參數(shù)資料
型號: LMH6560MTX
廠商: NATIONAL SEMICONDUCTOR CORP
元件分類: 緩沖放大器
英文描述: Quad, High-Speed, Closed-Loop Buffer
中文描述: BUFFER AMPLIFIER, PDSO14
封裝: TSSOP-14
文件頁數(shù): 15/23頁
文件大?。?/td> 1104K
代理商: LMH6560MTX
Application Notes
(Continued)
Z
0
can be calculated by knowing some of the physical di-
mensions of the pcb line, such as pcb thickness, width of the
trace and
e
r
, relative dielectric constant. The formula given in
transmission line theory for calculating Z
0
is as follows:
(1)
e
r
relative dielectric constant
h
pcb height
W trace width
th thickness of the copper
If we ignore the thickness of the copper in comparison to the
width of the trace then we have the following equation:
(2)
With this formula it is possible to calculate the line imped-
ance vs. the trace width.
Figure 5
shows the impedance
associated with a given line width. Using the same formula it
is also possible to calculate what happens when
e
varies
over a certain range of values. Varying the
e
over a range of
1 to 10 gives a variation for the Characteristic Impedance of
about 40
from 80
to 38
. Most transmission lines are
designed to have 50
or 75
impedance. The reason for
that is that in many cases the pcb trace has to connect to a
cable whose impedance is either 50
or 75
. As shown
e
r
and the line width influence this value.
Next, there will be a discussion of some issues associated
with the interaction of the transmission line at the source and
at the load.
Connecting a Load Using a Transmission Line
In most cases, it is unrealistic to think that we can place a
driver or buffer so close to the load that we don’t need a
transmission line to transport the signal. The pcb trace
length between a driver and the load may affect operation
depending upon the operating frequency. Sometimes it is
possible to do measurements by connecting the DUT directly
to the analyzer. As frequencies become higher the short
lines from the DUT to the analyzer become long lines. When
this happens there is a need to use transmission lines. The
next point to examine is what happens when the load is
connected to the transmission line. When driving a load, it is
important to match the line and load impedance, otherwise
reflections will occur and this phenomena will distort the
signal. If a transient is applied at T = 0 (
Figure 6
, trace A) the
resultant waveform may be observed at the start point of the
transmission line. At this point (begin) on the transmission
line the voltage increases to (V) and the wave front travels
along the transmission line and arrives at the load at T = 10.
At any point across along the line I = V/Z
, where Z
is the
impedance of the transmission line. For an applied transient
of 2V with Z
= 50
the current from the buffer output stage
is 40mA. Many vintage op amps cannot deliver this level of
current because of an output current limitation of about
20mA or even less. At T = 10 the wave front arrives at the
load. Since the load is perfectly matched to the transmission
line all of the current traveling across the line will be ab-
sorbed and there will be no reflections. In this case source
and load voltages are exactly the same. When the load and
the transmission line have unequal values of impedance a
different situation results. Remember there is another basic
which says that energy cannot be lost. The power in the
transmission line is P = V
2
/R. In our example the total power
is 2
2
/50 = 80mW. Assume a load of 75
. In that case a
power of 80mW arrives at the 75
load and causes a voltage
of the proper amplitude to maintain the incoming power.
20064240
FIGURE 4.
20064243
FIGURE 5.
L
www.national.com
15
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