# How does a telephone hybrid transformer work?

Suppose the following circuit:

I can see how an incoming signal (Kim's voice) wouldn't be coupled with the lower pair, but I can't see how an outgoing signal (Ron's voice) only goes through the lower pair.

When the signal from the upper pair reaches the hybrid then one part of it goes to the 2 wire telephone line (seeing a line impedance of Z1) and the other part (right inner inductors) does the same except it will 180° out of phase when it reaches the lower pair (because the impedance Z0 perfectly matches Z1). As such Kim's voice won't travel back.

On the other hand, I don't understand how a signal coming from the left phone would only travel through the lower pair.

• Kim's voice actually would be coupled into the lower pair. The trick lies in the circuit around Z0, and the dots at particular points on the coils. The dots tell you which way the coil is wound. If the dots on the two coils of a transformer are together then the output is in phase with the input. If they are at opposite ends, then the output is 180 degrees out of phase. I'll have to let someone with a better understanding explain the hybrid transformer in detial, though.
– JRE
Commented Sep 21, 2015 at 12:58

On the other hand, I don't understand how a signal coming from the left phone would only travel through the lower pair.

I've concluded that some of the signal from the telephone (Ron's voice) does reach the transmit amplifier (the source of Kim's voice) BUT this is just not that a big deal because, the important thing is that Kim's voice (port 2) does not get transmitted to port 3 (The receive path for Ron's voice).

Here is a better diagram showing the impedances not shown above: -

Also here is what wiki says about hybrid transformers: -

Double transformer hybrid

When both the 2-wire and the 4-wire circuits must be balanced, double transformer hybrids are used, as shown at right. Signal into port W splits between X and Z, but due to reversed connection to the windings, cancel at port Y. Signal into port X goes to W and Y. But due to reversed connection to ports W and Y, Z gets no signal. Thus the pairs, W & Y, X & Z, are conjugates.

Note 1 - Port Y is the balancing impedance shown as Z$$\_B\$$ in the top diagram in my answer.

Note 2 - Port W is the telephone line

Note 3 - Ports X or Z are interchangeable as the transmit output or the receiver input.

• Thanks. I guess that this sets it then, since it clearly says that a signal into W (1 in my figure) does appear in both X and Z (2 and 3 in my figure), as you said.
– 4nt
Commented Sep 21, 2015 at 14:33
• @Ant Yes I'm convinced now - I did the theory so long ago I struggled and was pleased to find the wiki page to confirm this. Commented Sep 21, 2015 at 14:34

This is a really old question and the accepted answer was provided almost a decade ago but I feel this is a really cool question and warrants a closer inspection. I'll reuse the amazing graph from Andy's answer but provide a somewhat modified answer that I believe expands somewhat on that topic.

First, terminology:

• RX side = "signal from far end" carrying Kim's voice
• TX side = "signal to far end" carrying Ryan's voice
• phone circuit = local phone circuit with Ryan on the phone listening to Kim and speaking to her

Second, the basic function: as the OP pointed out themselves, the voice originating from the RX side gets cancelled in the TX side because the impedance Z_B is very closely matched to the impedance of the phone circuit itself and therefore, almost identical opposing current with Kim's voice appears on T2 coils cancelling Kim's voice coming from the phone circuit.

Now, the interesting part: the voice originating from the phone circuit itself (Ryan's voice).

The key here is that from the point of view of the phone circuit (Ryan) the circuit is NOT symmetric:

• the RX side has higher impedance than the TX side
• there is still balancing network at play

Therefore,

1. the current induced by the Ryan's voice in the T2 windings on the TX side will have to be higher than any current induced by the Ryan's voice in the T1 windings on the RX side because the impedance of the TX side of the circuit is lower.

2. however, the current induced by the Ryan's voice cannot be simultaneously present on the TX side and on the RX side due to the presence of the balancing network trying to extinguish either one or the other.

3. the TX side has higher current induced by the Ryan's voice and wins in the magnetic battle reaching the equilibrium with almost all the current induced by Ryan's voice being on TX side.