Use a series inductor or a series capacitor, as To rotate the impedance zL via the constant VSWR circle until it intersects the Re=1 constant resistance circle. Will be zero reflection at the matching network. The above technique can be used for any impedance on the interior of the Smith Chart. The result is that z2 = 1, so that no powerįigure 6. The resulting impedance matching network is shown in Figure 6. Moving zL to z1 and then z2 with a tx line and series capacitor. The path taken by zL to reach the center of the Smith Chart is shown via the brown curve in Figure 5.įigure 5. We simply choose C such that z_cap = -i*2.228, and the Since z1 = 1 + i*2.228, we can exactly match this impedance to the center of the Smith Chart Read directly off the smith chart yielding z1 = 1 + i*2.228. The resultant impedance is z1 (shown in Figure 3), and the value for this can be Where the constant VSWR circle intersects the Re=1 constant resistance circle. In Figure 3, I've also drawn in green the constant reactance curve given by Im=2.228.
Effect of Tx line (length = 0.0632 wavelengths) on zL. The resultant rotation on the Smith Chart is illustrated via the brown curve in Figure 3:įigure 3. Hence, we can rotate the impedance zL to the Re=1 circle using a transmission line of length 0.0632 wavelengths. We can calculate the required length of the transmission line corresponding to 45.5 degrees: Since one wavelength corresponds to two complete rotations around the Smith Chart, we have: Using the degrees scale on the outer rim of the smith chart. This can be done in different ways for example,īy drawing the lines and then using a protracter to determine the angles, or reading off the angle difference In Figure 2, we have found the angle between the two lines to be 45.5 degrees.
Determining the Tx Line Length to Reach the Re=1 circle. Then find the angle between the two lines, as shown in Figure 2.įigure 2. To figure out the rotation needed, draw a line from the center of the Smith Chart through zL, and thenĭraw another line from the center of the Smith Chart to the location where the constant SWR circle intersects
Until it intersects the Re=1 circle, then we can exactly match any impedance that does not have a reflectionĬoefficient magnitude equal to 1 (that is, we can match anything on the interior of the Smith Chart). Now, if we use a transmission line section to rotate the impedance on the Smith Chart Recall that if an impedance is of the form z1 = 1 + iX, then we can exactly match it using a series Impedance zL on the Smith Chart along with its constant VSWR circle. To any location in the black circular ring of Figure 1:įigure 1. That is, given the load impedance zL, a transmission line section can relocate the impedance That a transmission line section will enable the load impedance to move in a circle about the center of Suppose we use a transmission line section to move the impedance. Let's look at example 2 from the previous page again.
#Smith chart online how to#
We'll show how to use transmission line sections and a series component to exactly match any load impedance. This method won't work unless the real part of the load impedance is 1 (that is, Re=1). How to use series inductors and capacitors to cancel out the reactance of a load. The Smith Chart - Impedance Matching with Tx Lines, Series Inductors and Capacitors Smith Chart Tutorial - Impedance Matching with Tx Lines, Series L and C Previous: Series L and C
0 kommentar(er)
