# Posts belonging to Category Electronics 101

## L-Network impedance matching

Impedance matching is an everyday problem for RF circuit designers. The L-Network is one of the easiest lossless ways of matching a low source impedance to a higher load impedance. This article shows how they work and how to design them.

Matching a transistor amplifier’s low output impedance with the higher impedance of an antenna (typically 50 or 75 Ohms) is just one everyday example of where an L-Network can be used. For this article, we are going to design an L-Network that matches a 75 Ohm source (function generator) with a 1 kΩ load (resistor).

To make it easier for you to understand how this network works, I will present the complete circuit first, explain how it works and then explain how the values are being calculated. Here is the finished L-Network:

L-Network matching 75 Ohms source impedance to 1000 Ohms load impedance

RS = Source Resistance (75 Ohms)
Ls = Series inductance (j263)
Cp = Parallel Capacitance (-j284)

Don’t be frightened by the little “j” in front of the values, in case you have never seen it. The letter “j” in electronics engineering stands for the imaginary unit. Adding “j” to a reactance is basically the so-called “vector notation.” A positive value indicates an inductive reactance and a negative value is always a capacitive reactance. If you are not used to it, you can just think of Ls as an inductor with a reactance of 263 Ohms and of Cp as a capacitor with a reactance of 284 Ohms.

Note that the reactance of a capacitor and an inductor is frequency dependent. Therefore, this circuit does not work well over a large bandwidth.

This circuit makes the 1000 Ohms load ‘look’ like a 75 Ohms load to the source. But how does it work? To answer this question, we have to look at the equivalent circuit. For that, we will reduce the overall circuit to simpler elements, step by step.

Let’s start with Cp and RL. They are in parallel. The overall impedance of two resistors is calculated as follows:

$R = \frac{R_1 * R_2}{R_1 + R_2}$

Likewise, the resulting complex impedance of a resistor and a capacitor can be calculated as follows:

$Z = \frac{Xc * R}{Xc + R}$

In our case, we get:

$Z = \frac{X_c * R_L}{Xc + R_L} = \frac{-j284 * 1000}{-j284 + 1000} = 75-j263$

75 is the real part in Ohms and -j263 is a capacitive reactance of 263 Ohms. That means the capacitor and resistor in parallel will act like a resistor with 75 Ohms and a capacitor with 263 Ohms reactance in series.

Equivalent circuit for Cp and RL

Now if you insert that back into the original circuit and have a close look at it, you may notice something. Look at the reactance of Cp and Ls.

Overall equivalent circuit after reducing Cp and RL to a series equivalent circuit

Correct, the reactance of both the capacitor and the inductor are the same, just with different vectors. If you have never worked with a complex impedance beforehand, you will now be amazed by how much easier it makes things. We can simply add the reactances together, and since 263 + -263 is zero, we only have the 75 Ohms load resistance left. Yes, it’s that simple, the source ‘sees’ the 1 kΩ load as a 75 Ω load.

Overall equivalent circuit of the L-Network

An often-welcomed side effect is that this L-Network acts as a low-pass filter. So, if it’s used to match a low transistor amplifier impedance directly to the higher antenna impedance, one suppresses harmonics at the same time.

If one wishes to have a high-pass characteristic, one can simply switch the positions of the inductor and the capacitor. It really changes nothing. The impedance values, of course, need to be properly adjusted in that case.

Let’s talk about how to design a circuit like this from scratch. We will use the above values, source impedance of 75 Ohms and load impedance of 1000 Ohms at a frequency of 16 MHz. The frequency doesn’t matter until the end when we have to pick actual values for Ls and Cp.

The first thing we have to calculate is the quality factor, or Q factor. It describes how under-damped an oscillator or resonator is, or equivalently, characterizes a resonator’s bandwidth relative to its center frequency. The Q of an L-Network is calculated as follow:

$Q = \sqrt{\frac{R_{Load}}{R_{Source}}-1}$

RSource = Source Resistance

With our values plugged into the equation, we get the following:

$Q = \sqrt{\frac{1000}{75}-1} = 3.512$

So 3.512 is our Q. The rest is just as simple. To get the needed impedance of the inductor, we use the following formula:

$X_L = Q*R_{Source} = 3.512 * 75 = 263$

That means the inductor needs to have an impedance of 263 Ohms at the desired frequency. We can write this as j263 to make sure it’s apparent that it is an inductive reactance. The impedance of the capacitor needs to be the load resistance divided by the Q factor.

$X_C = \frac{R_{Load}}{Q} = \frac{1000}{3.512} = 284$

Again, this means that the impedance of the capacitor at the desired frequency needs to be 284 Ohms. Using the vector notation, we write -j284 to indicate that it is a capacitive impedance.

Okay, now we need to find actual values for the inductor and the capacitor. To be able to calculate either one, we need to settle on an operating frequency. For this article we will use 16 MHz. Using the following formulas, we can find the actual values for the parts at the desired frequency:

$L = \frac{X_L}{\omega} = \frac{263}{2 \pi (16*10^6)} = 2.6 \mu H$

That means our inductor needs to have a value of 2.6 μH. I chose a value of 2.2 μH as it is a common value I had in stock. Since the circuitry itself in real life usually contains what’s called parasitic inductance, it won’t throw the circuit off by far. Next up is the capacitor:

$C = \frac{1}{\omega X_C} = \frac{1}{2 \pi (16*10^6) 284} = 35 pF$

Instead of 35 pF, I used a 33 pF capacitor. Not only is it a more common value, we can again assume a little bit of stray capacitance by the remaining circuit.

75 Ohms to 1000 Ohms L-Network characterized with a return loss bridge

I used a return loss bridge, a signal generator and an oscilloscope to verify the performance of this circuit. And sure enough, it acts just like a 75 Ohm resistive terminator at about 16 MHz. The exact frequency for best operation turned out to be roughly 16.1 MHz.

Confirming the performance of the 75 Ohms to 1000 Ohms L-Network using the Teledyne LeCroy HDO4024

Update (04/06/2013): I previously wrote that one can not match a high impedance source to a low impedance load. That is incorrect. Naturally, it works perfectly fine the other way around. One only has to switch the position of the load and the source.

KJ6QBA took out the time to simulate this circuit using LTspice. He also simulated the same circuit used in reverse to match a 1000 Ohm source to a 75 Ohm load. Here’s the result:

LTspice simulation of the L-Network provided by KJ6QBA

## Blown Capacitors – Repairing LCDs and TFTs

Blown capacitors have been around as long as capacitors themselves. Replacing defective capacitors is something every electronics engineer should be capable off. Let’s have a look how to turn a broken LC Display back into a functioning device.

Free TVs and computer screens are almost literally laying on the side of the road. About a quadzillion of Liquid Crystal Display (LCD) or Thin Film Transistor Liquid Crystal Display (TFT LCD) technology based TVs and computer screens are deemed defective and tossed away every day. A majority of them only have a minor flaw that can easily be fixed by novices.

Defective electrolytic capacitors usually stick out by a big bulge on the top. They are designed to burst exactly at that location. That is why there are several lines carved in the metal layer on top of the capacitor. If the pressure becomes too big, the metal will open along the carved lines and act as a pressure release valve.

Defective capacitors in a LCD TV power supply

The fix is simple: replace the capacitors. Do not only replace the obviously blown capacitors but all capacitors in that general region as well. I highly recommend replacing broke capacitors with capacitors with a higher voltage rating. Obviously the current voltage rating wasn’t enough, so why using the same value? If the different physical size of the higher voltage rated capacitor isn’t a problem, this will ensure that the LCD will now live a long life.

More blown capacitors

Blown capacitors are probably the most common reason for a LC Display failing. It is actually so common that one could speculate whether or not this is actually intended by the manufacturer to make sure you will soon buy a new TV or computer screen.

Looking on eBay for broken LCDs that just won’t turn on can be a good bargain source for a new TV.

## Why you should use high quality soldering tips

That a chain is only as strong as the weakest link should be common sense. Yet I see many people investing into extremely cheap soldering irons and then being surprised about their weak performance. Let’s compare tips of a budget soldering iron and a professional brand soldering iron.

A while back I read an article in QST [1] about preserving soldering iron tips. The author suggested to keep the tip of the iron in a tin bath to keep air from getting in touch with it. This is supposed to prevent corrosion.

My initial reaction was: “Are you freakin’ kidding me?!?” The Author actually tries to suggest a workaround for a simple problem, a low quality low-end solder tip. I got a different suggestion: Buy a high-end quality soldering tip!!!

Many people try to live by the rule “tight is right” when it comes to their budget. Unfortunately, few actually understand that investing a little bit more once is usually more economical that investing a little bit over and over and over again.

Here is a picture of a $5 soldering iron’s tip. The soldering iron is a cheap-shot from a local convenience store. After just being used for 2 times (maybe 30 minutes altogether), the tip looks like a scene of devastation: Tip of a cheap soldering iron ( ~$ 5) from a local convenience store after being used twice

Now here is a tip of an Ersa Multitip C15 iron that I have used intensively for 3 years:

Tip of a Ersa Multitip C15 soldering iron after three years of intensive use

Despite some discoloration, the tip is still in perfectly good shape.

The difference in quality is apparent. And trust me, the Multitip C15 has been tortured by me with weeks of operation without unplugging and acid flux core solder. The tip is still the same that was on it when I bought the soldering iron.

Of course the higher quality soldering iron and the tip are more expensive. The tip alone for the Multitip C15 costs about $10, so twice as much as the whole budget soldering iron. Yet the Multitip C15 with a price of about$ 40 is still considered entry-level and affordable.

When it comes to tools please don’t be tricked into the allegedly cheaper options. On the long run it will cost you so much more money, time and frustration to work with low-budget tools.