Aviation has been a love of mine since I was a very little person. Living in Nevada, seeing posters and ads for the Reno Air Races, specifically the Texans, transfixed me. I was hooked. As a teenager, I got an RC glider and dabbled in RC gliders and planes until my early 20’s. Later in …

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Sunday, October 6, 2024 9:00 PM Eastern Update:

Hurricane Milton was located just over 300 miles west-northeast of Progreso, Mexico, and 835 miles west-southwest of Tampa, Florida with winds of 100 miles per hour. Hurricane Milton was moving in an east-southeast direction at 6 mph and is forecasted to arrive near Tampa, Florida on Wednesday as at least a Category 3.

The Hurricane Watch Net is mak...

Constructors tend to copy popular designs, good or bad, and one of the components they see in pics online are the compensation capacitors connected across the 50Ω interface jack.

Single layer high voltage ceramic capacitors are popular, blue is the popular colour for high voltage ones, selected on specified capacitance and some very high voltage rating, often in the range of 3-6kV.

No, I didn’t forget Q, D, tan δ, or ESR… I left them out because constructors don’t seem to consider that part of the requirement.

So let’s review the sense of this.

Voltage rating

Q: What is the peak voltage in a 50 ohm load at 1kW?

A: 316Vpk.

So what is an appropriate capacitor rating with some safety margin?

1kV gives a 10dB safety margin at 1000W in 50Ω.

500V gives a 14dB safety margin at 100W in 50Ω.

So, you might ask why people seek out 6-20kV capacitors.

Q, D, tan δ, ESR

Q, D, tan δ, ESR are different metrics of the dielectric loss of the capacitor.

Let’s focus on Q. Q or quality factor is given by \(Q=\frac{|X|}{R}\). In admittance terms, we can also say \(Q=\frac{|B|}{G}\).

Power lost in the capacitor with a constant applied AC voltage can be calculated several ways, one is \(P=V^2 G=\frac{V^2 |B|}{Q}=\frac{V^2}{Q |X|}\)… the last can be convenient.

Another useful equation is \(P=\frac{V^2 2 \pi f C}{Q}\), so increasing voltage V, frequency f, and capacitance C increase dissipation; and increasing Q decreases dissipation.

So for example:

• at 100W in 50Ω, V=70.7V;
• if Ccomp was a 100pF 1kV capacitor and at 30MHz  Q=100, |X|=53; and
• P=0.943W.

1W dissipation will probably overheat a 8mm diameter disk ceramic capacitor, so whilst it easily withstands the applied voltage, it is not up to 100W continuous RF power.

For this scenario, Q needs to be perhaps 500 or more for capacitor temperature reasons.

RF capacitors

Capacitors are made with a wide range of dielectric chemistry, and within the broad ceramic capacitor category, dielectrics (and capacitors) are often denoted as Class 1 and Class 2. Class 1 dielectrics have much higher Q than Class 2, and Class 1 dielectric is often used for capacitors of 100pF or less, but Class 2 dielectric is used above that.

If you need say 120pF, it may be better to use two smaller capacitors in parallel as they are more likely to use Class 1 dielectric and you also get the benefit of a little more surface area to help with heat dissipation.

An example thermal analysis

Above is the internals of a common mode choke which uses 10pF 3kV 6mm ceramic capacitors for compensation of the coax pigtails. Measured capacitor Q is 500 @ 30MHz.

Above is a thermal pic of the internals with temperature stabilised running 100W @ 30MHz. The capacitor temperature reaches 20.2°, a rise of 5.7° at estimated dissipation 18.5mW. Note that these small capacitors have very low specific heat capacity, the temperature stabilizes in just 10s in this example.

Note that the dissipation in a 100pF capacitor of the same Q would be 10 times as much, ~0.2W, and even though a 100pF capacitor has more surface area, it may overheat at that power (especially inside an enclosure with other heat sources).

You don’t need a thermal camera to evaluate your own build, if you cannot comfortably touch the capacitor at some power level, it is too hot and that is excessive power.

Transmission line stub

A transmission line stub may offer a practical alternative to a capacitor.

People often speak of one of the properties of a certain coaxial line as a certain capacitance per length, pF/m, and would suggest that since RG58A/U is specified as 101pF/m, then a length of 990mm would be equivalent to 100pF. That is quite naive.

Above is a plot of impedance components of 790mm @ 30MHz, and |X|=53, the same reactance as a 100pF capacitor, but somewhat shorter. Observe that the reactance vs frequency curve is not a straight line… so the concept that the stub looks simply like 101pF/m is not sound in this case.

Above is a calculation from SimNEC of the power lost in the transmission line stub at 70.7V applied, the voltage due to 100W in 50Ω. Power lost is 0.7W which will result in a quite small increase in line temperature.

The example illustrates that such stubs are not equivalent to high Q capacitors, Q in this case is 130, not wonderful at all.

Let’s consider two shorter stubs in parallel.

Above is a calculation from SimNEC of the power lost in the transmission line stub at 70.7V applied, the voltage due to 100W in 50Ω. Power lost is 2*0.17=0.34W, half that of the previous case, which will result in a quite small increase in line temperature. Q in this case is 270, not wonderful, but better.

Now one advantage of the latter configuration is that it can be constructed by taking a single length of coax equal to the sum of the two stubs, forming into a shaped loop and connecting braid to braid and inner to inner at the ends to make a two terminal open circuit stub. There is no need to weatherproof the open ends that would otherwise exist.

Conclusions

For capacitors commonly used for compensation of RF transformers:

• heating is an important limitation, and one that is commonly ignored;
• extreme voltage rating is often uppermost in constructors minds, but that may be less a priority than thought, and may also lead to selection of a capacitor with poorer losses and higher operating temperature;
• replacing a failed capacitor with a higher voltage rated one might not be the best solution;
• transmission line stubs may provide a practical alternative; and
• mindless copying of published designs might not produce good results.

The conclusions of this article ring true, but I am not certain that changes to the Google algorithm, or the introduction of AI answers to Google queries explain the changes that are reflected in the above chart. 

Here's the article: 

As my ham radio journey takes me on a more formal study of electricity, my home laboratory grows. I remembering purchasing the Tektronix 2465 a few years back on a suggestion of a friend, not having a clue what I was doing with it. I just knew that real electronics enthusiasts had an oscilloscope.

Now that I am crawling my way through the Sedra/Smith’s Microelectronic Circuits tome, I have plenty of reasons to fire up my mushrooming pile of test equipment.

Let’s start with this circuit, the bridge rectifier.

The bridge rectifier is a fairly simple circuit to build, consisting of only four diodes (1N4001 in this case) and a load (10kΩ resistor). In the Multisim sketch above, the device on the left is a function generator producing a 5 volt peak-to-peak 60 Hz sine wave. On the positive half-wave of the signal cycle, the function generator’s positive terminal feeds into node D1/D2 of the four diodes forming the bridge rectifier. This positive signal forward biases D2, allowing the signal to continue on toward the load R1. D1 and D4, however, are reverse biased, and the signal is only able to conduct through D2. On the negative half-wave of the sine signal cycle, the function generator’s negative terminal now feeds a positive signal into node D3/D4. The signal now forward biases D4 and reverse biases D3 and D2. This half signal proceeds through the load R1 in the same direction as the positive half-wave portion of the signal did. All of the electricity conducting through the load in the same direction is considered direct current, as it does not have a component that flows in the reverse “backward” direction (as the original imputted sine wave did). Therefore, the bridge rectifier has done its job of “rectifying” alternating current into direct current.

By placing oscilloscope leads across the load resistor, the desired output of the bridge rectifier can be seen, at least in theory. Below is the Multisim oscilloscope simulation result. The bolded reference line is at zero volts. You will note that this “full-wave” rectifier has produced a waveform that is more or less continuously greater than zero volts. As we will see, this is in contrast to the “half-wave” rectifier, where there is no voltage seen when the originating sine wave enters its negative half.

In my home lab, I went ahead and set up a bridge rectifier on a breadboard, and set out to show its characteristics on my oscilloscope.

Part of the new fleet of test equipment are these two beauties, the Topward 8112 function generator producing a 60 Hz signal shown to be 5 volts peak-to-peak on the oscilloscope. And a Fluke 45 digital multimeter which will come in handy later.

After constructing the circuit exactly as shown in the schematic, I hooked up the oscilloscope across the resistor and found a trace that looked like a half-wave rectifier instead of a full-wave. This pretty much looks like a circuit consisting of an sinusoidal signal traversing through a signal forward-biased diode. There is no inversion of the negative portion of the sine wave, as seen with the full-wave rectifier, rather the negative portion of the wave just turns in to zero volts. In essence, unlike the full-wave rectifier, only half of the original sine wave is “rectified” into direct current. This is NOT what this circuit should look like on the oscilloscope.

Based on some earlier work, I expected though that this is what the oscilloscope would show without a bit of careful consideration. It turns out that this oscilloscope trace is not the result of the circuit malfunctioning, but rather a poor testing design creating a false result.

Remember the circuit diagram. The signal generator produces a signal that enters into the bridge rectifier at two distinct nodes D1/D2 and D3/D4. The signal generator needs to be electrically isolated from the output of the bridge rectifier, which occurs through nodes D2/D4 and D1/D3. Instead, however, the oscilloscope and function generator share a single ground, the earth ground in my house when I plug the unit into the wall. This makes both node D3/D4 and D1/D3 grounded which in essence creates a short circuit across diode D3. Without D3, the circuit now functions as a half-wave rectifier.

The multimeter was used to prove that indeed the oscilloscope and function generator grounds were shared. Everything was unplugged from each device and the digital multimeter was set to resistance. Checking the resistance between the outer shields of the BNC connectors of the oscilloscope and function generator showed, none, indicating a short circuit, and proving that this was the cause of the bridge rectifier’s apparent malfunction.

In order to fix this issue, I checked to see if my new TekPower TP-3003D-3 power supply shared a common ground as well. It did not! There were several megaohms of resistance between the ground terminal of the supply and the generator. So, I connected the oscilloscope probe ground and the function generator ground to one of the negative terminals of the power supply and checked the output waveform. It was still not correct.

I had one last trick up my sleeve. I declared the function generators ground as the common reference ground of the circuit. I attached one oscilloscope probe to one side of the load resistor (the top trace) and the other oscilloscope probe to the other side of the load resistor (the bottom trace). I attached the oscilloscope probe ground leads to the function generator ground. The results are shown in the oscilloscope trace below.

In essence, I can see that the top trace is displaying a half-wave rectifier. The bottom wave is showing another half-wave rectifier out of phase with the top trace. In fact, this bottom trace has been inverted, and this waveform now appears upside-down with the humps going upward, when really they would be going downward with negative voltages.

However, the inversion of the second signal is important, because now I can do math with the oscilloscope. Take a look at the third row of controls on the scope below. First I want to “Chop” the displayed signal as this will allow us to see the waveforms better. I have already inverted the signal. Now I want to add the first (upright) and second (inverted) channels together, essentially performing a subtraction function.

Revisit the image of the oscilloscope display above, and you can see the middle trace is now showing a full-rectifier. The middle waveform in the mathematical operation of Ch1 – Ch2 as performed by the oscilloscope.

And sure enough, when you take a voltage reading with the multimeter across the load resistor, you can see a 1.8 V output of direct current. Woo hoo!

I used the new AI image feature on this WordPress site to prompt a proud scientist overlooking the display of a full wave rectifier on his oscilloscope. So darn goofy! My favorite part is the Hollywood vanity lights just like in a professional lab! It is also hard to tell exactly why he is so proud given those results…

It is not surprising that I have a interest in test equipment. Learning the nuances, limitations, and intricacies of testing is paramount, or else you can make what I would refer to in my other life as iatrogenic errors. This little experiment is perfect for showing how something that works exactly as it should can be deemed malfunctioning due to misinterpretation of a lab test. While getting it wrong with this device in my little electronics laboratory is inconsequential, bad testing in the real world can be disastrous.

So the lesson here? Always look for that common ground.

So far away,

KM1NDY

I’ve heard about it before but only just looked into bunkers on the air (or UK BOTA). Turns out there are a lot of sites around and good way to get out and about, use the radio, explore your local region and maybe learn a little bit of history.

Here’s a list of useful sites to get you going:

• Main site
• Database
• Maps
  • The Google Map didn’t appear for me, but the OS map is nice, or use the combined on below:
• Combined awards map

The main rules are:

• Be within 1000 m of the bunker location (with the right permissions etc.)
• Qualify the bunker with 25 HF contacts or 10 VHF+ contacts

I’d recommend joining the WhatsApp group listed on the website, and you’ll get invited to the spotting group too.

I was surprised by the number of bunkers locally to me, especailly compared to doing SOTA activations and having to drive at least an hour to get anywhere!

I’ve not activated one yet but there appear to be plenty that have never been done near me.