# Low loss impedance matching without a transformer

## 2 answers

# 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).

#### 0 comments

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.

#### 6 comments

@Olin your image is not showing. I am going to add more sections to this answer so it might be that your answer "what Andy described" might not hit the mark in terms of the fullness of what I hope to complete!

@Andy: Strange. I see it in my browser. I'm using Edge on Win 10, and also just tested it with IE on Win 10. Both showed the image correctly. What browser and OS are you using?

@Olin I'm using google chrome and I still can't see it - bug alert!!!

And I can't see it Firefox on Android

@Andy: See the main meta question https://meta.codidact.com/questions/276272 and let them know which images you see.

## 0 comments