Common Mode and Differential Mode fundamentals:
An ideal differential amplifier, amplifies the
"difference" signal between its two inputs and totally rejects any voltage which
is common to both inputs. The voltage of interest is referred as the
differential voltage, this is the differential mode signal and is expressed as
VDM, where VDM=(V+input)-(V-input). VCM is the voltage which is common to both inputs
and is referred as the Common-Mode Voltage. The ideal differential amplifier
rejects all of the common-mode component, regardless of its amplitude and
frequency. To calculate with the imperfections of the amplifier, we use
CMRR(ratio) or CMRR(dB).
This is the relation in gain of ADM and ACM, see formulas below.
CMRR specification is an absolute value, and does not specify polarity (or
degrees of phase shift) of the error. Therefore, we can't simply subtract the
error from the displayed waveform. CMRR generally is highest (best) at DC and
degrades with increasing frequency of VCM. The best specification is a plot the
CMRR specification as a function of frequency.
|The transfer equation is:||Where Vout is referenced to ground.|
|CMRR(ratio):||Where CMRR is expressed in ratio notation of Differential Mode Gain versus Common Mode Gain|
|CMRR(dB):||Where CMRR is expressed in dB notation of Differential Mode Gain versus Common Mode Gain.|
CMRR dB versus Ratio:
|CMRR ratio||=||CMRR dB|
Suppose we want to measure the voltage of an emitter resistor of an Audio Power Amplifier. This by full load of 50Vtt at the output by 1kHz. The measured voltage across Re2, (0.1W resistor) is 10mVtt. We measure the voltage with an OpAmp with a CMRR of 10,000 at 1kHz. The following items could be calculated;
The differential amplifier we use has a CMRR(ratio) specification of 10,000 at 1 kHz or a CMRR(dB) of 80dB. With the amplifier driven to full power with a 1 kHz sine wave of 50Vtt, the voltage for the negative period is handled by T2 and will be 25Vtt(half periode). The output voltage of the OpAmp will be one ten thousandth of the common-mode signal will erroneously appear as VDM at the output of the differential amplifier, which would be 25 V/10,000 is 2.5 mV. The 2.5 mV represents up to a 25% error in the true 10mV signal measured across Re2!
To measure differential voltages in electronic circuits with an acceptable tolerance we have to use OpAmps with a good CMRR(dB) behaviour, otherwise this will occur to much errors at the output of the measuring OpAmp and will influence the measured values.
The inherent weakness of unbalanced interconnections is that the shield, which is also a signal conductor, is a path for power line related currents that always flows between equipment grounds. The voltage drop across the resistance of the shield and connectors adds directly to the signal, producing the familiar hum and buzz.
Circuit of a unbalanced system
When an output is connected to an input, a series circuit called a "voltage divider" is formed, where voltage drops are proportional to impedance (see Figure above). The input impedance of Device-2 is called a “load” on the output of Device-1. For typical equipment, Rout ranges from 1W to 100W and Rin ranges from 10 kW to 100 kW, transferring 99% to 99.9% of the available signal voltage. Using a Tube Pre-Amp (Rout=1kW), the divider ranges lies between 90%-99%!
Actual output impedance can be very important to know. For example, when an output drives a long cable, high output impedance can seriously degrade treble response. The capacitance of shielded audio cable, typically about 150 pF per meter, the actual cable capacitance forms a low-pass filter with the output of the Pre-Amp. If the output impedance is 1kW (Tube Pre-Amp), the maximum frequency response (-3dB) will be 106kHz for 10meter, 10.6kHz for 100 meter cable. If the output impedance were lower, say 100W, the effects would be insignificant better (factor 10x).
There are two major problems in connecting two balanced systems together, therefore we need to analyze these problems to come to a good design. The advantages are much greater than the disadvantages of balanced lines circuits.
1) Analogue audio designers must be aware of the noisy currents which flow through shields from one chassis to another, avoiding the "pin 1 problem" with a well reasoned internal ground system. Keeping in mind that external ground systems could be poisoned with incredible amount of pollution coming from the malicious outside world.
2) The goal of modern audio interfaces is to transfer maximum signal voltage while, of course, rejecting ground voltage differences and interference from external electrostatic and magnetic fields. In a basic balanced interface, the output impedances of the driver (R8, R9) and the input impedances of the receiver (device-2) effectively form a Wheatstone bridge. Rejection of the common-mode voltage is critically dependent on the ratio matching of the driver/receiver common-mode impedances in the two circuit sides. The bridge is most sensitive to small impedance changes in one of its inputs/outputs when all inputs/outputs have the same impedance. The circuit is least sensitive when the output and input have widely different impedances. Therefore, we can minimize the sensitivity of a balanced system to impedance imbalances by making common-mode impedances very low at one end of the line and very high at the other side.
Circuit of a balanced system
Continuing point 1,
To avoid ground loops in audio systems, we need to make a central ground point in the audio line-up. I've chosen for the Pre-Amp as central ground. This means that the analogue ground is connected to the earth. There are several ways to do this, see connection diagram below.
My favourite one is the circuit with two diodes anti parallel decoupled with a C. This coupling is only made in the Pre-Amp section. In the remaining units I've avoid any coupling to earth. Because of the safety rules of a double isolated mains connected equipment you have to deal with it in some cases. To make the best connection set S1 to AGND and S2 to AGND, keeping in mind that device-2 is not connected to earth.
Continuing point 2,
Making common-mode impedances very low at one end of the line and very high at the other side. In other words, make the output of the balanced line driver as low as possible. In practise this will be around 25W per output stage (R8, R9), lower that that could damage or destroy the output stage when it is short-circuit at full output voltage. For the input we have to deal with our goal, an impedance as high as possible but keep in mind that this will also increase the input noise of Device-2 (see Johnson noise in Noise chapter of this site). I've chosen for 20kW at each input for the best noise behaviour of device-2 and for a reasonably CMRR. Bill Whitlock of Jensen Transformers, Inc. has developed a circuit to optimise the CMRR of a balanced input. This is a patented design and described in the document below.
|Balanced Lines, Bill Whitlock|