Differential Signaling - Understanding Signal Integrity - page 168

subtracts from the common-mode voltage, making it a 50 Ω resistance in series with

a 0V source. This validates the model; the DO + and DO − pins have the correct

50 impedance, and with respect to ground, they switch between the correct volt-

age levels (0 and 1V).
With this circuit arraignment the voltage divider principle can be applied to
find the launched voltage. Since the driver and transmission line impedances are
both 50

Ω

Ω , the voltage on the DO + and DO − pins will each be half the switched

voltage. As shown, this is ± 0.25V when measured with respect to the common-

mode voltage.
Our analysis to this point is illustrated in Figure 13.8. For simplicity only the
DO+ signal is shown. As expected the launched voltage is 0.25V higher than the
common-mode voltage. Ohm's law can now be used as was done in Chapter 6 to
find that 5 mA is sourced by DO +. The DO analysis is the same, noting that the

pin is 0.25V lower than the common-mode voltage. That driver sinks 5 mA from
the common-mode supply.
The arrows in Figure 13.7 show that the currents from the DO + and DO

−

out-
puts flow in the opposite directions. Because they are the same value, this makes
the net current in the common-mode connection zero, and the connection can
be removed without disturbing the circuit operation. This has been done in Fig-
ure 13.9, where we see the voltages of the two outputs measured with respect to
ground rather than to the common-mode voltage. The DO + output is 0.25V higher
the 0.5-V common-mode voltage, making it 0.75V with respect to ground. The
DO

−

output is 0.25V lower than the common-mode voltage, making it 0.25V.
The voltage difference between the two outputs is the differential output volt-
age and is given in:

−

VV V
+

= − = 0.75 − 0.25 = 0.5V

diff DO DO −

(13.2)

All the background information is now available to find the differential imped-
ance; the differential voltage is 0.5V and the current is 5 mA. Ohm's law determines
the relationship between the transmission line impedance, voltage, and current. In
this case the differential impedance is 100

Ω , as given in:

V

0.5V diff

Z = = = 100 Ω

(13.3)

diff

I 5mA

diff

Z =50 Ω

oo

50 Ω

DO +

RI+

VDO +

5mA

RT

+

0.5V + 0.25V

I = 0.25V/50 = 5 mA Ω

Figure 13.8 A 0.25-V, 5-mA signal measured with respect to V cm is launched down a 50 Ω transmis-

sion line.