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I've built a high impedance buffer out of an op amp on a small piece of perf board. I'm going to attach the input to the first IF on an Icom IC-718 and the output will be pulled out to a jack that I've drilled into the case.

I know that I want to keep stuff as short as possible... but at the same time, there's no super good place to mount the board that I've created. How bad an idea is it to tap into the IF with a piece of RG-174 that's about 6 inches long between the IF and the board? I would tie the shield to ground as close to where I tap the IF as possible.

This is my first major modification to an expensive radio beyond repairs and I'm trying really hard to get it right the first time. :)

David Hoelzer
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  • At what frequency is that IF? In general, high input impedance is **not** what you want in radio frequency devices. – Marcus Müller Jun 21 '16 at 12:39
  • You normally need to impedance-match your sink to your signal source. Which is why tapping a line can have very undesirable effects. In fact, by adding a tap with a finite length of transmission line with a high impedance end, you'll build a filter. In fact, this is pretty close to the original concept of what is called "tapped delay line", a term that I can but recommend to Google :) – Marcus Müller Jun 21 '16 at 12:43
  • Also, sorry, as an European: how much is 6 inch in engineerable units? This is important, because it directly relates to the delay of signals traveling along that tap line – Marcus Müller Jun 21 '16 at 12:44
  • It seems very appropriate in this instance... I'm tapping the IF for a panadapter and don't want to inadvertently degrade or introduce signal into it. Am I thinking about this incorrectly? I believe that the first IF is 70 MHz – David Hoelzer Jun 21 '16 at 13:00
  • I don't care if the signals are delayed. I'm just tapping the line out to a pan adapter. I am most concerned about influencing the existing function of the receiver. – David Hoelzer Jun 21 '16 at 13:02
  • Here's an interesting tidbit from talk on wikipedia: https://en.wikipedia.org/wiki/Talk%3AShielded_cable I haven't looked further, but it claims the correct solution is to work out a proper ground rather than grounding only one end of the cable. – David Hoelzer Jun 21 '16 at 13:35
  • Just an idea: can you add a photograph of the trace you want to intercept? That would make answering easier :) – Marcus Müller Jun 23 '16 at 07:45
  • Sure.. If I have time I'll pop the radio open today. – David Hoelzer Jun 23 '16 at 10:51

1 Answers1

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So, you say:

I don't care if the signals are delayed. I'm just tapping the line out to a pan adapter. I am most concerned about influencing the existing function of the receiver

And that's exactly what you're going to do with this.

Your Opamp circuit has a high impedance, which translates into "looks like an open end". That means that your tapping coax has a very distinct feature: A reflection coefficient of 1.

This can easily be seen by the "mismatch" equation: The ratio of electric field (voltage!) going through the interface between two transmission lines with different impedances and getting reflect is defined as:

$\Gamma = \frac{E^-}{E^+} = \frac{Z_2 - Z_1}{Z_1 + Z_2} $

with $Z_1$ being the impedance of the transmission line from which the wave comes, and $Z_2$ being the impedance of the second transmission line or the sinking device.

In your case, $Z_2\rightarrow \infty$, and hence

$$\begin{align} \lim_{Z_2\rightarrow \infty} \Gamma &=\lim_{Z_2\rightarrow \infty} \frac{Z_2 - Z_1}{Z_1 + Z_2}\\ &= \lim_{Z_2\rightarrow \infty} \frac{\frac{Z_2}{Z_2} - \frac{Z_1}{Z_2}}{\frac{Z_1}{Z_2} + \frac{Z_2}{Z_2}}\\ &= \lim_{Z_2\rightarrow \infty} \frac{1 - \frac{Z_1}{Z_2}}{\frac{Z_1}{Z_2} + 1}\\ &=\frac{1-0}{0+1}\\ &=1 \end{align}$$

That means that all energy put into that tap line returns to the line you've tapped – which is pretty logical! Where should the energy you "steal" from the original coax go if you don't sink it somewhere? It's pretty clear you'll have a bad SWR in this situation.

So let's look at the wave from your IF line:

It meets the point where you've tapped the original line. Let's assume that you've build a reflection-free splitter here – that's highly unlikely, but it makes things both better for your use case, and easier to understand.

So, the wave enters your 6 inch (really, inches are a terrible unit for RF design – with meters, and hertz, everything can be calculated without converting) line. That's about $d=0.15\text{ m}$.

Now, RG-174 has a velocity factor of 0.66 – meaning that a wave travels with $\frac 23$ of the speed of light in vacuum, which is accurately enough $v_{RG-174}=\frac 23 \, 3\cdot 10^8\frac{\text m}{\text s} = 2\cdot 10^8\frac{\text m}{\text s}$.

So, for the distance between your tapping point, and the end of your RG174, and back, the wave needs

$\Delta t = \frac{2d}{v_{RG-174}} = \frac{0.3\text{ m}}{2\cdot 10^8\frac{\text m}{\text s}}$.

After that time, the wave will hit the tapping point – and interfere with the original IF!

By adding up a wave and a delayed version, you have built a frequency-selective device – a filter.

Now, what's relevant here is to know the phase with which the interference will happen. Let's have a look at that.

The phase $\varphi$ of a signal is the delay, measured in full periods $T$ of the signal (and period is related to frequency by $T=\frac1f$; from your comment, $f=70\text{ MHz}$); in other words:

$$\begin{align} \Delta \varphi &= 2\pi \frac{\Delta t}{T}\\ &= 2\pi \frac{\Delta t}{\frac1f}\\ &= 2\pi \frac{\frac{2d}{v_{RG-174}}}{\frac1f}\\ &= 2\pi \frac{2df}{v_{RG-174}} &\approx 2\pi \cdot 0.1 \end{align}$$

In other words, you're overlaying your original IF signal with a 0.1 waveperiods delayed version of itself. That will cancel out specific frequencies, while others will be amplified.

Now, we could derive the frequency response of the tapped delay line filter you've built – but since that will in practice depend a lot on things that are hard to know beforehand, like reflection coefficient of your tap (since that massively changes impedance of the line, depending on a lot of things), exact length, etc, this would be pretty much in vain.

The essence is: don't tap analog coax lines. Simple as that.

Get a proper splitter with the correct input impedance, and make sure your opamp circuit is impedance-matched to your transmission line. Everything else will end in significant reflections, interference and frequency-dependent cancelation of signal that is hard to predict and to measure.

Marcus Müller
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  • You seem to have assumed a particular value for $f$ in your last step — perhaps mention that, especially since as $f$ goes down the delay and therefore the effect is smaller? – Kevin Reid AG6YO Jun 21 '16 at 18:13
  • @KevinReidAG6YO true, used the 70MHz from OP's comment. – Marcus Müller Jun 21 '16 at 18:37
  • So then, what would be the proper way to tap the intermediate frequency in a transceiver? I'm not splicing a piece of coax, I was going to use a piece of coax to tap off of the board. – David Hoelzer Jun 21 '16 at 21:24
  • After more researching, would the proper way to tap the IF be with a coupling capacitor of about 10 picofarads? – David Hoelzer Jun 21 '16 at 22:42
  • Also, your answer flies in the face of http://ham.stackexchange.com/questions/5808/adding-a-panadapter-via-sdr making me wonder who I should rely upon. – David Hoelzer Jun 21 '16 at 23:21
  • The "closely coupled" line approach is actually a directional coupler/splitter, which is what my answer suggested that splitter might have different output impedances on its two outputs. I don't know what the author of the other answer had in mind exactly. My way to do this would be actually cutting the if, sinking it completely into a proper fan-out buffer IC in a configuration that is impedance matched to your IF source, and using one of the outputs to drive the original IF receiver, and the other whatever you want to observe. That way you get both buffering and a second output. – Marcus Müller Jun 22 '16 at 05:53
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    1st off, in your answer, you stated "don't tap analog coax lines. Simple as that." Since this question regards tapping an on-pcb IF trace in a transciever, I don't think that applies here. - With that said, I agree that tapping with a buffer-amp at the end of ~15cm of coax is a bad idea due to the delay of the coax involved. @DavidHoelzer I suggested the buffer amp based on an assumption that it would be installed "in-line" with the IF circuit. If you need to locate it a few inches away, then you'd need to cut the trace, detour the signal to the buffer amp, then send it back in a coax 'loop'. – Robherc KV5ROB Jun 22 '16 at 06:32
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    @RobhercKV5ROB I of course don't know the actual schematic/PCB here, but when routing RF/IF signals on PCB, you do things like microstrip lines, which, exactly as coax, are transmission lines with a wave impedance, and tapping those changes impedance, hence leads to reflections etc. Now, at 70 MHz things are still quite benign (I mean 10% phase shift due to 2x 15cm cable? Ok, that's something you can handle if you know what is happening), and you'd typically just drop in an arbitrary power splitter, and you're good to go. – Marcus Müller Jun 22 '16 at 06:36
  • I.E. Signal goes from pcb to coax (with proper grounding) at cut in trace (please impedance-match if your trace & coax aren't a close match already); then from coax to buffer-amp pcb (again matching impedance as closely as possible); then through a 2nd length of coax back to the IF trace on main pcb for your transciever to use it. - This method, so long as your trace/coax impedances are close, should result in a fairly negligible effect on your IF signal. – Robherc KV5ROB Jun 22 '16 at 06:36
  • @RobhercKV5ROB I can fully second that – Marcus Müller Jun 22 '16 at 06:37
  • Ok... Let's cut to the chase... Imagining that I'm not an EE for a second... :) Should I cut the trace, route to the buffer from the previous answer with capacitive coupling and route back with as short a coax as I can manage? – David Hoelzer Jun 22 '16 at 10:07
  • @DavidHoelzer if you don't add unwanted reflections anywhere, then there can be no interference, and then the length of the cable doesn't matter. You don't need capacitive coupling – you need **matching** the impedance of the line. – Marcus Müller Jun 22 '16 at 10:12
  • Sounds like I need to go back to the schematics and figure out the impedances. The diagrams don't specify it but I may be able to figure it out from the data sheets of the chips used in the mixer. – David Hoelzer Jun 22 '16 at 10:38
  • often, the receiving end of a coax line is fed into some resistor array, and high-impedance-input ICs "measure" the voltage across that. Look for these resistors! – Marcus Müller Jun 22 '16 at 11:56
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    You can physically measure the traces & the pcb, then input the measurements here: [microstrip/pcb trace impedance calculator](https://www.eeweb.com/toolbox/microstrip-impedance). Accurate measurements should give you the most accurate impedance calculation for tue trace itself that is achievable. Generally speaking, most HAM radio circuits should be built around the same 50 ohm impedance expected at the antenna terminal(s) (luckily, that's a perfect match for your rg174), so as long as the measurements agree, I expect you'll be able to use the buffer amp you already made (continued) – Robherc KV5ROB Jun 22 '16 at 12:57
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    ...and simply connect it by cutting the IF trace, then run coax from the cut trace to your buffer, then from the buffer back the the trace (on the other side of the cut, of course). Just be sure when you make the cut that you remove all of the trace copper from a section at **least** 1.5x as long as the pcb is thick (to avoid changing the trace's characteristic impedance & creating reflections...or not fully breaking the trace & getting no signal from your tap), and try to keep your soldeing as 'clean' as possible. @DavidHoelzer – Robherc KV5ROB Jun 22 '16 at 13:02
  • @RobhercKV5ROB This is precisely what I was suggesting. *whew* Ok, I'll try that out. I can always put a dot of solder onto the broken trace if I seriously attenuate or otherwise screw the signal up. I purchased the RG-174 because of the 50 ohm impedance and my expectation that everything in the right would be based on that. – David Hoelzer Jun 22 '16 at 13:12
  • @MarcusMüller I may not have been clear, but there is no existing coax. It's a short trace from one mixer to a filter. – David Hoelzer Jun 22 '16 at 13:13