Timeline for 2/4 Wire Conversion Circuit For UK Telephone
Current License: CC BY-SA 3.0
8 events
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| May 1, 2014 at 14:22 | comment | added | Matt Taylor | I thought though as in my second circuit I take the signal post X1 before removing the offset (which will be half supply) the NONIN_IN can be grounded through the terminating network of resistors and capacitors. That way even if there is no signal at V3, the non-inverting input to X3 is always 1.55 volts (due to the potential divider network). Maybe I have completely misunderstood....I am a little out of my depth here. | |
| May 1, 2014 at 13:45 | comment | added | Andy aka | Your circuit is wrong dude. You can't just wire an op-amp up to as you have done - it must be a differential amplifier as shown in your first circuit BUT with the ground point connected to mid-rail like I said a couple of comments earlier. | |
| May 1, 2014 at 13:43 | comment | added | Andy aka | Your waveform generator has to be applied correctly and it has to have the precise output impedance of the complex network you use. Your circuit does not indicate where you applied the signal so I can't say any more. | |
| May 1, 2014 at 13:25 | comment | added | Matt Taylor | Hi Andy, thanks so much for your help so far. As you can see from the original question I have added a modified circuit with the suggestions you made. It works fine when there is only one source generator, however if a signal is present from the other side of the line the output from X3 is still just my offset voltage...any ideas? | |
| Apr 30, 2014 at 16:58 | comment | added | Andy aka | Try R15 tied to your mid-rail generator's output (OFFSET). If the network line can be called R4 then you only have to make R1/R4 = R2/R3 (by whatever means within reason). | |
| Apr 30, 2014 at 16:17 | comment | added | Andy aka | Using a different impedance such as the complex one you list is not a problem. Use the impedance to drive to the line to form a 2:1 divider and matched resistors for the cancellation circuit and it'll work. | |
| Apr 30, 2014 at 15:24 | comment | added | Matt Taylor | Thanks for your answer Andy. I understand that, but the problem that I'm having is getting the diff-amp to operate on a single-rail supply, rather than a dual-rail. Also, would this work with impedances other than 600 Ohm? I'm building it for an EU phone line, which uses a termination of 370ohm + (620ohm||310nF). | |
| Apr 30, 2014 at 15:05 | history | answered | Andy aka | CC BY-SA 3.0 |