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tcm3105 how to connect single ended oscillator?

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neazoi

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Hello I have two TCM3105 chips and I need to connect them using a single crystal as the common clock. How can this be done?
 

The data sheet says if an external clock is used you should connect it to OSC1 and leave OSC2 disconnected. From this I would deduce that OSC2 is the output side of the oscillator amplifier so if you connect the crystal as usual to one of the TCM3105 ICs and connect it's OSC2 pin to OSC1 on the other IC it should work. Keep the wiring as short as possible as the loading on the first IC will be increased.

Brian.
 
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    neazoi

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The data sheet says if an external clock is used you should connect it to OSC1 and leave OSC2 disconnected. From this I would deduce that OSC2 is the output side of the oscillator amplifier so if you connect the crystal as usual to one of the TCM3105 ICs and connect it's OSC2 pin to OSC1 on the other IC it should work. Keep the wiring as short as possible as the loading on the first IC will be increased.
Brian.
So I will leave the OSC2 pin on the second IC disconnected?
 

That is correct.

I would guess that between OSC1 and OSC2 there is an inverting amplifier circuit and some kind of bias network so it forms an oscillator when the crystal is connected across them. The data sheet says to leave OSC2 disconnected if you drive it from an external clock so I'm reasoning that OSC1 is the input of the amplifier because driving a signal into OSC2 would backdrive the internal amplifier. By implication, if you drive OSC1 you would be able to take an inverted copy of the signal from OSC2 where it could be used in other circuits.

The only problem may be that a crystal only needs a small amount of signal to drive it so the output at OSC2 may be 'weak', that is why I recommend you take care not to place too much load on it. Also bear in mind that the the second IC will probably be working from an inverted clock compared to the one with the crystal although in that kind of circuit it probably wont make any difference to operation.

If it doesn't work, your other option is an external oscillator feeding OSC1 on both ICs with OSC2 left unconnected on both.

Brian.
 
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    neazoi

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That is correct.

I would guess that between OSC1 and OSC2 there is an inverting amplifier circuit and some kind of bias network so it forms an oscillator when the crystal is connected across them. The data sheet says to leave OSC2 disconnected if you drive it from an external clock so I'm reasoning that OSC1 is the input of the amplifier because driving a signal into OSC2 would backdrive the internal amplifier. By implication, if you drive OSC1 you would be able to take an inverted copy of the signal from OSC2 where it could be used in other circuits.

The only problem may be that a crystal only needs a small amount of signal to drive it so the output at OSC2 may be 'weak', that is why I recommend you take care not to place too much load on it. Also bear in mind that the the second IC will probably be working from an inverted clock compared to the one with the crystal although in that kind of circuit it probably wont make any difference to operation.

If it doesn't work, your other option is an external oscillator feeding OSC1 on both ICs with OSC2 left unconnected on both.

Brian.

Thanks Brian, I will try this.
I design a more advanced version of the baycom 1.2kbps, which I would like to produce as a kit for the hams. 9.6kbps if great but it requires mods to the radio.
 

That is correct.

I would guess that between OSC1 and OSC2 there is an inverting amplifier circuit and some kind of bias network so it forms an oscillator when the crystal is connected across them. The data sheet says to leave OSC2 disconnected if you drive it from an external clock so I'm reasoning that OSC1 is the input of the amplifier because driving a signal into OSC2 would backdrive the internal amplifier. By implication, if you drive OSC1 you would be able to take an inverted copy of the signal from OSC2 where it could be used in other circuits.

The only problem may be that a crystal only needs a small amount of signal to drive it so the output at OSC2 may be 'weak', that is why I recommend you take care not to place too much load on it. Also bear in mind that the the second IC will probably be working from an inverted clock compared to the one with the crystal although in that kind of circuit it probably wont make any difference to operation.

If it doesn't work, your other option is an external oscillator feeding OSC1 on both ICs with OSC2 left unconnected on both.

Brian.

Can I connect it directly on the telephone line like this https://freecircuits.files.wordpress.com/2011/01/picture131.jpg ?
 

The link shows a completely different device. The link is to a DTMF decoder while the TCM3105 is an FSK modem.

Note there is an error in the schematic, the LEDs should be returned to ground. In most countries the phone line wires are floating with a controlled impedance in both wires, if you ground one as shown it will stop the line working in most cases and the telco will see it like a phone permanently off-hook. It is far safer to use an isolating transformer, both for sgnal coupling and also for safety (a phone line can have more than 125V on it while ringing). To make it go off-hook a constant current load should be enabled across the line with an opto-coupler as isolation.

Brian.
 

The link shows a completely different device. The link is to a DTMF decoder while the TCM3105 is an FSK modem.

Note there is an error in the schematic, the LEDs should be returned to ground. In most countries the phone line wires are floating with a controlled impedance in both wires, if you ground one as shown it will stop the line working in most cases and the telco will see it like a phone permanently off-hook. It is far safer to use an isolating transformer, both for sgnal coupling and also for safety (a phone line can have more than 125V on it while ringing). To make it go off-hook a constant current load should be enabled across the line with an opto-coupler as isolation.

Brian.

Yes I know it is a different device, I showed it as an example of extracting (and transmitting) audio tines through tel lines. The standard 600:600 transformer is safe but the size cost and limitation in availability in all countries is an obstacle to the little dirty cheap modem project I am on. A high voltage capacitor (say 250v AC) in series with a resistor is by far easier and cheaper (but not safer) way. I thought so, that the GND connection there is just nonsense. The TCM3105 datasheet shows the use of an opamp and a transformer to do the job, but I would like to make it dirty scheap withour the opamp and the transformer (if it is possible).

I wonder if the zener diodes connected this way will limit the voltage to safe levels or not. I have also seen varistors connected directly accross the telephone line.

I have also found this little homemade telephone (without dialer) that might give an idea http://www.circuit-zone.com/ediy_blog/289/01.gif
and also this simple circuit http://www.talkingelectronics.com/projects/200TrCcts/101-200TrCcts.html#27
 
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I'm not sure that dialer will work on many lines, a real telephone alerts the switch (line exchange) equipment it has gone off-hook by passing a constant current between the lines. The current varies in different regions but can be up to 100mA. I can't see that circuit allowing enough to flow as it is. The classic circuit is a small power transistor held in fixed VB-E mode so it works as a constant current device.

I don't think the simple transformer or capacitor arrangement will work with the TCM3105. The op-amps are there for to isolate the transmit and receive signals, the equivalent of a 'hybrid' network in a voice circuit but with no sidetone leakage. It is important that the transmitted tones and received tones are kept apart as the modem sees them but are combined into one signal on the line interface. The op-amp is there to subtract the transmitted tones from the total on the line so all that remains is the received tone from the other end of the connection.

Brian.
 

I don't think the simple transformer or capacitor arrangement will work with the TCM3105. The op-amps are there for to isolate the transmit and receive signals, the equivalent of a 'hybrid' network in a voice circuit but with no sidetone leakage. It is important that the transmitted tones and received tones are kept apart as the modem sees them but are combined into one signal on the line interface. The op-amp is there to subtract the transmitted tones from the total on the line so all that remains is the received tone from the other end of the connection.
Brian.
You are talking about isolation of the two different frequency tones in a full duplex link?
If a half duplex link is used, will there be any need for the opamps?
 
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Not two different frequencies necessarily as the transmitted and received tones might be the same and simultaneous. They are for splitting the incoming and outgoing tones into different paths. If you are only using half duplex it shouldn't be a problem and you can simply capacitor couple the tone to the audio path.

Brian.
 

Not two different frequencies necessarily as the transmitted and received tones might be the same and simultaneous. They are for splitting the incoming and outgoing tones into different paths. If you are only using half duplex it shouldn't be a problem and you can simply capacitor couple the tone to the audio path.
Brian.

This is the little dirty cheap half duplex modem I am thinking of (apologize about the crappy schematic)

The dip switches in the dialer can be replaced by a microcontroller.
I have to find a way to electronically hook-off/on the line and two such circuits I have found are shown on the right (will the bridge act like a constant current?)

Any suggestions Brian?
*Keep in mind that this has to be dirty cheap
 

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I'm getting a little confused (easy for me - I've had lots of practice).

If it is going to be used as a radio modem you don't need the phone line interface at all. What you might have is a bandwidth problem as you need at least twice the bit rate (at least to 20KHz for 9600 bauds FSK) which is obviously a problem for most Ham radio bands.

If you are trying to push 9600 Bauds down a phone line you have no chance whatsoever, the maximum standard speed you will manage is 1200Bauds. Beyond that you have to use a multi-pointed constellation modulation such as QAM or QPSK. The impedance across a phone line is normally in the range 400 Ohms to 600 Ohms and you need to draw maybe 50mA DC or more from the line to 'seize' it and make the switch recognize it being off-hook. Because of the low impedance you would need much smaller series resistors on the transmit signal. For example, DTMF tones are normally sent at about -10dB line level (see attached file).

Brian.
 

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    neazoi

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I'm getting a little confused (easy for me - I've had lots of practice).

If it is going to be used as a radio modem you don't need the phone line interface at all. What you might have is a bandwidth problem as you need at least twice the bit rate (at least to 20KHz for 9600 bauds FSK) which is obviously a problem for most Ham radio bands.

If you are trying to push 9600 Bauds down a phone line you have no chance whatsoever, the maximum standard speed you will manage is 1200Bauds. Beyond that you have to use a multi-pointed constellation modulation such as QAM or QPSK. The impedance across a phone line is normally in the range 400 Ohms to 600 Ohms and you need to draw maybe 50mA DC or more from the line to 'seize' it and make the switch recognize it being off-hook. Because of the low impedance you would need much smaller series resistors on the transmit signal. For example, DTMF tones are normally sent at about -10dB line level (see attached file).

Brian.

Thanks Brian,
I'll change the dtmf output resistor to a much lower value, say 1k or 100R. The schematic attached is for a 1200kbps half duplex through telephone network (like the old BBS). But since I may use the same chip for an RF modem (or a second one for full duplex) I asked about the crystal at the beginning. I am sorry about the confusion.
So for the off-hook situation (answered), all I need to do is detect the ring on the line and then connecting the rest of the modem to the line using a relay or something (auto answer). Would that be enough to draw 50mA? (based on my attached schematic).
 

Basically yes. To detect the ring you need to isolate the DC across the line by using a series capacitor then rectify the signal and use the resulting DC as a 'ring indicator' signal. Typical telco specifications for the ringing voltage are 75V at 25Hz but the cadence (ring pattern) varies from one country to another.

You can simply use a resistor to load the line so it draws ~50mA but the drawback to that method is you are also loading the audio on the line with it. In old fashioned telephones a big inductor was used so the audio impedance was high while resistance was low but these days it is much easier to use a transistor load and feed the audio to it's base so it modulates the voltage across the line directly.

Things to be aware of:
1. Modern phones have low pass filters in the line wires so they do not load ADSL signals (f > 20KHz)
2. The ring detection circuit often has back-to-back Zener diodes (or a diac) in series with the capacitor to lower it's loading at low AC voltages where the audio itself might be seen as ringing voltage, and to make it less prone to interference being mistaken as ring signal.

Brian.
 

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