AndrewT:
Yes,yes, here are some answers for you.....
1.Changing all the R values to 10k would not change much other than give you an increase in input impedance ie:3.3kohm roughly.
2.Changing to 100R then is not a good idea,for obvious reasons!
3.The choice of 5k1 res.are well within the chosen IC's current/biasing requirements,but not measured values! but should be closely matched!....how am I doing then?
Q1:I notice that the o/p caps are reversed...is this correct?
Q2:Would it be better to use bipolar caps instead on the outputs?
Q3: Will I then be able to drive (both) an unbalanced o/p with this at the same time?
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doing OK.
the 5k1 need to be selected to suit what they do.
Some control the gain affecting signal level. These can vary either side of nominal value, maybe as far as +-5%.
Some are intrinsic to rejecting interference. These must be <0.1% to achieve worthwhile rejection
Walt J. does a nice diagram comparing the transmit and receive resistor/impedance values arranged in a bridge to show why balance across and round the bridge must be accurate to remove the common mode interference.
the 5k1 need to be selected to suit what they do.
Some control the gain affecting signal level. These can vary either side of nominal value, maybe as far as +-5%.
Some are intrinsic to rejecting interference. These must be <0.1% to achieve worthwhile rejection
Walt J. does a nice diagram comparing the transmit and receive resistor/impedance values arranged in a bridge to show why balance across and round the bridge must be accurate to remove the common mode interference.
Q1:I notice that the o/p caps are reversed...is this correct?
Q2:Would it be better to use bipolar caps (47uFs) instead on the outputs?
Q3: Will I then be able to drive (both) an unbalanced o/p with this at the same time?[/QUOTE]
Thank you...very enlightning stuff indeed!
I will now first try this on a breadboard & will post you the results,but before I rush & buy the wrong parts...just wanted to know if you'd please give me some advice on the above questions too? Couldn't find the answers anywhere!
I'll be immersing myself into Jung's articles tonight....you're indeed a very hard taskmaster lol!...(you must be an excellent teacher,but I hope not one..hee..hee)
Q2:Would it be better to use bipolar caps (47uFs) instead on the outputs?
Q3: Will I then be able to drive (both) an unbalanced o/p with this at the same time?[/QUOTE]
Thank you...very enlightning stuff indeed!
I will now first try this on a breadboard & will post you the results,but before I rush & buy the wrong parts...just wanted to know if you'd please give me some advice on the above questions too? Couldn't find the answers anywhere!
I'll be immersing myself into Jung's articles tonight....you're indeed a very hard taskmaster lol!...(you must be an excellent teacher,but I hope not one..hee..hee)
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there are no resistors or resistor ratios in that Neve circuit that require that level of accuracy or matching.
regards
After reading a few articles & test results by others,I now realize that there is an undeniable discrepency between circuit design & real life results due to so many variables involved.However,compromises aside it does help to adhere close to the theoritical calculations as possible which would surely help one to achieve more than the minimum 40dB CMRR!
Q:What do you think will be the result of this circuit if I chose the
10k resistors arbitrarily against choosing closely matched ones?
Is it not worth at all? Any thoughts on this?
Thanks.
Walt J. does a nice diagram comparing the transmit and receive resistor/impedance values arranged in a bridge to show why balance across and round the bridge must be accurate to remove the common mode interference.[/QUOTE]
AndrewT,I can't find the article by W.Jung you've mentioned.Could you kindly give me link please?
Thanks
AndrewT,I can't find the article by W.Jung you've mentioned.Could you kindly give me link please?
Thanks
Andrew has lured you onto the wrong path. The way the two diff. amps are connected to single ended source converts one to a noninverting and one to an inverting buffer. Common mode suppression is no issue here.
The ouput impedances don't need to be matched too, because they are low in value compared to the expected high impedance inputs of the supposingly connected line receivers.
regards
Have a third look.
The circuit is entirely accurate. Each half is a differential in - single ended out amplifier. The reason that the inputs are swapped around is so that you get one differential amp with an inverted output and one differential amp with a non-inverted output from the original single ended (unbalanced) input.
There doesn't appear to be any advantage to close tolerance resistors as these would only improve the common mode rejection ratio (more reading..) of each individual differential amplifier. The input is a single ended (a.k.a. unbalanced) source and the noise would be predominantly differential mode, so high CMRR would be wasted.
http://waltjung.org/PDFs/Op_Amps_in_Line_Driver_and_Receiver_Circuits_P2.pdf
Juergen look at fig8. R* and the same advice in Kester's paper fig8.59.
http://waltjung.org/PDFs/WTnT_Practical_Circuits_for_Quiet_Audio_Transmission.pdf
Fig1 shows the bridge equivalency that is discussed in detail in the paper.
Juergen look at fig8. R* and the same advice in Kester's paper fig8.59.
http://waltjung.org/PDFs/WTnT_Practical_Circuits_for_Quiet_Audio_Transmission.pdf
Fig1 shows the bridge equivalency that is discussed in detail in the paper.
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yes thats the circuit in question. Although the output impedances should be equal, they don't need to be exactly equal i.e. matched.
For example it would require a source impedance mismatch of say 10R to imbalance a perfect line receiver with inbuilt 10k resistors to 0.1%.
The usual tolerance of 1% is just fine for the 120R resistors. Even the 5% carbon films should do in most cases.
regards
I can easily select matches of 0r1 resistors or 0r33 resistors to better than 1% and with a little more effort to better than 0.05%. That is a matching accuracy in the region of 0.1milliohms. What's unrealistic in matching to 120milliohms?
Jung is looking for CMRR of 80-100dB. In this case accurate matching is necessary, but difficult to achieve.
Is such a good CMRR really necessary in all circumstances? Only if the unwanted common-mode signal is likely to be equal in magnitude to the wanted differential signal. If the cable run is not too long, the cable is well constructed, and the electromagnetic environment is not too severe (e.g. no hum loops, and we are not recording in a power station or electric train!) then a lower CMRR should be OK. If we are simply trying to do a bit better than a normal unbalanced cable then CMRR of 20-30dB should be fine. The circuit achieves this, provided the right pairs of resistors are matched to within 1%.
Just because it is not perfect does not mean it is useless!
Is such a good CMRR really necessary in all circumstances? Only if the unwanted common-mode signal is likely to be equal in magnitude to the wanted differential signal. If the cable run is not too long, the cable is well constructed, and the electromagnetic environment is not too severe (e.g. no hum loops, and we are not recording in a power station or electric train!) then a lower CMRR should be OK. If we are simply trying to do a bit better than a normal unbalanced cable then CMRR of 20-30dB should be fine. The circuit achieves this, provided the right pairs of resistors are matched to within 1%.
Just because it is not perfect does not mean it is useless!
Andrew, you are correct that the overall matching must be maintained. The reason both Jung and Rane say that because it is a common engineering issue, EVERYBODY says that.
But if the 10k resistors are matched at 0.1% that's a 10 ohms tolerance. If you match the 120 ohms at 1% that's a tolerance of 1.2 ohms. Tolerances add vectorially. It's no use to try to match the 120 ohms to better than 1%, you're still no better than 0.1% overall.
It's not so much a matter of 'interpreting Jung' but common engineering sense.
jd
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