How does this telephone circuit modulate the current?

Question

I've been reading on how old landline telephones work (POTS, as they call it) and found out it's quite a bit more involved than I had expected.

This circuit comes from the wikipedia article "Telephone", and I have a few questions.

First, what I've got so far. Please tell me if I got anything wrong:

1. There is a pretty high DC voltage (48 V) between both wires (called tip and ring). I still don't know which one is higher though. Wikipedia says the voltage is negative with respect to earth to avoid corrotion but I don't know which one is connected to ground, if tip or ring.

2. The voices are modulated on the AC current, not on the voltage. This means that the current along the loop should be i = I_DC + k1 * x1 + k2 * x2. In that equation I_DC is the current resulting from the DC offset on the line, x1 and x2 are the voice signals of each user (they could be air pressure at the microphone) and k1 and k2 are constants to convert to current.

With that in mind, I'm trying to understand how this circuit works. I have a few questions:

• At audio frequencies, the impedance of the speaker should be fairly resistive so the fact that it's modulated current makes no difference at all. The current drives the speaker just as well as a R_speaker * i voltage would. Is this correct?

• You get feedback on the speaker, right? The microphone signal goes through the transformer, I think I should hear myself but I never payed attention to that, so I don't remember (also it's been a while since I used a landline phone).

• Related to the previous questions, Wikipedia mentions that A3 is a hybrid coil, so that the outgoing signal doesn't overpower the incoming signal. What I think this means is that A3 adds up both signals, but since the current from A2 goes into the middle tap, it's level is reduced by the turns ratio on the transformer, since the incoming signal uses the whole coil. Is this correct?

• A2 is the microphone, represented as a variable resistance, which I don't understand. My first guess would be that the variable resistance generates AC current due to the DC voltage applied to the resistor. Is this correct? However, most modern phones use electret microphones, which just generate a small voltage proportional to the sound pressure. How is this implemented on a modern phone? Do they feed a current source with the amplified voltage from the mic? And how is it implemented on an old phone? What kind of microphone works as a variable resistance?

• What are the typical amplitudes of the AC currents?

These are all the questions I can think of right know. I will update if I think of anything else.

Answer 1

There is a pretty high DC voltage (48 V) between both wires (called tip and ring). I still ... don't know which one is connected to ground, if tip or ring.

The ring is about -48 volts with respect to earth ground (just remember the phone system emergency power is the equivalent to 4 car batteries and you will easily remember this voltage). The tip is about zero volts with respect to earth ground. By the way (BTW), this is the on-hook voltage. As soon as you go off-hook the voltage drops as tip and ring now carry sounds (dial tone).

The voices are modulated on the AC current, not on the voltage...

If you have voltage and a load you have current. They are all related:

V  = I x R

...so both current and voltage change when a voice is being sent over tip and ring.

At audio frequencies,

Phone audio is sampled at 8K bytes per second so the upper frequency cut-off is 4KHz. Nothing higher gets through a normal phone connection.

the impedance of the speaker should be fairly resistive ...

As said before, if you measured both the voltage between tip and ring and the current through tip and ring you will find both are changing according to the audio being sent over the tip and ring.

You get feedback on the speaker, right? The microphone signal goes through the transformer... so that the outgoing signal doesn't overpower the incoming signal...

A POTS (Plan Old Telephone Service) line is a 2 wire interface. But it essentially carries a full duplex conversation. To accomplish this the local hand set needs to remove the local sound from the local ear piece. This is done using a cleverly wound transformer which subtracts the local sound before sending the signal to the local ear piece.

A2 is the microphone, represented as a variable resistance, which I don't understand...

The off-hook voltage is about 3 to 9 volts. This is supplied by the phone company. Old phone microphones were made using carbon and would change resistance when air pressure waves hit the diaphragm. Again, as the resistance changes so does the voltage and current as given by the equation:

V = I x R

...most modern phones use electric microphones, which just generate a small voltage proportional to the sound pressure. How is this implemented on a modern phone? ...

No matter how a modern phone is designed, it always need to emulate the old POTS standard. That is, all phones must appear the same to the phone company's equipment. So it does not matter how the microphone works in a modern phone. As long as the audio modulates the voltage and current over the tip and ring the same as the old POTS standard.

What are the typical amplitudes of the AC currents?

Assuming you are referring to the ringer signal, it is about 90 volts at 20Hz.