Electronic – What does the 25 Ω termination resistor do in a 0/180-degree hybrid power splitter

balunRFtransformer

The circuit in the picture is a 3dB power splitter using a 0/180-degree hybrid transformer/balun.

A 180-degree hybrid transformer

I see the impedance ratio of the transformer is \$ (14/10) ^2 \approx 2 \$, thus the equivalent output impedance of a 50 Ω input is 100 Ω into a differential load, or 50 Ω into two single-ended load, which makes sense.

But I don't understand the function of the 25 Ω termination resistor at the center-tap. I understand that the center-tap of a transformer is often connected to ground to establish a common ground reference, but what does the resistor do? How do I calculate the effect of a 25 Ω resistor to the impedance at each port of this hybrid?

Best Answer

The 25 Ω resistor doesn't do anything when the hybrid is acting as a power splitter, in this case, the input is Port A and the outputs are Port B and Port C. The center-tap can be terminated with any impedance, shorted to ground, or kept as an open-circuit. Similarly, when the circuit is being driven as a power combiner with a pair of ideal 0-degree and 180-degree differential signals at Port B and Port C, 100% of the power is delivered to Port A with matched impedance, again, the resistor is not doing anything.

The resistor provides termination when (1) the hybrid is used as a power combiner, and (2) Port B and Port C are being driven by an in-phase common mode signal. When the power combiner is driven by two in-phase inputs, no power is delivered to Port A because common-mode signal is suppressed and goes to Port D, the center-tap. Because the two 50 Ω signal sources are in-phase, they are effectively identical voltage sources in parallel, two 50 Ω parallel resistors give the output impedance of 25 Ω - here, the 25 Ω resistor at the center-tap acts as a dummy load and provides impedance termination for any common-mode signal/noise presented in the differential input, ensuring power is never reflected.

Source of the circuit is Practical Radio-frequency Handbook by Ian Hickman.

A hybrid can divide the input signal power between two outputs with negligible loss, each output being 3 dB down on the input. The basic hybrid circuit is shown in Figure 4.2a. If a signal is applied at port A, it will be divided equally between ports B and C whilst no power is delivered to port D (which could therefore be loaded with any termination from a short to an open circuit) as can be seen from the symmetry of the circuit, given that ports B and C are both terminated in 50 Ω. The outputs at ports B and C are in antiphase and the arrangement is known as a 180° hybrid (port D is often terminated internally in 25 Ω and only ports A, B and C made available to the user).

The corollary is that if two identical signals of equal amplitude but 180° out of phase are applied to ports B and C, all of the available power is combined and delivered to port A, port D again being isolated. If, however, the two identical signals were in phase, the currents in the centre tapped winding would produce no net flux on the core, so that port A is isolated and all the power is delivered to port D. If this is terminated with a 25 Ω load, then since ports B and C each supply half of the power, each will ‘see’ a 50 Ω termination.

I was looking for a suitable implementation for power splitter and first saw this picture in the Power dividers and directional couplers from Wikipedia with no explanation given, and I kept thinking the circuit as a power splitter and missed the explanation in the book.

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