LMH6560 datasheet(15/23 Pages) NSC

15 / 23 page

Application Notes (Continued)

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)

relative dielectric constant

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

r varies

over a certain range of values. Varying the e

r over a range of

1 to 10 gives a variation for the Characteristic Impedance of

about 40

Ω from 80Ω to 38Ω. Most transmission lines are

designed to have 50

Ω or 75Ω impedance. The reason for

that is that in many cases the pcb trace has to connect to a

cable whose impedance is either 50

Ω or 75Ω. As shown e

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 lineI=V/Z

0, where Z0 is the

impedance of the transmission line. For an applied transient

of 2V with Z

0 =50

Ω 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

Ω. In that case a

power of 80mW arrives at the 75

Ωload and causes a voltage

of the proper amplitude to maintain the incoming power.

20064240

FIGURE 4.

20064243

FIGURE 5.

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