As an alternative to doing the math, as detailed in Andy's answer, you can use a graphical aid called a Smith chart:

These were used routinely before computers to match transmitters to antennas. The math behind them is what Andy described. See the Wikipedia page for details, which is where the image above was copied from.

posted over 4 years ago

CC BY-SA 4.0

4y ago

Olin Lathrop‭

7964 reputation 18 327 951 466

Low-pass impedance transformation

Theory

Input Impedance: -

$$Z_{IN} = j\omega L + \dfrac{\dfrac{R_L}{j\omega C}}{R_L + \dfrac{1}{j\omega C}} = j\omega L +\dfrac{R_L}{1 + j\omega R_L C} = \dfrac{j\omega L - \omega^2 R_L LC + R_L}{1 + j\omega R_L C}$$

Multiply numerator and denominator by the denominator's complex conjugate to get: -

$$Z_{IN} = \dfrac{(j\omega L - \omega^2 R_L LC + R_L)(1 - j\omega R_L C)}{1 + \omega^2 R_L^2 C^2}$$

$$Z_{IN} = \dfrac{j\omega L -\omega^2 R_L LC + R_L + \omega^2 R_L LC +j\omega^3 R_L^2 LC^2 - j\omega R_L^2 C}{1 + \omega^2 R_L^2 C^2}$$

$$\boxed{Z_{IN} = \dfrac{j\omega L + R_L + j\omega^3 R_L^2 LC^2 - j\omega R_L^2 C}{1 + \omega^2 R_L^2 C^2}}$$

To find $\color{red}{\boxed{\omega}}$ that produces a real impedance, equate the numerator's imaginary terms to zero: -

$$\omega L + \omega^3 R_L^2 LC^2 - \omega R_L^2 C = 0$$

$$\color{red}{\boxed{\omega = \sqrt{\dfrac{R_L^2 C - L}{R_L^2 LC^2}} = \sqrt{\dfrac{1}{LC} - \dfrac{1}{R_L^2 C^2}}}}$$

$\color{red}{\boxed{\omega}}$ is the operating frequency of the impedance transformer. We use this relationship below: -

With imaginary terms at zero, $Z_{IN}\rightarrow R_{IN}$ hence: -

$$\boxed{R_{IN} = \dfrac{\cancel{j\omega L} + R_L + \cancel{j\omega^3 R_L^2 LC^2} - \cancel{j\omega R_L^2 C}}{1 + \omega^2 R_L^2 C^2}} = \dfrac{R_L}{1 + \omega^2 R_L^2 C^2}$$

If we substitute the formula for $\color{red}{\boxed{\omega}}$ in the above equation, we can drill down to this: -

$$\boxed{R_{IN} = \dfrac{L}{R_L C}} \text{ and therefore } \boxed{L = R_{IN} R_L C}$$

We can now substitute for L in the $\color{red}{\boxed{\omega}}$ equation hence: -

$$\color{red}{\omega = \sqrt{\dfrac{1}{LC} - \dfrac{1}{R_L^2 C^2}}}\color{black}{ \Longrightarrow \dfrac{1}{C}\sqrt{\dfrac{1}{R_{IN} R_L} - \dfrac{1}{R_L^2}} = \dfrac{1}{R_L C}\sqrt{\dfrac{R_L}{R_{IN}}-1}}$$

We can now solve for C: -

$$\boxed{ C = \dfrac{1}{\omega R_L}\sqrt{\dfrac{R_L}{R_{IN}}-1} }$$

So, if the frequency is 10 MHz, C = 118.62 pF

Given C, we can calculate L = 1.779 μH

Low-pass simulation

Circuit

Results

Conclusion

What the results tell us at 10 MHz is this: -

• The voltage amplification to the load from the source is 1.761 dB
• A 50 Ω to 50 Ω interface will naturally attenuate by 6 dB hence, the result is good
• The phase angle of current into L1 is leading the applied voltage by a fraction of 1 degree (0.0147°) i.e. the circuit input impedance can be regarded as resistive
• The input impedance magnitude is 50.003 Ω i.e. as desired
• The usable bandwidth (± 10°) is about 2 MHz producing a 63 Ω to 46 Ω impedance magnitude change across the bandwidth

High-pass impedance transformation

The alternative to the above circuit is to swap L and C like this: -

Without going through the math again we get: -

$$\boxed{R_{IN} = \dfrac{L}{R_L C}} \text{ and therefore } \boxed{L = R_{IN} R_L C}\text{ (as previously) but,}$$

$$\boxed{ C = \dfrac{1}{\omega R_{IN}}\sqrt{\dfrac{1}{\dfrac{R_L}{R_{IN}}-1}} }$$

So, if the frequency is 10 MHz, C = 142.35 pF

Given C, we can calculate L = 2.135 μH and we see this AC response: -

High-pass simulation

Circuit

Results

Conclusion

• The voltage amplification is 1.761 dB (as per low-pass version)
• The phase angle of current into L1 is leading the applied voltage by a fraction of 1 degree (0.000493°) i.e. the circuit input impedance can be regarded as resistive
• The input impedance magnitude is 49.995 Ω i.e. as desired

Why choose one over the other?

Both circuits work just fine for resistive loads but, if the load is partially reactive (as is often the case), the load can be broken into its parallel equivalent circuit. Then, you choose the top circuit for a load that has parallel capacitance and, you choose the lower circuit for a load that has parallel inductance. Then make the appropriate numerical adjustments to C (low-pass) or L (high-pass).

posted over 4 years ago

CC BY-SA 4.0

4y ago

Andy aka‭

2299 reputation 19 75 266 282