage gain, where they act to resonate against or with Cp. Bandwidth is maximized when the peak and valley are kept shallow (meaning, $$\\sqrt{\frac{L_L}{C_S}} \sim R_S\$$; they will differ by a factor depending on the relative value of Cp, which also limits how well damped this response can be).

3. Load termination (source short-circuit)

We can ignore Cp (it's shorted out!), so Zo and Fc are given by LL and Cs alone. Of course if we're using Zo mismatched to RL, we also have to consider if one or the other element dominates, and may want to address the gain peak (say by adding a snubber network).

As for what dimensions these parameters correspond to, in a real transformer design: it seems that would be well scoped to a separate question.

As a short hint: for transmission line transformer (TLT) type designs, the upper cutoff is simply the electrical length of the windings. So, for say 30MHz bandwidth, less than 10m of wire must be used, and preferably less than 2.5m (maximum length of either winding). The lower cutoff is simply turns and inductivity of the core ($$\L_M = N^2 A_L\$$).

• This is excellent, thank you. Commented Mar 17, 2023 at 17:19

I'm building an RF transformer for a custom application (~500KHz to ~5MHz) and my band edges are bumping into the roll off region on the transformer. It's not the end of the world, but it would be nice to increase the bandwidth to avoid this. Is there any practical way to increase the bandwidth of a transformer?

$$\color{red}{\boxed{\text{ Your analysis is flawed }}}$$

If you plot $$\V_{OUT}\$$ you'll see that you have a 3 dB bandwidth from 75 kHz to about 15 MHz and not the 500 kHz to 5 MHz that you previously believed. So, to increase the bandwidth, all you need to do is plot the correct signal and not confuse yourself with what you originally plotted: -

The image was taken from OP's final image and was doctored to make it easy to read and focusses on the only relevant waveform (Vout). Points to note: -

• The pass-band (1 MHz in the middle) is at a level of -0 dBV
• If it were a 1:1 transformer (rather than 1:2) the required load would be 50 Ω and the output would be at a level of -6 dBV
• But, because it's a 2:1 step-up transformer, the output is 6 dB higher at 0 dbV into 200 Ω (matched)

Hopefully this shows the importance of plotting the right signal; originally the OP thought that Vout/Vin was the correct way of analysing it but, because of the very low input impedance at low frequencies and labelling the incorrect node, the result gave a false impression of a miraculous low frequency response whilst giving a pessimistic view above 5 MHz.